THERMODYNAMIC PROPERTIES AND CALCULATION

Academic Resource Center THERMODYNAMIC PROPERTIES

A quantity which is either an attribute of an entire system or is a function of position which is continuous and does not vary rapidly over microscopic distances, except possibly for abrupt changes at boundaries between phases of the system; examples are temperature, pressure, volume, concentration, surface tension, and viscosity. Also known as macroscopic property. BASIC CONCEPTS-1

 First Law of Thermodynamic:  Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms.  ∆(Energy of the system) + ∆(Energy of surroundings) = 0  ∆Ut = Q + W → ∆(nU) = Q + W  dUt = dQ + dW → d(nU) = dQ + dW  There exists a form of energy, known as internal energy U. BASIC CONCEPTS-2

 PV diagram

 Virial Equations of State PV = a + bP + cP2 + ...... →

 Ideal gas: Z=1 or PV = RT  Van Der Waals Equation of State For Ideal Gas: Equation for Calculation Heat capacity:

dQ + dW = CvdT dW = – PdV dQ = CvdT + PdV Let V=RT/P : BASIC CONCEPTS-3

Statements of the Second Law: Statement 1: No apparatus can operate in such a way that its only effect (in system and surroundings) is to convert heat absorbed by a system completely into work done by the system.

Statement 2: No process is possible which consists solely in the transfer of heat from one temperature level to a higher one PRIMARY THERMODYNAMIC PROPERTIES— P, V, T, S & U Combining the first and second laws in reversible process The only requirements are that the system be closed and that the change occur between equilibrium states.  H = U + PV  A = U – TS  G = H – TS d(nU) = Td(nS) – Pd(nV) d(nH) = Td(nS) + (nV)dP d(nA) = – Pd(nV) – (nS)d d(nG) = (nV)dP – (nS)dT dU = TdS – PdV dH = TdS + VdP dA = – PdV – SdT dG = VdP – SdT

Maxwell’s equation EXAMPLE 1  Air at 1 bar and 298.15K (25℃) is compressed to 5 bar and 298.15K by two different mechanically reversible processes:  (a) Cooling at constant pressure followed by heating at constant volume.  (b) Heating at constant volume followed by cooling at constant pressure.  Calculate the heat and work requirements and ΔU and ΔH of the air for each path. The following heat capacities for air may be assumed independent of temperature: -1 -1 CV= 20.78 and CP=29.10 J mol K Assume also for air that PV/T is a constant, regardless of the changes it undergoes. At 298.15K and 1 bar the molar volume of air is 0.02479 m3 mol-1. KEYS SOLUTIONS:

 In suche case take the system as 1 mol of air contained in an imaginary piston/cyclinder arrangement. Since the processes considered are mechanically reversible, the piston is imagined to move in the cylinder withour friction. The final volume is

 (a) During the first step the air is cooled at the constant pressure of 1 bar until the final volume of 0.004958 m3 is reached. The temperature of the air at the end of this cooling step is: SOLUTIONS

 Also, 푄 = ∆퐻 − ∆ 푃푉 = ∆퐻 − 푃∆푉 = −6,941 − 1 × 105 0.004958 − 0.02479 = −4,958J  During the second step the volume is held constant at while the air is heated to its final state.

 The complete process represents the sum of its steps. Hence, Q = -6,941 + 4,958 = -1,938J ΔU = -4,958 + 4,958 = 0J  Since the first law applies to the entire process, , and therefore, 0 = -1,983 + W W = 1,983J SOLUTIONS

 ,  also applies to the entire process. But

 and therefore,

 Hence ,

 and SOLUTIONS

 Two different steps are used in this case to reach the same final state of the air. In the first step the air is heated at a constant volume equal to its initial valve until the final pressure of 5 bar is reached. The air temperature at the end of this step is:

 For this step the volume is constant, and

 During the second step the air is cooled at the constant pressure of 5 bar to its final state: SOLUTIONS

 Also, ∆푈 = ∆퐻 − ∆ 푃푉 = ∆퐻 − 푃∆푉 = −34,703 − 5 × 105 0.004958 − 0.02479 = −24,788J  For the two steps combined, Q =24,788-34,703 = -9,915J ΔU =24,788-24,788=0J

 and as before  During the second step the volume is held constant at V2 while the air is heated to its final state.  The property changes and calculated for the given change in state are the same for both paths. On the other hand the answers to parts (a) and (b) show that Q and W depend on the path. EXAMPLE 2

 Air is compressed from an initial condition of 1 bar and 25℃ to a final state of 5 bar and 25℃ by three different mechanically reversible processes in a closed system:  (a) Heating at constant volume followed by cooling at constant pressure.  (b) Isothermal compression.  (c)Adiabatic compression followed by cooling at constant volume.  Assume air to be an ideal gas with the constant heat

capacities, CV= (5/2)R and CP = (7/2)R. Calculate the work required, heat transferred, and the changes in internal energy and enthalpy of the air for each process. KEYS SOLUTIONS

 Choose the system as 1 mol of air, contained in an imaginary frictionless piston/cylinder arrangement. For R=8.314 J mol-1K-1, -1 -1 CV=20.785 CP=29.099J mol K  The initial and final conditions of the air are identical with those of Ex.1, where the molar volumes are given as:

V1=0.02479 V2=0.004958m3  Moreover, since the initial and final temperatures are the same, then for all parts of the problem: SOLUTIONS

 (a) The heat transferred, from Ex.1(b) is Q=- 9.915J. Thus by the first law applied to the entire process:

 (b) Equation for the isothermal compression of an ideal gas applies here:

 (c) The intial adiabatic compression of the air takes it to its final volume of 0.004958m3. The temperature and pressure at this point are: SOLUTIONS

γ−1 V1 0.02479 0.4 T2 = T1 = 298.15 ( ) = 567.57K V2 0.004958 γ V1 0.02479 1.4 P2 = P1 = 1 ( ) = 9.52bar V2 0.004958 For this step Q=0, and

W = CV ∆T = 20.785 567.57 − 298.15 = 5,600J For the second step at constant V, W=0. For the overall process, W = 5,600 + 0 = 5,600J SOLUTIONS

 Moreover, , and by the first law, Q = ∆푈 − 푊 = 0 − 5,600 = −5,600J

 Although the property changes and are zero for each process, Q and W are path-dependent. Since the work for each of these mechanically reversible processes is given by , the work for each process is proportional to the total area below the paths on the PV diagram representing the process. The relative sizes of these areas correspond to the numerical values of W. EXAMPLE 3

 Determine the enthalpy and entropy changes of liquid water for a change of state from 1 bar and 25 ℃ to 1,000 bar and 50℃. The following data for water are availbale: -1 - 3 - -1 t/℃ P/bar CP/J mol K V/cm mol β/K 1 1 25 1 75.305 18.071 256×10-6 25 1,000 ...... 18.012 366×10-6 50 1 75.314 18.234 458×10-6 50 1,000 ...... 18.174 568×10-6 KEYS SOLUTIONS

 For application to the change of state described, equations require integration. Since enthalpy and entropy are state functions, the path of integration is arbitrary. Since the data indicate

that CP is a weak function of T and that both V and β are weak functions of P, integration with arithmetic means is satisfactory. The integrated forms of Eqs that result are: SOLUTIONS

 For P=1 bar,

 and for t=50℃ SOLUTIONS

 Substitution of these numerical values into the equation for gives: 1 − 513 × 10−6 323.15 18.204 (1,000 − 1) ∆H = 75.310 323.15 − 298.15 + 10cm3bar J−1

 Similarly for , 323.15 513 × 10−6 18.204 (1,000 − 1) ∆S = 75.310ln − 298.15 10cm3bar J−1

 Note that the effect of a pressure change of almost 1,000 bar on the enthalpy and entropy of liquid water is less than that of a temperature change of only 25℃ THERMODYNAMIC PROPERTIES AND THEIR CHARACTERISTICS

Unit 33 Objectives ElectrElectrochemistrochemistryy After studying this Unit, you will be able to • describe an electrochemical cell and differentiate between galvanic Chemical reactions can be used to produce electrical energy, and electrolytic cells; conversely, electrical energy can be used to carry out chemical • apply Nernst equation for reactions that do not proceed spontaneously. calculating the emf of galvanic cell and define standard potential of the cell; Electrochemistry is the study of production of • derive relation between standard electricity from energy released during spontaneous potential of the cell, Gibbs energy chemical reactions and the use of electrical energy to of cell reaction and its equilibrium bring about non-spontaneous chemical constant; • define resistivity (ρ), conductivity transformations. The subject is of importance both κ ¥ for theoretical and practical considerations. A large ( ) and molar conductivity ( m) of ionic solutions; number of metals, sodium hydroxide, chlorine, • differentiate between ionic fluorine and many other chemicals are produced by (electrolytic) and electronic electrochemical methods. Batteries and fuel cells conductivity; convert chemical energy into electrical energy and are • describe the method for used on a large scale in various instruments and measurement of conductivity of devices. The reactions carried out electrochemically electrolytic solutions and can be energy efficient and less polluting. Therefore, calculation of their molar conductivity; study of electrochemistry is important for creating new • justify the variation of technologies that are ecofriendly. The transmission of conductivity and molar sensory signals through cells to brain and vice versa conductivity of solutions with and communication between the cells are known to change in their concentration and have electrochemical origin. Electrochemistry, is Λ° therefore, a very vast and interdisciplinary subject. In define m (molar conductivity at zero concentration or infinite this Unit, we will cover only some of its important dilution); elementary aspects. • enunciate Kohlrausch law and learn its applications; • understand quantitative aspects of electrolysis; • describe the construction of some primary and secondary batteries and fuel cells; • explain corrosion as an electrochemical process. 3.1 Electrochemical In Class XI, Unit 8, we had studied the construction and functioning of Daniell cell (Fig. 3.1). This cell converts the chemical energy liberated Cells during the redox reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (3.1) to electrical energy and has an electrical potential equal to 1.1 V when concentration of Zn2+ and Cu2+ ions is unity ( 1 mol dm–3)*. Such a device is called a galvanic or a voltaic cell. If an external opposite potential is applied [Fig. 3.2(a)] and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V [Fig. 3.2(b)] when, the reaction stops altogether and no current flows through the cell. Any further increase in the external potential again starts the reaction but in the opposite direction [Fig. 3.2(c)]. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions. Both types of cells are quite important and we shall study some of their salient features in the following Fig. 3.1: Daniell cell having electrodes of zinc and copper dipping in the solutions of their pages. respective salts.

*Strictly speaking activity should be used instead of concentration. It is directly proportional to concentration. In dilute solutions, it is equal to concentration. You will study more about it in higher classes.

Chemistry 64 Fig. 3.2: Functioning of Daniell cell when external voltage Eext opposing the cell potential is applied.

As mentioned earlier (Class XI, Unit 8) a galvanic cell is an 3.2 Galvanic Cells electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. In this device the Gibbs energy of the spontaneous redox reaction is converted into electrical work which may be used for running a motor or other electrical gadgets like heater, fan, geyser, etc. Daniell cell discussed earlier is one such cell in which the following redox reaction occurs. Zn(s) + Cu2+(aq) → Zn2+ (aq) + Cu(s) This reaction is a combination of two half reactions whose addition gives the overall cell reaction: (i) Cu2+ + 2e– → Cu(s) (reduction half reaction) (3.2) (ii) Zn(s) → Zn2+ + 2e– (oxidation half reaction) (3.3) These reactions occur in two different portions of the Daniell cell. The reduction half reaction occurs on the copper electrode while the oxidation half reaction occurs on the zinc electrode. These two portions of the cell are also called half-cells or redox couples. The copper electrode may be called the reduction half cell and the zinc electrode, the oxidation half-cell. We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combinations of different half-cells. Each half- cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge as shown in Fig. 3.1. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we don’t require a salt bridge. 65 Electrochemistry At each electrode-electrolyte interface there is a tendency of metal ions from the solution to deposit on the metal electrode trying to make it positively charged. At the same time, metal atoms of the electrode have a tendency to go into the solution as ions and leave behind the electrons at the electrode trying to make it negatively charged. At equilibrium, there is a separation of charges and depending on the tendencies of the two opposing reactions, the electrode may be positively or negatively charged with respect to the solution. A potential difference develops between the electrode and the electrolyte which is called electrode potential. When the concentrations of all the species involved in a half-cell is unity then the electrode potential is known as standard electrode potential. According to IUPAC convention, standard reduction potentials are now called standard electrode potentials. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to the solution. Thus, there exists a potential difference between the two electrodes and as soon as the switch is in the on position the electrons flow from negative electrode to positive electrode. The direction of current flow is opposite to that of electron flow. The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half-cell on the right hand side minus the potential of the half-cell on the left hand side i.e.

Ecell = Eright – Eleft This is illustrated by the following example: Cell reaction: Cu(s) + 2Ag+(aq) ⎯→ Cu2+(aq) + 2 Ag(s) (3.4) Half-cell reactions: Cathode (reduction): 2Ag+(aq) + 2e– → 2Ag(s) (3.5) Anode (oxidation): Cu(s) → Cu2+(aq) + 2e– (3.6) It can be seen that the sum of (3.5) and (3.6) leads to overall reaction (3.4) in the cell and that silver electrode acts as a cathode and copper electrode acts as an anode. The cell can be represented as: Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)

and we have Ecell = Eright – Eleft = EAg+⎥Ag – ECu2+⎥Cu (3.7) 3.2.1 Measurement The potential of individual half-cell cannot be measured. We can measure of Electrode only the difference between the two half-cell potentials that gives the emf Potential of the cell. If we arbitrarily choose the potential of one electrode (half-

Chemistry 66 cell) then that of the other can be determined with respect to this. According to convention, a half-cell called standard hydrogen electrode ⎥ ⎥ + (Fig.3.3) represented by Pt(s) H2(g) H (aq), is assigned a zero potential at all temperatures corresponding to the reaction 1 H+ (aq) + e– → H (g) 2 2 The standard hydrogen electrode consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen is maintained at unity (Fig. 3.3). This implies that the pressure of hydrogen gas is one bar and the concentration of hydrogen ion in the solution is one molar. At 298 K the emf of the cell, standard hydrogen electrode ⎜⎜second half-cell constructed by taking standard hydrogen electrode as anode (reference half-cell) and the other half-cell as cathode, gives the reduction potential of the other half-cell. If the concentrations of the oxidised and the reduced forms of the species in the right hand half-cell are unity, then the cell potential is equal to standard electrode  potential,E R of the given half-cell.    E = E R - E L  As E L for standard hydrogen electrode is zero.    Fig. 3.3: Standard Hydrogen Electrode (SHE). E = E R – 0 = E R The measured emf of the cell : ⎥ ⎥ + ⎜⎜ 2+ ⎥ Pt(s) H2(g, 1 bar) H (aq, 1 M) Cu (aq, 1 M) Cu is 0.34 V and it is also the value for the standard electrode potential of the half-cell corresponding to the reaction : Cu2+ (aq, 1M) + 2 e– → Cu(s) Similarly, the measured emf of the cell : ⎥ ⎥ + ⎜⎜ 2+ ⎜ Pt(s) H2(g, 1 bar) H (aq, 1 M) Zn (aq, 1M) Zn is -0.76 V corresponding to the standard electrode potential of the half-cell reaction: Zn2+ (aq, 1 M) + 2e– → Zn(s) The positive value of the standard electrode potential in the first case indicates that Cu2+ ions get reduced more easily than H+ ions. The reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu (or alternatively we can say that hydrogen gas can reduce copper ion) under the standard conditions described above. Thus, Cu does not dissolve in HCl. In nitric acid it is oxidised by nitrate ion and not by hydrogen ion. The negative value of the standard electrode potential in the second case indicates that hydrogen ions can oxidise zinc (or zinc can reduce hydrogen ions). 67 Electrochemistry In view of this convention, the half reaction for the Daniell cell in Fig. 3.1 can be written as: Left electrode : Zn(s) → Zn2+ (aq, 1 M) + 2 e– Right electrode: Cu2+ (aq, 1 M) + 2 e– → Cu(s) The overall reaction of the cell is the sum of above two reactions and we obtain the equation: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)

0 0 0 Emf of the cell = Ecell = E R - E L = 0.34V – (– 0.76)V = 1.10 V Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in the reaction but provide their surface for oxidation or reduction reactions and for the conduction of electrons. For example, Pt is used in the following half-cells: + Hydrogen electrode: Pt(s)|H2(g)| H (aq) + – → With half-cell reaction: H (aq)+ e ½ H2(g) – Bromine electrode: Pt(s)|Br2(aq)| Br (aq) – → – With half-cell reaction: ½ Br2(aq) + e Br (aq) The standard electrode potentials are very important and we can extract a lot of useful information from them. The values of standard electrode potentials for some selected half-cell reduction reactions are given in Table 3.1. If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species. It can be seen that the standard electrode potential for fluorine is the

highest in the Table indicating that fluorine gas (F2) has the maximum tendency to get reduced to fluoride ions (F–) and therefore fluorine gas is the strongest oxidising agent and fluoride ion is the weakest reducing agent. Lithium has the lowest electrode potential indicating that lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 3.1 the standard electrode potential decreases and with this, decreases the oxidising power of the species on the left and increases the reducing power of the species on the right hand side of the reaction. Electrochemical cells are extensively used for determining the pH of solutions, solubility product, equilibrium constant and other thermodynamic properties and for potentiometric titrations.

Intext Questions 3.1 How would you determine the standard electrode potential of the system Mg2+|Mg? 3.2 Can you store copper sulphate solutions in a zinc pot? 3.3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Chemistry 68 Table 3.1 The standard electrode potentials at 298 K

Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s. Reaction (Oxidised form + ne– → Reduced form) E/V

– → – F2(g) + 2e 2F 2.87 Co3+ + e– → Co2+ 1.81 + – → H2O2 + 2H + 2e 2H2O 1.78 – + – → 2+ MnO4 + 8H + 5e Mn + 4H2O 1.51 Au3+ + 3e– → Au(s) 1.40 – → – Cl2(g) + 2e 2Cl 1.36 2– + – → 3+ Cr2O7 + 14H + 6e 2Cr + 7H2O 1.33 + – → O2(g) + 4H + 4e 2H2O 1.23 + – → 2+ MnO2(s) + 4H + 2e Mn + 2H2O 1.23 – → – Br2 + 2e 2Br 1.09 – + – → NO3 + 4H + 3e NO(g) + 2H2O 0.97 2+ – → 2+ 2Hg + 2e Hg2 0.92 Ag+ + e– → Ag(s) 0.80 Fe3+ + e– → Fe2+ 0.77 + – → O2(g) + 2H + 2e H2O2 0.68 – → – I2 + 2e 2I 0.54 Cu+ + e– → Cu(s) 0.52 Cu2+ + 2e– → Cu(s) 0.34 AgCl(s) + e– → Ag(s) + Cl– 0.22 AgBr(s) + e– → Ag(s) + Br– 0.10 + – → 2H + 2e H2(g) 0.00 Pb2+ + 2e– → Pb(s) –0.13 Sn2+ + 2e– → Sn(s) –0.14 2+ – Increasing strength of oxidising agent Ni + 2e → Ni(s) Increasing strength of reducing agent –0.25 Fe2+ + 2e– → Fe(s) –0.44 Cr3+ + 3e– → Cr(s) –0.74 Zn2+ + 2e– → Zn(s) –0.76 – → – 2H2O + 2e H2(g) + 2OH (aq) –0.83 Al3+ + 3e– → Al(s) –1.66 Mg2+ + 2e– → Mg(s) –2.36 Na+ + e– → Na(s) –2.71 Ca2+ + 2e– → Ca(s) –2.87 K+ + e– → K(s) –2.93 Li+ + e– → Li(s) –3.05

 + 1. A negative E means that the redox couple is a stronger reducing agent than the H /H2 couple.  + 2. A positive E means that the redox couple is a weaker reducing agent than the H /H2 couple.

69 Electrochemistry 3.3 Nernst We have assumed in the previous section that the concentration of all the species involved in the electrode reaction is unity. This need not be Equation always true. Nernst showed that for the electrode reaction: Mn+(aq) + ne–→ M(s) the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:

= V RT [M] E()()nn++E – ln M/M M/M nF [Mn+ ] but concentration of solid M is taken as unity and we have

= V RT 1 E()()nn++E – ln n+ (3.8) M/M M/M nF [M ]

V E + ()M/Mn has already been defined, R is gas constant (8.314 JK–1 mol–1), F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn+. In Daniell cell, the electrode potential for any given concentration of Cu2+ and Zn2+ ions, we write For Cathode: 1 V RT + E()2+ = E()2+ – ln ⎡⎤2 (3.9) Cu/Cu Cu / Cu 2F ⎣⎦Cuaq() For Anode: 1 V RT + E()2+ = E()2+ – ln 2 (3.10) Zn/Zn Zn / Zn 2F ⎣⎦⎡⎤Znaq()

E + E + The cell potential, E(cell) = ()Cu/Cu2 – ()Zn/Zn2

1 1 V RT V RT = E()2+ – ln 2+ - E()2+ + ln 2+ Cu / Cu 2F ⎣⎦⎡⎤Cu(aq) Zn / Zn 2F ⎣⎦⎡⎤Zn(aq)

V V RT 11 = E()2+ – E()2+ – ln – ln Cu / Cu Zn / Zn 2F 2+ 2+ ⎣⎦⎣⎦⎡⎤⎡⎤Cuaq() Znaq()

2 + V RT []Zn E = E() – ln (3.11) (cell) cell 2F []Cu2 + 2+ It can be seen that E(cell) depends on the concentration of both Cu and Zn2+ ions. It increases with increase in the concentration of Cu2+ ions and decrease in the concentration of Zn2+ ions. By converting the natural logarithm in Eq. (3.11) to the base 10 and substituting the values of R, F and T = 298 K, it reduces to + 0.[] 059 Zn2 V log E = E() – + (3.12) (cell) cell 2 []Cu2 We should use the same number of electrons (n) for both the electrodes and thus for the following cell

Chemistry 70 Ni(s)⎥ Ni2+(aq) ⎥⎥ Ag+(aq)⎥ Ag The cell reaction is Ni(s) + 2Ag+(aq) →Ni2+(aq) + 2Ag(s) The Nernst equation can be written as

V RT [Ni2+ ] E(cell) = E() – ln + cell 2F [Ag ]2 and for a general electrochemical reaction of the type:

a A + bB ⎯→⎯⎯ne cC + dD Nernst equation can be written as:

V RT E = E – 1nQ (cell) ()cell nF

cd V RT [C] [D] = E – ln ab (3.13) ()cell nF [A] [B]

Represent the cell in which the following reaction takes place Example 3.1 Mg(s) + 2Ag+(0.0001M) → Mg2+(0.130M) + 2Ag(s) V Calculate its E(cell) if E()cell = 3.17 V. The cell can be written as Mg⎥Mg2+(0.130M)⎥⎥Ag+(0.0001M)⎥Ag Solution

2+ V RT ⎣⎦⎡⎤Mg E()= E ()–ln cell cell + 2 2F ⎣⎦⎡⎤Ag

0. 059V 0.130 = 3.17 V – log 2 = 3.17 V – 0.21V = 2.96 V. 2 (.0 0001)

If the circuit in Daniell cell (Fig. 3.1) is closed then we note that the 3.3.1 Equilibrium reaction Constant Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) (3.1) from Nernst takes place and as time passes, the concentration of Zn2+ keeps Equation on increasing while the concentration of Cu2+ keeps on decreasing. At the same time voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as:

2+ V 2.303RT [Zn ] E = 0 = E – log 2+ (cell) ()cell 2F [Cu ]

2+ V 2.303RT [Zn ] or E() = log cell 2F [Cu2+ ] But at equilibrium,

71 Electrochemistry []Zn2+ 2 = K for the reaction 3.1 []Cu + c and at T = 298K the above equation can be written as

V 0. 059 V V E = log K = 1.1 V ( E = 1.1V) ()cell 2 C ()cell (1.1V × 2) log K = = 37.288 C 0.059 V 37 KC = 2 × 10 at 298K. In general,

V 2.303RT E() = log K (3.14) cell nF C Thus, Eq. (3.14) gives a relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding E value of the cell.

Example 3.2 Calculate the equilibrium constant of the reaction: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s) V E()cell = 0.46 V

V 0. 059 V Solution E()= log K = 0.46 V or cell 2 C 046. V ×2 log K = = 15.6 C 0. 059 V 15 KC = 3.92 × 10

3.3.2 Electro- Electrical work done in one second is equal to electrical potential chemical multiplied by total charge passed. If we want to obtain maximum Cell and work from a galvanic cell then charge has to be passed reversibly. The Gibbs reversible work done by a galvanic cell is equal to decrease in its Gibbs Energy of energy and therefore, if the emf of the cell is E and nF is the amount Δ the Reaction of charge passed and rG is the Gibbs energy of the reaction, then Δ rG = – nFE(cell) (3.15)

It may be remembered that E(cell) is an intensive parameter but Δ rG is an extensive thermodynamic property and the value depends on n. Thus, if we write the reaction Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s) (3.1) Δ rG = -2FE(cell) but when we write the reaction 2 Zn (s) + 2 Cu2+(aq) ⎯→2 Zn2+(aq)+2Cu(s)

Chemistry 72 Δ rG = –4FE(cell)

If the concentration of all the reacting species is unity, then E(cell) V = E()cell and we have

Δ V rG = –nF E(cell) (3.16)

V Thus, from the measurement of E()cell we can obtain an important Δ thermodynamic quantity, rG , standard Gibbs energy of the reaction. From the latter we can calculate equilibrium constant by the equation: Δ rG = –RT ln K.

Example 3.3 The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu2+(aq) ⎯→ Zn2+(aq) + Cu(s)

Δ V Solution rG = – nF E(cell) n in the above equation is 2, F = 96487 C mol –1 and V E()cell = 1.1 V Δ –1 Therefore, rG = – 2 × 1.1V × 96487 C mol = –21227 J mol–1 = –21.227 kJ mol–1

Intext Questions 3.4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. 3.5 Calculate the emf of the cell in which the following reaction takes place Ni(s) + 2Ag+ (0.002 M) → Ni2+ (0.160 M) + 2Ag(s) V Given that E(cell) = 1.05 V 3.6 The cell in which the following reaction occurs: 32+−()+→ () + () +() 0 2Feaq 2I aq 2Fe aq I2 s has Ecell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

3.4 Conductance It is necessary to define a few terms before we consider the subject of of Electrolytic conductance of electricity through electrolytic solutions. The electrical resistance is represented by the symbol ‘R’ and it is measured in ohm (Ω) Solutions which in terms of SI base units is equal to (kg m2)/(s3 A2). It can be measured with the help of a Wheatstone bridge with which you are familiar

73 Electrochemistry from your study of physics. The electrical resistance of any object is directly proportional to its length, l, and inversely proportional to its area of cross section, A. That is, l l R ∝ or R = ρ (3.17) A A The constant of proportionality, ρ (Greek, rho), is called resistivity (specific resistance). Its SI units are ohm metre (Ω m) and quite often its submultiple, ohm centimetre (Ω cm) is also used. IUPAC recommends the use of the term resistivity over specific resistance and hence in the rest of the book we shall use the term resistivity. Physically, the resistivity for a substance is its resistance when it is one metre long and its area of cross section is one m2. It can be seen that: 1 Ω m = 100 Ω cm or 1 Ω cm = 0.01 Ω m The inverse of resistance, R, is called conductance, G, and we have the relation:

1 AA G = = = κ (3.18) R ρl l The SI unit of conductance is siemens, represented by the symbol ‘S’ and is equal to ohm–1 (also known as mho) or Ω–1. The inverse of resistivity, called conductivity (specific conductance) is represented by the symbol, κ (Greek, kappa). IUPAC has recommended the use of term conductivity over specific conductance and hence we shall use the term conductivity in the rest of the book. The SI units of conductivity are S m–1 but quite often, κ is expressed in S cm–1. Conductivity of a material in S m–1 is its conductance when it is 1 m long and its area of cross section is 1 m2. It may be noted that 1 S cm–1 = 100 S m–1.

Table 3.2 The Material Conductivity/ Material Conductivity/ values of S m–1 S m–1 Conductivity of some Selected Conductors Aqueous Solutions 3 –5 Materials at Sodium 2.1×10 Pure water 3.5×10 3 298.15 K Copper 5.9×10 0.1 M HCl 3.91 Silver 6.2×103 0.01M KCl 0.14 Gold 4.5×103 0.01M NaCl 0.12 Iron 1.0×103 0.1 M HAc 0.047 Graphite 1.2×10 0.01M HAc 0.016 Insulators Semiconductors Glass 1.0×10–16 CuO 1×10–7 Teflon 1.0×10–18 Si 1.5×10–2 Ge 2.0

It can be seen from Table 3.2 that the magnitude of conductivity varies a great deal and depends on the nature of the material. It also depends on the temperature and pressure at which the measurements

Chemistry 74 are made. Materials are classified into conductors, insulators and semiconductors depending on the magnitude of their conductivity. Metals and their alloys have very large conductivity and are known as conductors. Certain non-metals like carbon-black, graphite and some organic polymers* are also electronically conducting. Substances like glass, ceramics, etc., having very low conductivity are known as insulators. Substances like silicon, doped silicon and gallium arsenide having conductivity between conductors and insulators are called semiconductors and are important electronic materials. Certain materials called superconductors by definition have zero resistivity or infinite conductivity. Earlier, only metals and their alloys at very low temperatures (0 to 15 K) were known to behave as superconductors, but nowadays a number of ceramic materials and mixed oxides are also known to show superconductivity at temperatures as high as 150 K. Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on (i) the nature and structure of the metal (ii) the number of valence electrons per atom (iii) temperature (it decreases with increase of temperature). As the electrons enter at one end and go out through the other end, the composition of the metallic conductor remains unchanged. The mechanism of conductance through semiconductors is more complex. We already know (Class XI, Unit 7) that even very pure water has small amounts of hydrogen and hydroxyl ions (~10–7M) which lend it very low conductivity (3.5 × 10–5 S m–1). When electrolytes are dissolved in water, they furnish their own ions in the solution hence its conductivity also increases. The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance. The conductivity of electrolytic (ionic) solutions depends on: (i) the nature of the electrolyte added (ii) size of the ions produced and their solvation (iii) the nature of the solvent and its viscosity (iv) concentration of the electrolyte (v) temperature (it increases with the increase of temperature). Passage of direct current through ionic solution over a prolonged period can lead to change in its composition due to electrochemical reactions (Section 3.4.1).

* Electronically conducting polymers – In 1977 MacDiarmid, Heeger and Shirakawa discovered that acetylene gas can be polymerised to produce a polymer, polyacetylene when exposed to vapours of iodine acquires metallic lustre and conductivity. Since then several organic conducting polymers have been made such as polyaniline, polypyrrole and polythiophene. These organic metals, being composed wholly of elements like carbon, hydrogen and occasionally nitrogen, oxygen or sulphur, are much lighter than normal metals and can be used for making light-weight batteries. Besides, they have the mechanical properties of polymers such as flexibility so that one can make electronic devices such as transistors that can bend like a sheet of plastic. For the discovery of conducting polymers, MacDiarmid, Heeger and Shirakawa were awarded the Nobel Prize in Chemistry for the year 2000.

75 Electrochemistry 3.4.1 We know that accurate measurement of an unknown resistance can be Measurement of performed on a Wheatstone bridge. However, for measuring the resistance the Conductivity of an ionic solution we face two problems. Firstly, passing direct current of Ionic Solutions (DC) changes the composition of the solution. Secondly, a solution cannot be connected to the bridge like a metallic wire or other solid conductor. The first difficulty is resolved by using an alternating current (AC) source of power. The second problem is solved by using a specially designed vessel called conductivity cell. It is available in several designs and two simple ones are shown in Fig. 3.4.

Fig. 3.4: Two different types of conductivity cells.

Basically it consists of two platinum electrodes coated with platinum black (finely divided metallic Pt is deposited on the electrodes electrochemically). These have area of cross section equal to ‘A’ and are separated by distance ‘l’. Therefore, solution confined between these electrodes is a column of length l and area of cross section A. The resistance of such a column of solution is then given by the equation: l l R = ρ = (3.17) A κ A The quantity l/A is called cell constant denoted by the symbol, G*. It depends on the distance between the electrodes and their area of cross-section and has the dimension of length–1 and can be calculated if we know l and A. Measurement of l and A is not only inconvenient but also unreliable. The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations (Table 3.3) and at different temperatures. The cell constant, G*, is then given by the equation: l G* = = R κ (3.18) A

Chemistry 76 Molarity Concentration Conductivity Molar Conductivity Table 3.3 Conductivity and mol L–1 mol m–3 S cm–1 S m–1 S cm2mol–1 S m2 mol–1 Molar conductivity 1.000 1000 0.1113 11.13 111.3 111.3×10–4 of KCl solutions at 298.15K 0.100 100.0 0.0129 1.29 129.0 129.0×10–4 0.010 10.00 0.00141 0.141 141.0 141.0×10–4

Once the cell constant is determined, we can use it for measuring the resistance or conductivity of any solution. The set up for the measurement of the resistance is shown in Fig. 3.5. Fig. 3.5: Arrangement for measurement of It consists of two resistances R3 resistance of a and R4, a variable resistance R1 and the conductivity cell having the solution of an electrolyte. unknown resistance R2. The Wheatstone bridge is fed by an oscillator O (a source of a.c. power in the audio frequency range 550 to 5000 cycles per second). P is a suitable detector (a headphone or other electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions:

R14 R Unknown resistance R2 = (3.19) R3 These days, inexpensive conductivity meters are available which can directly read the conductance or resistance of the solution in the conductivity cell. Once the cell constant and the resistance of the solution in the cell is determined, the conductivity of the solution is given by the equation: cell constant G* κ= = (3.20) RR The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the ions in which they dissociate, the concentration of ions or ease with which the ions move under a potential gradient. It, therefore, becomes necessary to define a physically more meaningful quantity called molar Λ conductivity denoted by the symbol m (Greek, lambda). It is related to the conductivity of the solution by the equation: κ Molar conductivity = Λ = (3.21) m c In the above equation, if κ is expressed in S m–1 and the concentration,

77 Electrochemistry –3 Λ 2 –1 c in mol m then the units of m are in S m mol . It may be noted that: 1 mol m–3 = 1000(L/m3) × molarity (mol/L), and hence κ (S m−1 ) 2 –1 Λ (S m mol ) = −−3 1 m 1000 L m × molarity (mol L ) If we use S cm–1 as the units for κ and mol cm–3, the units of 2 –1 concentration, then the units for Ëm are S cm mol . It can be calculated by using the equation:

−13 − κ (S cm )× 1000( cm /L) L (S cm21 mol ) = m molarity (mol/L) Both type of units are used in literature and are related to each other by the equations: 1 S m2mol–1 = 104 S cm2mol–1 or 1 S cm2mol–1 =10–4 S m2mol–1.

Example 3.4 Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L–1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 S/m. Solution The cell constant is given by the equation: Cell constant = G* = conductivity × resistance = 1.29 S/m × 100 Ω = 129 m–1 = 1.29 cm–1 Conductivity of 0.02 mol L–1 KCl solution = cell constant / resistance G * 129 m–1 = = = 0.248 S m–1 R 520 Ω Concentration = 0.02 mol L–1 = 1000 × 0.02 mol m–3 = 20 mol m–3 κ Λ = Molar conductivity = m c

248 × 10–3– S m 1 = 20 mol m–3 = 124 × 10–4 S m2mol–1

1.29 cm–1 Alternatively, κ = 520 Ω = 0.248 × 10–2 S cm–1 Λ κ 3 –1 –1 and m = × 1000 cm L molarity

Chemistry 78 0.248×10–2–13– S cm ×1000 cm L 1 = 0.02 mol L–1 = 124 S cm2 mol–1

The electrical resistance of a column of 0.05 mol L–1 NaOH solution Example 3.5 of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. A = π r2 = 3.14 × 0.52 cm2 = 0.785 cm2 = 0.785 × 10–4 m2 Solution l = 50 cm = 0.5 m

ρ l RA 5.55×Ω× 103 0.785cm2 R = or ρ == = 87.135 Ω cm A l 50cm 1 κ 1 –1 Conductivity = = = ( ) S cm ρ 87.135 = 0.01148 S cm–1 κ × 1000 Λ 3 –1 Molar conductivity, m = cm L c 0.01148 S cm–13– ×1000 cm L 1 = 0.05 mol L–1 = 229.6 S cm2 mol–1 If we want to calculate the values of different quantities in terms of ‘m’ instead of ‘cm’,

ρ = RA l

5.55 × 103 Ω × 0.785×10–4 m2 = 0.5 m = 87.135 ×10–2 Ω m 1 κ 100 –1 = = Ω m = 1.148 S m ρ 87.135

κ –1 Λ 1.148 S m and m = = c 50 mol m –3 = 229.6 × 10–4 S m2 mol–1.

Both conductivity and molar conductivity change with the 3.4.2 Variation of concentration of the electrolyte. Conductivity always decreases with Conductivity decrease in concentration both, for weak and strong electrolytes. and Molar This can be explained by the fact that the number of ions per unit Conductivity volume that carry the current in a solution decreases on dilution. with The conductivity of a solution at any given concentration is the Concentration conductance of one unit volume of solution kept between two 79 Electrochemistry platinum electrodes with unit area of cross section and at a distance of unit length. This is clear from the equation: κ A G = = κ (both A and l are unity in their appropriate units in l m or cm) Molar conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore, κA Λ = = κ m l Since l = 1 and A = V ( volume containing 1 mole of electrolyte) Λ κ m = V (3.22) Molar conductivity increases with decrease in concentration. This is because the total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in κ on dilution of a solution is more than compensated by increase in its Λ volume. Physically, it means that at a given concentration, m can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration approaches zero, the molar conductivity is known as limiting molar ° conductivity and is represented by the symbol Ë m. The variation in Λ m with concentration is different (Fig. 3.6) for strong and weak electrolytes. For strong electrolytes, Λ increases slowly with dilution and can be Strong Electrolytes represented by the equation: Λ ° ½ m = Ë m – A c (3.23) It can be seen that if we plot (Fig. Λ 3.12) m against c1/2, we obtain a straight line with Fig. 3.6: Molar intercept equal to ° conductivity versus Ë m and slope equal c½ for acetic acid to ‘–A’. The value of (weak electrolyte) the constant ‘A’ for and potassium a given solvent and chloride (strong temperature electrolyte) in depends on the aqueous solutions. type of electrolyte i.e., the charges on the cation and anion produced on the dissociation of the electrolyte in

Chemistry 80 the solution. Thus, NaCl, CaCl2, MgSO4 are known as 1-1, 2-1 and 2- 2 electrolytes respectively. All electrolytes of a particular type have the same value for ‘A’.

The molar conductivity of KCl solutions at different concentrations at Example 3.6 298 K are given below: –1 Λ 2 –1 c/mol L m/S cm mol 0.000198 148.61 0.000309 148.29 0.000521 147.81 0.000989 147.09 1/2 Show that a plot between Ëm and c is a straight line. Determine the values ° of Ë m and A for KCl. Taking the square root of concentration we obtain: Solution 1/2 –1 1/2 Λ 2 –1 c / (mol L ) m/S cm mol 0.01407 148.61 0.01758 148.29 0.02283 147.81 0.03145 147.09 Λ 1/2 A plot of m( y-axis) and c (x-axis) is shown in (Fig. 3.7). It can be seen that it is nearly a straight line. From the intercept (c1/2 = 0), ° 2 –1 we find that Ë m= 150.0 S cm mol and A = – slope = 87.46 S cm2 mol–1/(mol/L–1)1/2.

Fig. 3.7: Variation

of Ëm against c½.

81 Electrochemistry ° Kohlrausch examined Ë m values for a number of strong electrolytes ° and observed certain regularities. He noted that the difference in Ë m of the electrolytes NaX and KX for any X is nearly constant. For example at 298 K: ° ° ° ° Ë m (KCl) – Ë m (NaCl)= Ë m (KBr) – Ë m (NaBr) ° °  2 –1 = Ë m (KI) – Ë m (NaI) 23.4 S cm mol and similarly it was found that ° ° ° °  2 –1 Ë m (NaBr)– Ë m (NaCl)= Ë m (KBr) – Ë m (KCl) 1.8 S cm mol On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. λ λ – Thus, if °Na+ and °Cl are limiting molar conductivity of the sodium and chloride ions respectively, then the limiting molar conductivity for sodium chloride is given by the equation: ° λ0 + λ0 – Ë m (NaCl) = Na + Cl (3.24)

In general, if an electrolyte on dissociation gives v+ cations and v– anions then its limiting molar conductivity is given by: ° ν λ0 ν λ0 Ë m = + + + – – (3.25) 0 0 Here,λ + andλ − are the limiting molar conductivities of the cation and anion respectively. The values of λ0 for some cations and anions at 298 K are given in Table 3.4.

Ion λ 0/(S cm2mol–1) Ion λ 0/(S cm2 mol–1)

H+ 349.6 OH– 199.1 Table 3.4 Na+ 50.1 Cl– 76.3 Limiting molar K+ 73.5 Br– 78.1 conductivity for some ions in Ca2+ 119.0 CH COO– 40.9 3 water at 298 K − 2+ 2 Mg 106.0 SO4 160.0

Weak electrolytes like acetic acid have lower degree of dissociation at Weak electrolytes Λ higher concentrations and hence for such electrolytes, the change in m with dilution is due to increase in the degree of dissociation and consequently the number of ions in total volume of solution that contains

1 mol of electrolyte. In such cases Ëm increases steeply (Fig. 3.12) on ° dilution, especially near lower concentrations. Therefore, Ë m cannot be Λ obtained by extrapolation of m to zero concentration. At infinite dilution (i.e., concentration c → zero) electrolyte dissociates completely (α =1), but at such low concentration the conductivity of the solution is so low ° that it cannot be measured accurately. Therefore, Ë m for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions (Example 3.8). At any concentration c, if α is the degree of dissociation then it can be approximated to the ratio of molar conductivity Ë at the ° m concentration c to limiting molar conductivity, Ë m. Thus we have:

Chemistry 82 Λ α = m Λ ° (3.26) m But we know that for a weak electrolyte like acetic acid (Class XI, Unit 7),

cα2 ccΛΛ22 K == mm = a ()1 −α ⎛⎞Λ ΛΛοο()− Λ Λ ο 2 ⎜⎟1 − m mm m (3.27) m Λ ο ⎝⎠m Using Kohlrausch law of independent migration of ions, it is possible to Applications of ° λo calculate Ë m for any electrolyte from the of individual ions. Moreover, Kohlrausch law for weak electrolytes like acetic acid it is possible to determine the value ° Λ of its dissociation constant once we know the Ë m and m at a given concentration c.

° Calculate Ë m for CaCl2 and MgSO4 from the data given in Table 3.4. Example 3.7 We know from Kohlrausch law that Solution Λ οοο=λ + λ 2 –1 2 –1 2+2 – m ()CaCl2 CaCl= 119.0 S cm mol + 2(76.3) S cm mol = (119.0 + 152.6) S cm2 mol–1 = 271.6 S cm2 mol–1

Λ οοο=λ + λ 2 –1 2 –1 m () 2+ 2– = 106.0 S cm mol + 160.0 S cm mol MgSO4 Mg SO4 = 266 S cm2 mol–1 .

° 2 –1 Example 3.8 Ë m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm mol respectively. Calculate Λo for HAc.

Λ οοο=λ + λ =λοοοοοο + λ +λ +λ −λ −λ m ()HAc H+–Ac H+––+–+Cl Ac Na Cl Na Solution =+ΛΛοο − Λ ο mm()HCl ( NaAc ) m ( NaCl ) = (425.9 + 91.0 – 126.4 ) S cm2 mol –1 = 390.5 S cm2 mol–1 .

The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S Example 3.9 –1 ° 2 –1 cm . Calculate its dissociation constant if Ë m for acetic acid is 390.5 S cm mol .

−−51 3 κ 4. 95× 10 Scm 1000cm − Solution Λ == × =4815. Scm21 mol m c 0. 001028 mol L−1 L

Λ 48.15 Scm21 mol− α=m = =0.1233 Λο−21 m 390.5 Scm mol

2 –1 2 cα 0.(.) 001028mol L× 0 1233 − –1 K== =×178. 105 mol L ()1.−−α 1 0 1233

83 Electrochemistry Intext Questions

3.7 Why does the conductivity of a solution decrease with dilution? Λ ° 3.8 Suggest a way to determine the m value of water. 3.9 The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given λ0(H+) = 349.6 S cm2 mol–1 and λ0(HCOO–) = 54.6 S cm2 mol–1

3.5 Electrolytic In an electrolytic cell external source of voltage is used to bring about a chemical reaction. The electrochemical processes are of great importance Cells and in the laboratory and the chemical industry. One of the simplest electrolytic Electrolysis cell consists of two copper strips dipping in an aqueous solution of copper sulphate. If a DC voltage is applied to the two electrodes, then Cu 2+ ions discharge at the cathode (negatively charged) and the following reaction takes place: Cu2+(aq) + 2e– → Cu (s) (3.28) Copper metal is deposited on the cathode. At the anode, copper is converted into Cu2+ ions by the reaction: Cu(s) → Cu2+(s) + 2e– (3.29) Thus copper is dissolved (oxidised) at anode and deposited (reduced) at cathode. This is the basis for an industrial process in which impure copper is converted into copper of high purity. The impure copper is made an anode that dissolves on passing current and pure copper is deposited at the cathode. Many metals like Na, Mg, Al, etc. are produced on large scale by electrochemical reduction of their respective cations where no suitable chemical reducing agents are available for this purpose. Sodium and magnesium metals are produced by the electrolysis of their fused chlorides and aluminium is produced (Class XII, Unit 6) by electrolysis of aluminium oxide in presence of cryolite. Quantitative Michael Faraday was the first scientist who described the quantitative Aspects of aspects of electrolysis. Now Faraday’s laws also flow from what has Electrolysis been discussed earlier. Faraday’s Laws After his extensive investigations on electrolysis of solutions and melts of Electrolysis of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two laws of electrolysis: 1. First Law The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). 2. Second Law The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).

Chemistry 84 There were no constant current sources available during Faraday’s times. The general practice was to put a coulometer (a standard electrolytic cell) for determining the quantity of electricity passed from the amount of metal (generally silver or copper) deposited or consumed. However, coulometers are now obsolete and we now have constant current (I) sources available and the quantity of electricity Q, passed is given by Q = It Q is in coloumbs when I is in ampere and t is in second. The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction: Ag +(aq) + e– → Ag(s) (3.30) One mole of the electron is required for the reduction of one mole of silver ions. We know that charge on one electron is equal to 1.6021× 10–19C. Therefore, the charge on one mole of electrons is equal to: –19 23 –1 –19 NA × 1.6021 × 10 C = 6.02 × 10 mol × 1.6021 × 10 C = 96487 C mol–1 This quantity of electricity is called Faraday and is represented by the symbol F. For approximate calculations we use 1F  96500 C mol–1. For the electrode reactions: Mg2+(l) + 2e– ⎯→ Mg(s) (3.31) Al3+(l) + 3e– ⎯→ Al(s) (3.32) It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively. The charge passed through the electrolytic cell during electrolysis is equal to the product of current in amperes and time in seconds. In commercial production of metals, current as high as 50,000 amperes are used that amounts to about 0.518 F per second.

A solution of CuSO4 is electrolysed for 10 minutes with a current of Example 3.10 1.5 amperes. What is the mass of copper deposited at the cathode? t = 600 s charge = current × time = 1.5 A × 600 s = 900 C Solution According to the reaction: Cu2+(aq) + 2e– = Cu(s) We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu. For 900 C, the mass of Cu deposited = (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0.2938 g.

Products of electrolysis depend on the nature of material being 3.5.1 Products of electrolysed and the type of electrodes being used. If the electrode is Electrolysis inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert

85 Electrochemistry electrodes.The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these don’t seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur. For example, if we use molten NaCl, the products of electrolysis are + sodium metal and Cl2 gas. Here we have only one cation (Na ) which is reduced at the cathode (Na+ + e– → Na) and one anion (Cl–) which is –→ – oxidised at the anode (Cl ½Cl2+e ) . During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2. In this case besides Na+ and Cl– ions we also have H+ and OH– ions along with

the solvent molecules, H2O. At the cathode there is competition between the following reduction reactions: + – → V Na (aq) + e Na (s) E()cell = – 2.71 V + – → V H (aq) + e ½ H2 (g) E()cell = 0.00 V The reaction with higher value of E is preferred and, therefore, the reaction at the cathode during electrolysis is: + – → H (aq) + e ½ H2 (g) (3.33) + but H (aq) is produced by the dissociation of H2O, i.e., → + – H2O (l ) H (aq) + OH (aq) (3.34) Therefore, the net reaction at the cathode may be written as the sum of (3.33) and (3.34) and we have – → – H2O (l ) + e ½H2(g) + OH (3.35) At the anode the following oxidation reactions are possible: – → – V Cl (aq) ½ Cl2 (g) + e E()cell = 1.36 V (3.36) → + – V 2H2O(l ) O2 (g) + 4H (aq) + 4e E()cell = 1.23 V (3.37) The reaction at anode with lower value of E is preferred and therefore, water should get oxidised in preference to Cl– (aq). However, on account of overpotential of oxygen, reaction (3.36) is preferred. Thus, the net reactions may be summarised as: HO NaCl (aq) ⎯→⎯⎯⎯2 Na+ (aq) + Cl– (aq) – → – Cathode: H2O(l ) + e ½ H2(g) + OH (aq) – → – Anode: Cl (aq) ½ Cl2(g) + e Net reaction: → + – NaCl(aq) + H2O(l) Na (aq) + OH (aq) + ½H2(g) + ½Cl2(g) The standard electrode potentials are replaced by electrode potentials given by Nernst equation (Eq. 3.8) to take into account the concentration effects. During the electrolysis of sulphuric acid, the following processes are possible at the anode: → + – V 2H2O(l) O2(g) + 4H (aq) + 4e E()cell = +1.23 V, (3.38)

Chemistry 86 2– → 2– – V 2SO4 (aq) S2O8 (aq) + 2e E()cell = 1.96 V (3.39) For dilute sulphuric acid, reaction (3.38) is preferred but at higher concentrations of H2SO4 process, reaction (3.39) is preferred.

Intext Questions 3.10 If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire? 3.11 Suggest a list of metals that are extracted electrolytically. 3.12 Consider the reaction: 2– + – → 3+ Cr2O7 + 14H + 6e 2Cr + 8H2O What is the quantity of electricity in coulombs needed to reduce 1 mol of 2– Cr2O7 ?

Any battery (actually it may have one or more than one cell connected 3.6 Batteries in series) or cell that we use as a source of electrical energy is basically a galvanic cell where the chemical energy of the redox reaction is converted into electrical energy. However, for a battery to be of practical use it should be reasonably light, compact and its voltage should not vary appreciably during its use. There are mainly two types of batteries. In the primary batteries, the reaction occurs only once and after use 3.6.1 Primary over a period of time battery becomes dead and cannot be reused Batteries again. The most familiar example of this type is the dry cell (known as Leclanche cell after its discoverer) which is used commonly in our transistors and clocks. The cell consists of a zinc container that also acts as anode and the cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide and carbon (Fig.3.8). The space between the electrodes is filled by a moist paste of ammonium chloride

(NH4Cl) and zinc chloride (ZnCl2). The electrode reactions are complex, but they can be written approximately as follows : Anode: Zn(s) ⎯→ Zn2+ + 2e– + –⎯→ Cathode: MnO2+ NH4 + e MnO(OH) + NH3 In the reaction at cathode, manganese is reduced from the + 4 oxidation state to the +3 state. Ammonia produced in the reaction forms a complex with Zn2+ to give [Zn 2+ (NH3)4] . The cell has a potential of nearly 1.5 V. Mercury cell, (Fig. 3.9) suitable for low current devices like hearing aids, watches, etc. consists of zinc – mercury amalgam as anode and a paste of HgO and carbon as the cathode. The electrolyte is a paste of KOH and ZnO. The electrode reactions for the cell are given below: Fig. 3.8: A commercial dry cell consists of a graphite (carbon) Anode: Zn(Hg) + 2OH– ⎯→ ZnO(s) + H O + 2e– 2 cathode in a zinc container; the – ⎯→ – Cathode: HgO + H2O + 2e Hg(l) + 2OH latter acts as the anode.

87 Electrochemistry The overall reaction is represented by Zn(Hg) + HgO(s) ⎯→ ZnO(s) + Hg(l) The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.

3.6.2 Secondary Batteries A secondary cell after use can be recharged by passing current through it in the opposite direction so that it can be used again. A good Fig. 3.9: Commonly used mercury cell. secondary cell can undergo a large number of The reducing agent is zinc and the discharging and charging cycles. The most oxidising agent is mercury (II) oxide. important secondary cell is the lead storage battery (Fig. 3.10) commonly used in automobiles and invertors. It consists of a lead anode and a grid of

lead packed with lead dioxide (PbO2 ) as cathode. A 38% solution of sulphuric acid is used as an electrolyte. The cell reactions when the battery is in use are given below: 2– → – Anode: Pb(s) + SO4 (aq) PbSO4(s) + 2e 2– + – → Cathode: PbO2(s) + SO4 (aq) + 4H (aq) + 2e PbSO4 (s) + 2H2O(l) i.e., overall cell reaction consisting of cathode and anode reactions is: → Pb(s)+PbO2(s)+2H2SO4(aq) 2PbSO4(s) + 2H2O(l)

On charging the battery the reaction is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2, respectively.

Fig. 3.10: The Lead storage battery.

Chemistry 88 Another important secondary cell is the nickel- cadmium cell (Fig. 3.11) which has Fig. 3.11: A rechargeable longer life than the nickel-cadmium cell in a lead storage cell but jelly roll arrangement and more expensive to separated by a layer manufacture. We soaked in moist sodium or shall not go into potassium hydroxide. details of working of the cell and the electrode reactions during charging and discharging. The overall reaction during discharge is: → Cd (s)+2Ni(OH)3 (s) CdO (s) +2Ni(OH)2 (s) +H2O(l ) Production of electricity by thermal plants is not a very efficient method 3.7 Fuel Cells and is a major source of pollution. In such plants, the chemical energy (heat of combustion) of fossil fuels (coal, gas or oil) is first used for converting water into high pressure steam. This is then used to run a turbine to produce electricity. We know that a galvanic cell directly converts chemical energy into electricity and is highly efficient. It is now possible to make such cells in which reactants are fed continuously to the electrodes and products are removed continuously from the electrolyte compartment. Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells.

One of the most successful fuel cells Fig. 3.12: Fuel cell using H2 and O produces electricity. uses the reaction of hydrogen with oxygen 2 to form water (Fig. 3.12). The cell was used for providing electrical power in the Apollo space programme. The water vapours produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalysts like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions. The electrode reactions are given below: –⎯→ – Cathode: O2(g) + 2H2O(l ) + 4e 4OH (aq) – ⎯→ – Anode: 2H2 (g) + 4OH (aq) 4H2O(l) + 4e

89 Electrochemistry Overall reaction being: ⎯→ 2H2(g) + O2(g) 2 H2O(l ) The cell runs continuously as long as the reactants are supplied. Fuel cells produce electricity with an efficiency of about 70 % compared to thermal plants whose efficiency is about 40%. There has been tremendous progress in the development of new electrode materials, better catalysts and electrolytes for increasing the efficiency of fuel cells. These have been used in automobiles on an experimental basis. Fuel cells are pollution free and in view of their future importance, a variety of fuel cells have been fabricated and tried.

3.8 Corrosion Corrosion slowly coats the surfaces of metallic objects with oxides or other salts of the metal. The rusting of iron, tarnishing of silver, development of green coating on copper and bronze are some of the examples of corrosion. It causes enormous damage to buildings, bridges, ships and to all objects made of metals especially that of iron. We lose crores of rupees every year on account of corrosion. In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of iron (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex but it may be considered essentially as an electrochemical phenomenon. At a particular spot (Fig. 3.13) of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction V ⎯→ 2+ – E Anode: 2 Fe (s) 2 Fe + 4 e (Fe2+ /Fe) = – 0.44 V Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+ (which is believed to be

available from H2CO3 Oxidation: Fe (s)→ Fe2+ (aq) +2e– formed due to + – → Reduction: O2 (g) + 4H (aq) +4e 2H2O(l) dissolution of carbon Atomospheric dioxide from air into 2+ → + oxidation : 2Fe (aq) + 2H2O(l) + ½O2(g) Fe2O3(s) + 4H (aq) water. Hydrogen ion in water may also be Fig. 3.13: Corrosion of iron in atmosphere. available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction V + – Cathode: O (g) + 4 H (aq) + 4 e ⎯→ 2 H O (l) E + || =1.23 V 2 2 HO 22 H O The overall reaction being: + ⎯→ 2 + V 2Fe(s)+O2(g) + 4H (aq) 2Fe (aq)+ 2 H2O (l) E(cell) =1.67 V

Chemistry 90 The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide

(Fe2O3. x H2O) and with further production of hydrogen ions. Prevention of corrosion is of prime importance. It not only saves money but also helps in preventing accidents such as a bridge collapse or failure of a key component due to corrosion. One of the simplest methods of preventing corrosion is to prevent the surface of the metallic object to come in contact with atmosphere. This can be done by covering the surface with paint or by some chemicals (e.g. bisphenol). Another simple method is to cover the surface by other metals (Sn, Zn, etc.) that are inert or react to save the object. An electrochemical method is to provide a sacrificial electrode of another metal (like Mg, Zn, etc.) which corrodes itself but saves the object.

Intext Questions 3.13 Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging. 3.14 Suggest two materials other than hydrogen that can be used as fuels in fuel cells. 3.15 Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

The Hydrogen Economy At present the main source of energy that is driving our economy is fossil fuels such as coal, oil and gas. As more people on the planet aspire to improve their standard of living, their energy requirement will increase. In fact, the per capita consumption of energy used is a measure of development. Of course, it is assumed that energy is used for productive purpose and not merely wasted. We are already aware that carbon dioxide produced by the combustion of fossil fuels is resulting in the ‘Greenhouse Effect’. This is leading to a rise in the temperature of the Earth’s surface, causing polar ice to melt and ocean levels to rise. This will flood low-lying areas along the coast and some island nations such as Maldives face total submergence. In order to avoid such a catastrope, we need to limit our use of carbonaceous fuels. Hydrogen provides an ideal alternative as its combustion results in water only. Hydrogen production must come from splitting water using solar energy. Therefore, hydrogen can be used as a renewable and non polluting source of energy. This is the vision of the Hydrogen Economy. Both the production of hydrogen by electrolysis of water and hydrogen combustion in a fuel cell will be important in the future. And both these technologies are based on electrochemical principles.

Summary An electrochemical cell consists of two metallic electrodes dipping in electrolytic solution(s). Thus an important component of the electrochemical cell is the ionic conductor or electrolyte. Electrochemical cells are of two types. In galvanic cell, the chemical

91 Electrochemistry energy of a spontaneous redox reaction is converted into electrical work, whereas in an electrolytic cell, electrical energy is used to carry out a non-spontaneous redox reaction. The standard electrode potential for any electrode dipping in an appropriate solution is defined with respect to standard electrode potential of hydrogen electrode taken as zero. The standard potential of the cell can be obtained by taking the difference V V V of the standard potentials of cathode and anode ( E()cell = E cathode – E anode). The V V standard potential of the cells Vare related to standard Gibbs energy (ÄrG = –nF E()cell ) and equilibrium constant (ÄrG = – RT ln K) of the reaction taking place in the cell. Concentration dependence of the potentials of the electrodes and the cells are given by Nernst equation. The conductivity, κ, of an electrolytic solution depends on the concentration of the

electrolyte, nature of solvent and temperature. Molar conductivity, Ëm, is defined by = κ /c where c is the concentration. Conductivity decreases but molar conductivity increases with decrease in concentration. It increases slowly with decrease in concentration for strong electrolytes while the increase is very steep for weak electrolytes in very dilute solutions. Kohlrausch found that molar conductivity at infinite dilution, for an electrolyte is sum of the contribution of the molar conductivity of the ions in which it dissociates. It is known as law of independent migration of ions and has many applications. Ions conduct electricity through the solution but oxidation and reduction of the ions take place at the electrodes in an electrochemical cell. Batteries and fuel cells are very useful forms of galvanic cell. Corrosion of metals is essentially an electrochemical phenomenon. Electrochemical principles are relevant to the Hydrogen Economy.

Exercises 3.1 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn. 3.2 Given the standard electrode potentials, K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V Arrange these metals in their increasing order of reducing power. 3.3 Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq) →Zn2+(aq)+2Ag(s) takes place. Further show: (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell. (iii) Individual reaction at each electrode. 3.4 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s) Δ Calculate the rG and equilibrium constant of the reactions. 3.5 Write the Nernst equation and emf of the following cells at 298 K: (i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s)

Chemistry 92 2+ + (ii) Fe(s)|Fe (0.001M)||H (1M)|H2(g)(1bar)| Pt(s) 2+ + (iii) Sn(s)|Sn (0.050 M)||H (0.020 M)|H2(g) (1 bar)|Pt(s) – + (iv) Pt(s)|Br2(l)|Br (0.010 M)||H (0.030 M)| H2(g) (1 bar)|Pt(s). 3.6 In the button cells widely used in watches and other devices the following reaction takes place: → 2+ – Zn(s) + Ag2O(s) + H2O(l) Zn (aq) + 2Ag(s) + 2OH (aq) Δ Determine r G and E for the reaction. 3.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. 3.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity. 3.9 The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10–3 S cm–1. 3.10 The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 102 × κ/S m–1 1.237 11.85 23.15 55.53 106.74 Λ Λ ½ Calculate m for all concentrations and draw a plot between m and c . Find the Λ0 value of m . 3.11 Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar Λ0 2 –1 conductivity and if m for acetic acid is 390.5 S cm mol , what is its dissociation constant? 3.12 How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al. (ii) 1 mol of Cu2+ to Cu. – 2+ (iii) 1 mol of MnO4 to Mn . 3.13 How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl2.

(ii) 40.0 g of Al from molten Al2O3. 3.14 How much electricity is required in coulomb for the oxidation of

(i) 1 mol of H2O to O2.

(ii) 1 mol of FeO to Fe2O3.

3.15 A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

3.16 Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? 3.17 Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I–(aq)

93 Electrochemistry (ii) Ag+ (aq) and Cu(s) (iii) Fe3+ (aq) and Br– (aq) 3+ (iv) Ag(s) and Fe (aq) 2+ (v) Br2 (aq) and Fe (aq). 3.18 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes.

(ii) An aqueous solution of AgNO3with platinum electrodes.

(iii) A dilute solution of H2SO4with platinum electrodes.

(iv) An aqueous solution of CuCl2 with platinum electrodes.

Answers to Some Intext Questions

3.5E(cell) = 0.91 V Δ=−V −1 7 3.6 rG 45.54 kJ mol , Kc = 9.62 ×10 3.90.114, 3.67 ×10–4 mol L–1

Chemistry 94

Unit: Corrosion Science

Important Questions with Hints

1. What is corrosion of metals? Hints: Corrosion is a process of gradual deterioration and degradation of a metal surface by chemical or electrochemical reaction of metal with environment. 2. What is Pilling-Wedworth rule? Hints: This rule states that a metal is protective or non-porous if the volume of metal oxide layer is at least equal or greater than the volume of metal from which it is formed. The metal oxide layer is non protective and porous if the volume of oxide is less than the volume of the metal. 3. Iron corrodes faster than aluminium, even though iron is placed below aluminium in the electrochemical series. Why? Hints: When aluminium is exposed to environment, it forms a non-porous protective oxide

film of Al 2O3. This oxide film of aluminium surface does not permit corrosion to occur. So, aluminium is not corroded easily. 4. Rusting of iron is quicker in saline water than in ordinary water. Why? Hints: Saline water contains common salt, sodium chloride, which leads conductivity of water. When saline water comes in contact with iron, corrosion current increases and rusting of iron takes place rapidly. 5. How does rain effects on corrosion? Hints: Rain may have following important effects on corrosion: i. It may remove corrosion product as in the case on nickel. ii. Its presence may develop a product having protective quantities as in the case of copper. iii. The rate of attack may be increased as a result of break down or removal of an otherwise protective coating. 6. Why does corrosion occur in steel pipe connected to copper plumbing? Hints: Steel is an alloy of iron, which is higher in the electrochemical series. So, it forms the anode. Hence, iron undergoes oxidation and gets corroded when steel pipe is connected to copper plumbing is exposed to a corroding environment. 7. What do you mean by galvanic corrosion? Hints: When two dissimilar metals such as zinc and copper are electrically connected and exposed to an electrolyte, the metal higher in electrochemical series, i.e. zinc in the present

case, undergoes corrosion and the type of corrosion is called galvanic corrosion. For example, corrosion in steel pipe connected to copper plumbing. 8. Why does corrosion of water filled steel tanks occur below the waterline? [2012] Hints: The areas of steel tank above the water line is more oxygenated and acts as cathode, whereas the areas below water line is not exposed to air and act as anode. Thus, the anodic area below water line undergoes oxidation and gets corroded. This is called differential aeration corrosion. 9. What is meant by tem passivity? [2014] Hints: Passivity is the phenomenon in which a metal or an alloy shows high corrosion resistance due to formation of highly protective and very thin surface film, which is about 0.0004 mm thick. This film is insoluble, non-porous, and self-healing in nature. The metal such as Ti, Cr, Al, etc. and a wide verity stainless steel alloys containing Cr show high corrosion resistance due to formation of protective oxide film of these metals easily. Any damage to this film is automatically repaired in oxidising environments. In reducing environment, the passive metals become chemically active and undergo corrosion. 10. Describe the significances of Pilling-Bedworth rule. Hints: (i) If volume of metal is far greater than that of metal, the oxide film is non-porous. These metals are the least susceptible to oxidation corrosion. (ii) If the volume of oxide layer is far less than that of metal, oxide layer are sufficiently stressed or stained, leading to the formation of pores or cracks. These metals are highly susceptible to corrosion.

(iii) Corrosion by other gases e.g. chlorine (Cl 2), hydrogen sulphide (H 2S), etc.: (a) Silver (Ag) undergoes corrosion in presence of Cl 2 according to the following reaction:

2Ag + Cl 2 = 2AgCl However, AgCl film is protective and prevents further attack of chlorine on silver.

(b) Tin (Sn) reacts with chlorine to form SnCl 4, which is volatile and hence accelerates the corrosion of tin metal.

Sn + 2Cl 2 = SnCl 4 11. Presence of acidic oxides like CO 2, SO 2 and NO 2 accelerates the rusting of iron. Why? Hints: Iron in presence of acidic oxides readily forms the corresponding salts. The portion of iron surface, not in touch with the acidic oxides, forms the anode. The iron surface eroded by the acidic oxides forms the cathodes, accelerating the rusting of iron. 12. Why does impure metal corrode faster than pure metal? Hints: The impurities in metal cause heterogeneity in metal surface and form tiny electrochemical cells. The anodic areas of these electrochemical cells get corroded readily. 13. Crevice corrosion is accelerated by deposition of dust or impurities in crevice area. Why? Hints: These impurities restrict the supply of oxygen underneath the coved portion of metal. The coved crevice area act as anode and suffers further corrosion. 14. What is rusting of iron? Hints: When a droplet of water containing dissolved oxygen comes in contact with iron metal, an electrochemical cell is set up. At the anodic areas, iron dissolves as Fe 2+ ions with

the liberation of electrons. The liberated electrons flow from anodic to cathodic areas through the iron metal, where the electrons are intercepted by the dissolved oxygen, At cathode, Fe Fe 2+ + 2e - 2 + - - At anode, O + 4e + 2H 2O 4OH

2+ - 2Fe + O 2 + 2H 2O 2Fe + 4OH 2+ - Fe ions at anode and OH ions at cathode diffuse and combine to form Fe(OH)2 In presence of excess oxygen, Fe(OH) 2 oxidised to Fe(OH) 3 4Fe(OH) 2 + 2H 2O + O 2 4Fe(OH) 3 The yellow product is called rust, Fe(OH) 3, which has the actual composition as Fe 2O3.H 2O 15. How does waterline corrosion ships prevented? Hints: Waterline corrosion is prevented by painting the sides of ships by special antifouling paints. 16. Pure copper is immune to stress corrosion than Brass. Why? Hints: Brass is an alloy of copper and zinc. So, in corroding environment containing traces 2+ of ammonia or amines, copper and zinc present in brass form complexes [Cu(NH 3)4] and 2+ [Zn(NH 3)4] . This results in dissolution of brass leading to formation of fissures, which propagate and lead to development of cracks. 17. What do you mean by the terms galvanization and tinning? Hints: Coating of zinc on iron by hot dipping is called galvanization and coating of tin on iron is called tinning. 18. What is Sheradizing? Where it is used? Hints: The cementation coating of iron articles with zinc power is termed as Sheradizing. It is preferred by treatment of iron in zinc powder for two to three hours at 350-370 OC. Sheradizing is used for protecting screws, bolts, nuts, washers, etc. 19. Why does part of a nail inside the wood undergo corrosion easily? Hints: The part of nail inside the wood is not exposed to atmospheric condition and becomes deoxygenated thereby behaves as anode. So, due to differential aeration, the part of nail inside the wood acting as anode, undergoes oxidation and gets corroded. 20. A copper equipment should not possess a small steel bolt. Why? Hints: Steel contains iron, which is more anodic than copper. So, steel bolt acts as anode and gets corroded. The copper equipment acts as cathode and is protected. Since anodic area is very small, so it is completely corroded in the due course of time and action of bolt gets finished. 21. What do you mean by electrochemical corrosion? Explain anodic and cathodic electrochemical corrosion. Hints: Corrosion which takes place when a conducting metal is in contact with the metal or when two different metals or alloys are partially or completely immersed in a solution. Metals, which have non-uniform surface behaves like small electrochemical cell in presence of water containing carbon dioxide and dissolved oxygen (DO). This corrosion is an electrochemical phenomenon, which involves the flow of electron between the anodic and cathodic area. Mechanism of Electrochemical corrosion: Electrochemical corrosion may occur in two different processes

(a) Absorption of oxygen (b) Evaluation of hydrogen

Absorption of oxygen mechanism: Let M be the metal, which release electrons and are captured at cathodic area by oxygen. Anode: M → Mn+ + ne - (oxidation) - - Cathode: O2 + 4e + H 2O → 4OH (reduction) The hydroxide ions thus formed combine with metal ions to form corrosion products (figure 1)

Figure 1

Evaluation of hydrogen mechanism: When the environment is acidic then oxidation occurs in anode where loss of electron takes place. There electrons are captured at cathode area by H + resulting in the formation of H 2

Anode: M → Mn+ + ne - (oxidation) + - Cathode: 2H + 2e → H2 (reduction) + n+ Overall: 2M + 2nH → 2M + nH 2

Figure 2

Thus, hydrogen ions once displace from the acidic solution by metal ions and metal gradually dissolves at the anode. Hydrogen are liberated at the cathode. This process is continuously takes place and hence metal surface leads to corrosion.

22. What do you mean by cathode protection? Define the terms sacrificial anode protection and impressed current cathode protection. Hints: Cathode protection is a technique which is used to control the corrosion of a metal surface by making it the cathode of an electrochemical cell. Or it can be defined as the technique to prevent corrosion by converting all the anode sites on the metal surface to cathode sites by supplying electrical current from an alternating source. Sacrificial anode is one type of cathodic protection system, where anode is made from a metal alloy with a more active voltage than the metal of the structure it is protecting. The difference in potential between the two metals means the sacrificial anode material corrodes in preference to the structure. This effectively stops the oxidation reactions on the metal of the structure being protected. Metals which are used as sacrificial anode include magnesium, zinc and aluminium. Typical uses are for the hulls of the ships and boats, offshore, pipelines and production platforms, in salt-water cool marine engines, on small boat propellers and rudders etc.

The advantage of sacrificial anode systems over others are they need no external power source, easy to install, the low voltage and current between the anode and the surface it is protecting frequently generates stray current. Impressed current cathodic protection is a type of system applied where there are elevated current requirements for protection against corrosion. This system works by delivering controlled amount of DC current to the surfaces immersed in water with the lid of ultra- reliable zinc electrodes as well as combined anodes of metal oxide. This system helpful to protect against corrosion for a broad range of metallic materials in different settings such as pipelines, boat hulls, storage tanks etc. 23. What are inhibitors? Explain about anodic and cathodic inhibitors.[2012C, 2014, 2014C, 2015] Hints: Inhibitors are substances which when added in small quantity to the aqueous medium, decreases the rate of corrosion of the metal. They may be organic or inorganic small molecules that dissolve in the corroding medium but don’t form any protective film or scale either in cathodic or anodic area. Anodic inhibitors are generally salts react with metal ions to form insoluble compounds. They form a thin film on the anodic surface and prevent further corrosion. Examples of anodic inhibitors include silicates, phosphates, chromates of alkali and transition metals. Cathodic inhibitors: In acidic medium, corrosion takes place due to cathodic reaction, where hydrogen is liberated. + - H + 2e → H2 Thus, corrosion at anode can be controlled by (a) decreasing the evaluation of hydrogen at the cathode, (b) Increasing the over voltage of hydrogen at the anode The evaluation of hydrogen can be controlled by addition of organic inhibitors such as amines, heavy metal soaps, substituted ureas and thioureas. Antimony and arsenic oxides or salts such as meta arsenate are also used as inhibitors. In neutral solutions, the cathodic reactions is due to absorption ofoxygen and formed OH - ions - → - H2O 1/2O 2 +2e 2OH

This can be eliminated by adding reducing agents such as sodium sulphite (Na2SO3) or deaeration. Diffusion to the cathodic areas can be retarded by adding inhibitors like Zn, Ni or Mg salts to the corroding medium. 24. What are the different factors which affect corrosion? [2011C, 2014] Hints: The different factors which affect the corrosion are: 1. Nature of metal: purity, physical state of metal 2. Nature of oxide film: volume of oxide (pilling-Bedworth rule) 3. Nature of corrosion products 4. Passivity and passivation 5. Over voltage of hydrogen (Details in standard books) 25. What is rust? How much rust will be formed when 100kg of iron have completely rusted away?

Hints: Chemically rust is hydrated iron oxide, Fe 2O3.3H 2O The molecular weight of rust = 214 214 g rust contains = 112 g Fe 112 g iron produces rust = 214 g 100 kg iron produces rust = (214 x 100)/112 = 191.07 kg

Practice yourself

1. Explain about the following with suitable mechanism (i) Pitting corrosion [2011, 2011C, 2013C, 2015] (ii) Crevice corrosion (iii) Stress corrosion (iv) Water-line corrosion (v) Intergranular corrosion (vi) Concentration cell corrosion 2. What is electrochemical corrosion? Explain the electrochemical corrosion in terms of hydrogen evaluation and oxygen absorption mechanism. [2014, 2015] 3. Explain the various factors influencing the rate of corrosion 4. Discuss the role metal oxide formed in oxidation corrosion with the help of Pilling- Bedworth rule. 5. How corrosion is protected by catholic protection? 6. How corrosion can be protected by proper designing of metal articles? 7. Explain the mechanism of wet corrosion. 8. Write short notes on: (a) Cathodic protection (b) Galvanization and tinning (c) Corrosion by differential aeration (d) Corrosion inhibitors

(e) Anodic protection (f) Metallic coating for corrosion prevention (g) Metal cladding for corrosion prevention (h) Pilling Bed-Worth rule [2014, 2015] 9. How can galvanic corrosion be prevented? 10. Explain the difference in the use of anodic and cathodic coatings for corrosion protection. 11. Suggest the preventive measures to control corrosion [2014] 12. What is differential aeration corrosion [2013C, 2014C] 13. The rate of metallic corrosion increases with increase in temperature, Give reason. [2014C] 14. Explain why corrosion takes place at anode but corroded product forms at cathode or nearly cathode area. [2013] 15. How is cathode protection of iron different from its galvanization? [2013] 16. Discuss the influence of the following factors on the rate of metallic corrosion (any two) [2012] (i) Nature of corrosion product (ii) pH (iii) Nature of metal

Materials collected and prepared from: J. Borah etal , Chemistry for Polytechnic, Vol 2, Kalyani Publishers P.C. Jain etal , Engineering Chemistry, Dhanpat Rai Publishing Company(P) Ltd. M.G. Fontana, Corrosion Engineering, McGraw-Hill

Jyotishmoy Borah

Ellingham Diagrams Definitions The Gibbs free energy (∆G) of a reaction is a measure of the thermodynamic driving force that makes a reaction occur. A negative value for ∆G indicates that a reaction can proceed spontaneously without external inputs, while a positive value indicates that it will not. The equation for Gibbs free energy is: ∆G = ∆HT– ∆S where ∆H is the enthalpy, T is absolute temperature, and ∆S is entropy. The enthalpy (∆H) is a measure of the actual energy that is liberated when the reaction occurs (the “heat of reaction”). If it is negative, then the reaction gives off energy, while if it is positive the reaction requires energy. The entropy (∆S) is a measure of the change in the possibilities for disorder in the products compared to the reactants. For example, if a solid (an ordered state) reacts with a liquid (a somewhat less ordered state) to form a gas (a highly disordered state), there is normally a large positive change in the entropy for the reaction.

Construction of an Ellingham Diagram An Ellingham diagram is a plot of ∆G versus temperature. Since ∆H and ∆S are essentially constant with temperature unless a phase change occurs, the free energy versus temperature plot can be drawn as a series of straight lines, where ∆S is the slope and ∆H is the y-intercept. The slope of the line changes when any of the materials involved melt or vaporize. Free energy of formation is negative for most metal oxides, and so the diagram is drawn with ∆G=0 at the top of the diagram, and the values of ∆G shown are all negative numbers. Temperatures where either the metal or oxide melt or vaporize are marked on the diagram. The Ellingham diagram shown is for metals reacting to form oxides (similar diagrams can also be drawn for metals reacting with sulfur, chlorine, etc., but the oxide form of the diagram is most common). The oxygen partial pressure is taken as 1 atmosphere, and all of the reactions are normalized to consume one mole of O2. The majority of the lines slope upwards, because both the metal and the oxide are present as condensed phases (solid or liquid). The reactions are therefore reacting a gas with a condensed phase to make another condensed phase, which reduces the entropy. A notable exception to this is the oxidation of solid carbon. The line for the reaction

C+O2 ==> CO2 is a solid reacting with a mole of gas to produce a mole of gas, and so there is little change in entropy and the line is nearly horizontal. For the reaction

2C+O2 ==> 2CO we have a solid reacting with a gas to produce two moles of gas, and so there is a substantial increase in entropy and the line slopes rather sharply downward. Similar behavior can be seen in parts of the lines for lead and lithium, both of which have oxides that boil at slightly lower temperatures than the metal does. There are three main uses of the Ellingham diagram: 1. Determine the relative ease of reducing a given metallic oxide to metal; 2. Determine the partial pressure of oxygen that is in equilibrium with a metal oxide at a given temperature; and 3. Determine the ratio of carbon monoxide to carbon dioxide that will be able to reduce the oxide to metal at a given temperature.

Ease of Reduction The position of the line for a given reaction on the Ellingham diagram shows the stability of the oxide as a function of temperature. Reactions closer to the top of the diagram are the most “noble” metals (for example, gold and platinum), and their oxides are unstable and easily reduced. As we move down toward the bottom of the diagram, the metals become progressively more reactive and their oxides become harder to reduce. A given metal can reduce the oxides of all other metals whose lines lie above theirs on the diagram. For example, the 2Mg + O2 ==> 2MgO line lies below the Ti + O2 ==> TiO2 line, and so magnesium can reduce titanium oxide to metallic titanium.

Since the 2C + O2 ==> 2CO line is downward-sloping, it cuts across the lines for many of the other metals. This makes carbon unusually useful as a reducing agent, because as soon as the carbon oxidation line goes below a metal oxidation line, the carbon can then reduce the metal oxide to metal. So, for example, solid carbon can reduce chromium oxide once the temperature exceeds approximately 1225°C, and can even reduce highly-stable compounds like silicon dioxide and titanium dioxide at temperatures above about 1620°C and 1650°C, respectively. For less stable oxides, carbon monoxide is often an adequate reducing agent.

Equilibrium Partial Pressure of Oxygen

The scale on the right side of the diagram labelled “Po2” is used to determine what partial pressure of oxygen will be in equilibrium with the metal and metal oxide at a given temperature. The significance of this is that, if the oxygen partial pressure is higher than the equilibrium value, the metal will be oxidized, and if it is lower than the equilibrium value then the oxide will be reduced. To use this scale, you will need a straightedge. First, find the temperature you are interested in, and find the point where the oxidation line of interest crosses that temperature. Then, line up the straightedge with both that point, and with the point labelled “0” that is marked with short radiating lines (upper left corner of the diagram). Now, with the straightedge running through these two points, read off the oxygen partial pressure (in atmospheres) where the straightedge crosses the “Po2” scale, and this is the equilibrium partial pressure. It is possible to reach the equilibrium oxygen partial pressure by use of a hard vacuum, purging with an inert gas to displace the oxygen, or using a scavenger chemical to consume the oxygen. Ratio of CO/CO2 Needed for Reduction

When using carbon as a reducing agent, there will be a minimum ratio of CO to CO2 that will be able to reduce a given oxide. The harder the oxide is to reduce, the greater the proportion of CO needed in the gases.

To determine the CO/CO2 ratio to reduce a metal oxide at a particular temperature, use the same procedure as for determining the equilibrium pressure of oxygen, except line up the straightedge with the point marked “C” (center of the left side of the diagram), and read the ratio off of the scale marked “CO/CO2”. Chemistry 201

Lecture 17

Solubility equilibria

NC State University Solubility Equilibria

• What is the role of metal ions in solution pH?

• How do solution conditions affect solubility?

• Can precipitation be used to separate ions? Text : Sections 8.1 - 8.3 Role of metal ions

Metal ions in solution can act as weak acids as Shown in the Figure. These are called complex ions. Role Hydrated metal ions Acidity constants

Hydrated metal ion Ka 3+ -3 Fe(H2O)6 6 x 10 3+ -5 Al(H2O)6 1 x 10 2+ -8 Cu(H2O)6 3 x 10 2+ -9 Zn(H2O)6 1 x 10 2+ -10 Ni(H2O)6 1 x 10 Solubility Equilibria

Consider the solubility of NaCl in water. The ions Na+ and Cl- are in equilibrium with the solid NaCl(s).

The equilibrium constant is K = 36. This applies only to the ions and not to the solid. The solid does not have concentration. Solubility Equilibria

Since only the ions contribute to the equilibrium constant, we can write,

The equilibrium constant is product of two solubilities, and hence has the name solubility product. Since K = 36, this means that we have 6 M [Na+] and [Cl-] in a saturated solution of NaCl. Calculating Solubility Equilibria

Given these Ksp’s: – AgBr : 5.0 x 10-13 -50 – Ag2S : 6.3 x 10 -18 – Ag3PO4 : 2.6 x 10

What is the molar solubility of each compound? Calculating Solubility Equilibria Assuming a solid of AgBr is in equilibrium with the ions in solution, calculate the concentration + - -13 of Ag and Br . Ksp= 5.0 x 10 Solution: the solubility equilibrium is and the product is

The reaction leads to x moles of Ag+ and Br-. Calculating Solubility Equilibria The concentrations are: Calculating Solubility Equilibria

Assuming a solid of Ag2S is in equilibrium with the ions in solution, calculate the concentration + 2- -50 of Ag and S . Ksp= 6.3 x 10 Solution: the solubility equilibrium is and the product is

The reaction leads to x moles of Ag+ and S2-. Calculating Solubility Equilibria The concentrations are: Calculating Solubility Equilibria

Assuming a solid of Ag3PO4 is in equilibrium with the ions in solution, calculate the concentration + 3- -18 of Ag and PO4 . Ksp= 2.6 x 10 Calculating Solubility Equilibria

Assuming a solid of Ag3PO4 is in equilibrium with the ions in solution, calculate the concentration + 3- -18 of Ag and PO4 . Ksp= 2.6 x 10 Solution: the solubility equilibrium is and the product is

+ 3- The reaction leads to x moles of Ag and PO4 . Calculating Solubility Equilibria The concentrations are: Calculating Solubility Equilibria in the presence of a common ion What is the solubility of AgBr in 0.02 M aqueous NaBr? Calculating Solubility Equilibria in the presence of a common ion What is the solubility of AgBr in 0.02 M aqueous NaBr? Solution: the solubility equilibrium is

Make a reaction table Species Ag+ Br- Initial 0.0 0.02 Final x 0.02+x

Substitute into the solubility product expression and solve for x. Calculating Solubility Equilibria in the presence of a common ion

What is the solubility of AgBr in 0.02 M aqueous NaBr? Substitute into the solubility product expression and solve for x. Calculating Solubility Equilibria of metal hydroxides Will 0.01 M Mg2+ precipitate at pH 4? Step 1. Write down the solubility equilibrium.

Step 2. Determine the hydroxide ion concentration.

Step 3. Substitute into the solubility product expression.

Step 4. Compare the value to the tablulated Ksp. Calculating Solubility Equilibria of metal hydroxides Will 0.01 M Mg2+ precipitate at pH 4?

-12 For this reaction Ksp = 5.61 x 10 . Since the product of the concentrations is less than this value the hydroxide will not precipitate. Here is a related, but harder problem.

At what pH will 0.01 M Mg2+ precipitate?

Justification for treatment of Ksp The comparison of concentrations in the solubility product with the tabulated Ksp is an example of a free energy relationship.

At equilibrium we have a saturated solution so

Thus, if Q > Ksp then DG > 0 and the solution reaction, is not spontaneous.

Calculating Solubility Equilibria of metal hydroxides Can 0.01 M Pb2+ be separated from 0.01 M Mg2+ at pH 9? The answer to this question resides in the relative

Ksp for the hydroxides of each of these ions.

If one is completely insoluble then they can be separated. Calculating Solubility Equilibria of metal hydroxides Can 0.01 M Pb2+ be separated from 0.01 M Mg2+ at pH 9? Calculating Solubility Equilibria of metal hydroxides Can 0.01 M Pb2+ be separated from 0.01 M Mg2+ at pH 9? Step 1. Calculate the [OH-].

Calculating Solubility Equilibria of metal hydroxides Can 0.01 M Pb2+ be separated from 0.01 M Mg2+ at pH 9? Step 1. Calculate the [OH-].

Step2. Calculate the solubility products.

Calculating Solubility Equilibria of metal hydroxides Can 0.01 M Pb2+ be separated from 0.01 M Mg2+ at pH 9? Step 1. Calculate the [OH-].

Step2. Calculate the solubility products.

The answer in this case is yes, Pb(OH)2 can be separated since it will precipitate, while Mg(OH) 2 remains in solution. pH-dependent equilibria

For polyprotic ions, the solubility product will depend on the state of ionization of the ions.

For example, CaCO3 in the ocean is sparingly 2- soluble. However, if the CO3 concentration is reduced by the equilibrium,

then CaCO3 will begin to dissolve. This equilibrium is particularly important since coral reefs and diatoms are composed of CaCO3.

Precipitation of CaCO3

-9 CaCO3 has a solubility product of Ksp = 3.36 x 10 .

This is a moderately small Ksp, which means that CaCO3 should precipitate if there is sufficient 2+ 2- Ca and CO3 .

The equilibrium is

pH-dependent equilibria

2+ 2- What is the concentration of Ca and CO3 in a saturated solution at pH 7?

Step 1. Write the Ksp

Step2. Calculate the solubility.

Calcium carbonate in the oceans

2+ 2- The concentration of Ca and CO3 in the world’s oceans are 0.01 M and 3.1 x 10-4 M, respectively.

Calcium carbonate in the oceans

2+ 2- The concentration of Ca and CO3 in the worlds oceans are 0.01 M and 3.1 x 10-4 M, respectively.

Will CaCO3 precipitate?

Step 1. Calculate the standard free energy

Step2. Calculate the free energy of CaCO3 in sea.

Calcium carbonate in the oceans

2+ 2- The concentration of Ca and CO3 in the worlds oceans are 0.01 M and 3.1 x 10-4 M, respectively.

Will CaCO3 precipitate? Step 2. Calculate the free energy (cont’d)

Since DG > 0 the reaction is NOT spontaneous as written, formation of solution. Therefore, the spontaneous process is precipitation.

Calcium carbonate in the oceans

2+ 2- The concentration of Ca and CO3 in the worlds oceans are 0.01 M and 3.1 x 10-4 M, respectively.

Will CaCO3 precipitate? Simpler approach. Just calculate the reaction quotient and compare it to the solubility product.

The conclusion is the same. CaCO3 will precipitate. But, wait a minute… why hasn’t it already done so?

Using solubility data to determine Ksp Goals

• Quantify solubility of ionic compounds. • Determine pH conditions for precipitation and separation of hydroxide compounds. Water & Its treatment

1). Like air, water is one of the few basic materials which is of prime importance for the preservation of life on this earth.

2). All are aware of the uses of water for drinking, cooking, cooking, bathing & for farming etc.

3). But few know the importance of water as an engineering matrial.

4). As an engineering material water is used for producing steam, in boilers to generate hydro-electric power, furnishing steam for engines, for construction of concrete structures for manufacturing purposes & as a solvent in chemical process.

1 UNIT: III WATER & Its Treatment Sources of Water:

The Main Sources Of Water Are:

1). Rain water 2). River water 3). Spring or well water 4). Sea water

1). Rain water: Rain water is the purest form of natural water.

However, it dissolves considerable amount of gases (CO2 – SO2 - NO – NO2 …etc) and suspended solid particles from atmosphere, during its journey through it and becomes polluted.

2 UNIT: III WATER & Its Treatment 2). River water: River are formed by rain and spring waters. During its flow over the surfacee of land, it dissolves minerals of the soil such as chlorides, sulfates, bicarbonates of sodium, calcium, magnesium ions etc..

3). Spring or well water or Lake water: it contains cosnstant chemical composition. The minerals present in the lake water in the form dissolved form and high quantity of organic matter.

4). Sea water: It is the most impure form of natural water. It contains larger percentage of the dissolved salts (above 3.5%) out of which about 2.6% is NaCl. The NaCl which is present in the dissolved form in sea water will come out as NaCl crystals due to evaporation of sea water.

The other salts present in the sea water are sulphates of sodium, bicarbonates of potassium, magnesium, calcium, bromides of potassium, magnesium etc.

3 UNIT: III WATER & Its Treatment Underground water: Spring & well waters are the underground water sources. They are in general clearer in appearance due to the filtering action of the soil. They contain more of the dissolved salts generally, underground water is of high organic purity.

Types of impurities in water:

The impurities present in water are classified as:

1). Dissolved impurities: dissolved impurities may organic or inorganic. Inorganic impurities: the carbonates, bicarbonates, sulphates, chlorides of calcium, magnesium, iron potassium and aluminium. Organic impurities: Organic water products, amino acids, proteins, etc.

Gases: O2 , CO2 , Oxides of nitrogen and sulphur, H2S etc.

4 UNIT: III WATER & Its Treatment 2). Suspended impurities: It is of two types:

1. Inorganic - sand & clay; 2. Organic – vegetable and animal matter.

3) Biological Impurities: Micro-Organisms like Pathogenic bacteria, fungi, algae etc

5 UNIT: III WATER & Its Treatment DISADVANTAGES OF HARDWATER / CAUSES OF HARDNESS:

The following are the disadvantages when hard water is used for various purpose:

(i) DOMESTIC USE:

(a) Washing and Bathing : Hard water does not form lather easily with soap is wasted

(b) Drinking : Hard water causes bad effects on our digestive system. Sometimes, stone formation takes place in kidneys

(c) Cooking : The boiling point of water is increased due to the presence of salts. Hence, more fuel and time are required for cooking.

6 UNIT: III WATER & Its Treatment (ii) INDUSTRIAL USE:

(a) Textile Industry : Hard water causes wastage of soap. Precipitates of calcium and magnesium soap adhere to the fabrics and cause problem

(b) Paper Industry : Calcium and Magnesium salts in water may effect the quality of paper.

(c) Sugar Industry : Water containing sulphates, carbonates, nitrates affects the crystallisation of sugar.

(d) Pharmaceutical Industry : Hard water may form some undesirable products while preparation of pharmaceutical products.

7 UNIT: III WATER & Its Treatment (iii) STEAM GENERATION IN BOILERS: For steam generation, boilers are employed. If hard water is used in boilers, It may lead to the following troubles

(a) Boiler Corrosion

(b) Scale and Sludge formation.

(c) Priming and Foaming

(d) Caustic embrittlement Pharmaceutical industry

8 UNIT: III WATER & Its Treatment HARDNESS OF WATER (OR) HARDWATER AND SOFT WATER:

Hard Water : Those water which does not produce lather (or) very little lather with soap is called Hard Water

Soft Water : Soft water readily produce a lot of lather when mixed with little soap.

The Hardness of water is caused by the presence of dissolved salts such as Bicarbonates, Sulphates, Chlorides and Nitrates of bivalent metal ions like Ca+2 & Mg+2

9 UNIT: III WATER & Its Treatment Soap is sodium/potassium salt of higher fatty acids like stearic, oleic and palmetic acids.

When soap is mixed with soft water lather is produced due to stearic acid and sodium stearate

 Na – Stearate + H2O  NaOH + Stearic Acid [C17H35COOH]

 Stearic Acid + Na-Stearate  Formatioin of lather.

10 UNIT: III WATER & Its Treatment When soap comes in contact with HARD WATER, Sodium stearate will react with dissolved calcium and magnesium salts and produce calcium stearate or magnesium stearate which is white precipitate

 2Na – Stearate + Ca+2  Ca – Stearate ↓ + 2Na+

+2 + [2C17H35COONa] + Ca  [(C17H35COO)2 Ca] ↓ + 2Na (Soap) (Soluble) (Insoluble) (Soluble)

+2 +  [2C17H35COONa] + Mg  [(C17H35COO)2 Mg] ↓ + 2Na (Soluble) (Soluble) (Insoluble) (Soluble)

11 UNIT: III WATER & Its Treatment Different Type Of Water have different degree of hardness. The different types of water are commercially classified on the basis of degree of hardness as follows:

Hardness Name Of Water

0-70 mg/litre Soft Water

70-150 mg/litre Moderate Hard Water

150-300 mg/litre Hard Water

> 300 mg/litre Very Hard Water

12 UNIT: III WATER & Its Treatment TYPES OF HARDNESS

The hardness of water is of two types

(1) Temporary hardness (or) Carbonate hardness (2) Permanent hardness (or) Non-Carbonate hardness

(3) Temporary Hardness: Temporary hardness is caused by two

dissolved bicarbonate salts Ca(HCO3) and Mg(HCO3). The hardness is called “Temporary Hardness”

Because it can be removed easily by means of boiling.

Heating Ca(HCO3)2  CaCo3 ↓ + H2O + CO2 ↑ Heating Mg(HCO3)2  Mg(OH)2 ↓ + 2CO2 ↑

13 UNIT: III WATER & Its Treatment (2) PERMANENT HARDNESS:

Permanent hardness of water is due to the dissolved chlorides, sulphates and nitrates of calcium and magnesium.

These salts are CaCl2, CaSo4, Ca(NO3)2, MgCl2, MgSo4, Mg(No3)2 These Hardness cannot be removed easily by boiling. Hence it is called “Permanent Hardness”. Only chemical treatment can remove this hardness.

Total Hardness Of Water = Temporary Hardness + Permanent Hardness

14 UNIT: III WATER & Its Treatment DEGREE OF HARDNESS:

• The Concentration of hardness as well as non-hardness constituting ions are, usually expressed in the term of “Equivalent amount of

CaCo3”

• Since this mode permits the multiplication and division concentration, when required. The choice of CaCo3 in particular is due to its molecular weight (m.wt) is “100” (Equivalent wt = 50), and moreover, It is unsoluble salt that can be precipitated in water treatment.

15 UNIT: III WATER & Its Treatment • Therefore 100 parts by weight of CaCo3 hardness must be equivalent to

1 162 parts by weight of Ca(HCo3)2 hardness

2 146 parts by weight of Mg(HCo3)2 hardness

3 136 parts by weight of CaSo4 hardness

4 111 parts by weight of CaCl2 hardness

5 164 parts by weight of Ca(No3)2 hardness

6 120 parts by weight of MgSo4 hardness

7 146 parts by weight of MgCl2 hardness

8 136 parts by weight of Mg(No3)2 hardness

16 UNIT: III WATER & Its Treatment The method of calculating degree of hardness will be clear from the following formula

•Hardness causing in salt in terms of CaCo3

Amount of the hardness causing salt x 100 Molecular weight of hardness causing salt UNITS OF HARDNESS:

These are 4 different units in which the hardness of water is expressed as given below

(1) Parts per million (PPM): PPM is the number of parts of CaCo3 equivalent hardness per 106 parts of water.

6 i.e., 1 PPM = 1 part of CaCo3 equivalent hardness in 10 parts of water.

17 UNIT: III WATER & Its Treatment (2) Milli grams Per Litre (mg/litre): mg/L is the number of milligrams of

CaCo3 equivalent hardness present per litre of water.

i.e., 1 mg/L = 1 mg of CaCo3 equivalent hardness of 1 L of water.

But 1 L water weights = 1 kg of water 1 kg = 1000 gms = 1000 x 1000 mg = 106 mg

6 ∴ 1 mg/L = 1 mg of CaCo3 equivalent per 10 mg of water 6 = 1 part of CaCo3 equivalent per 10 parts of water

∴ 1 mg/L = 1 ppm

18 UNIT: III WATER & Its Treatment Degree Of Clark (ocl) :

o cl is number of grains (1/7000 lb) of CaCo3 equivalent hardness per gallon (10 lb) of water.

(or)

It is defined as the number of parts of CaCo3 equivalent hardness per 70,000 parts of water.

o ∴ 1 cl = 1 grain of CaCo3 eq. hardness per gallon of water.

(or)

o 1 cl = 1 part of CaCo3 eq. hardness per 70,000 parts of water

∴ 1 ppm = 0.07ocl

19 UNIT: III WATER & Its Treatment Degree Of French (oFr) :

o 5 Fr is the number of parts of CaCo3 equivalent hardness per 10 parts of water.

o 5 1 Fr = 1 part of CaCo3 equivalent hardness per 10 parts of water

∴ 0.1o Fr = 1 ppm

Note: The hardness of water can be converted into all the four units by making use of the following interconversion formula

1 ppm = 1mg/L = 0.07ocl = 0.1oFr 1ocl = 1.43oFr = 14.3 ppm = 14.3 mg/L

20 UNIT: III WATER & Its Treatment PROBLEM:

(1) A sample of water gives an analysis 13.6 mg/L of CaSO4. 7.3 mg/L of Mg(HCO3)2. Calculate the total hardness and permanent hardness.

Sol:

Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3

CaSO4 13.6 136 13.6x100 = 10 136

Mg(HCo3)2 7.3 146 7.3 x 100 = 5 146

The Total hardness of H2O = Temporary hardness + Permanent Hardness = 5 + 10 = 15 mg/L

Permanent hardness = 10 ppm (or) 10 mg/L

21 UNIT: III WATER & Its Treatment PROBLEM

(2) Calculate the total hardness of 1000 litre of a sample of water

containing the following impurities 16.2 mg/L of Ca(HCO3), 11.1 mg/L of CaCl2, 60 mg/L of MgSo4 and Ca(HCO3)2, 11.1 mg/L of CaCl2, 60 mg/L of MgSo4 and 19 mg/L if MgCl2. Salt Quantity Present (mg/L) M.Wt Eq. of CaCo Sol: 3 Ca(HCO3)2 16.2 162 16.2x100 = 10 162

CaCl2 11.1 111 11.1x100 = 10 111

MgSO4 60 120 60x100 = 50 120

MgCl2 19 95 19x100 = 20 95

Total hardness of H2O = Temporary hardness + Permanent Hardness = 10 + 10 + 50 + 20 = 90 mg/L

Total hardness for 1000 litres 22 UNIT: III WATER & Its Treatment = 90 x 1000 = 90,000 mg/L PROBLEM

(3) A Sample of hard water contains the following dissolved salts per litre.

CaCl2 = 111 mgs, CaSO4 = 1.36 mgs, Ca(HCO3)2 = 16.2 mgs, Mg(HCO3)2 = 14.6 mgs, Silica = 40 gms, Turbidity = 10 mgs. Calculate the temporary, permanent and total hardness of water in ppm, Ocl & OFr

Sol: Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3

CaCl2 111 mg/L 111 111x100 =100 111

CaSO4 1.36 mg/L 136 1.36x100 = 1 136

Ca(HCO3)2 16.2 mg/L 162 16.2x100 = 10 162

Mg(HCO3)2 14.6 mg/L 146 14.6x100 = 10 146

Note: Si & Turbidity must not be considered because they do not cause hardness to water.

23 UNIT: III WATER & Its Treatment Total hardness of H2O = Hardness of Ca(HCO3)2+ Mg(HCO3)2 interms of CaCO3 equivalents

= 10 + 10 = 20 mg/L

Permanent hardness = Hardness of CaCl2+ CaSO4 interms of CaCO3 equivalents

= 100 + 1 = 101 mg/L

Conversion of hardness:

1 ppm = 1mg/L = 0.07ocl = 0.1o Fr Total hardness of the sample of water = 121 ppm = 121 mg/L = 121 x 0.07 = 8.47ocl and = 121 x 0.1 = 12.1o F

Permanent hardness = 101 mg/L, 101 ppm, 7.07ocl, 10.1o Fr Total hardness = 20 mg/L, 20 ppm, 1.4ocl and 2o Fr

24 UNIT: III WATER & Its Treatment PROBLEM

(04) 1 litre of water from an underground reservoir in Tirupati town in A.P showed the following analysis for contents:

Mg(HCO3)2 = 42 mg; Ca(HCO3)2 = 146 mg; CaCl2 = 71 mg; NaOH = 40 mg; MgSO4 = 48 mg; Organic impurities = 100 mg;

Calculate temporary and permanent and total hardness of the sample of 10,000 lit. of water

Note: NaOH & Organic impurities donot causes any hardness. The Bicarbonate salt cause temp hardness,

While other are responsible for Permanent hardness.

Mg(HCO3)2 = 42 mg in 10,000 (or) 42/10,000 ppm Its CaCO3 eq’s are = [42/10000] x [100/146]

Similarly for Ca(HCO3)2 it is [146/10000] x [100/162] For CaCl2 it is [71/10000] x [100/111] For MgSO it is [48/10000] x [100/120] 4 25 UNIT: III WATER & Its Treatment Temporary Hardness = [42/10000] x [100/146] + [146/10000] x [100x162] Permanent Hardness = [71/10000] x [100/111] + [48/10000] x [100x120]

Total Hardness = Temporary Hardness + Permanent Hardness = ...... + ……… = ……... ppm

26 UNIT: III WATER & Its Treatment PROBLEM

(5) Calculate the temporary & permanent hardness of water in ocl,

containing the following dissolved salts. CaCO3=50 mg/L, MgCl2=9.5 Mg/L, CaCl2=2.2 mg/L and MgSO4=12 mg/L

Note: CaCO3 is an insoluble salt. It does not cause hardness. If CaCO3 is given as H.C.S, It must be considered as Ca(HCO3)2 whose hardness is expressed in the term of CaCO3 equivalent

Sol: Salt Quantity Present (mg/L) M.Wt Eq. of CaCo3 CaCO3 50 mg/L 100 50 mg/L

MgCl2 9.5 mg/L 95 9.5x100 = 10 95

MgSO4 12 mg/L 120 12x100 = 10 12

CaCl2 22.2 mg/L 111 22.2x100 = 20 111

27 UNIT: III WATER & Its Treatment Temporary Hardness Of Water = [Hardness Of CaCO3] = 50 ppm/50 mg/L = 50 x 0.7 = 3.5 ocl

Permanent Hardness Of Water = Hardness of MgCl2 + MgSO4 + CaCl2 = 10 + 10 + 20 = 40 mg/L = 40 x 0.7 = 2.8o cl

28 UNIT: III WATER & Its Treatment DETERMINATION OF HARDNESS OF WATER BY EDTA METHOD:-

1. This is a Complexometric method where Ethlene Diamine Tetra Acetic Acid (EDTA) is the reagent

2. EDTA forms complexes with different metal ions at different pH.

3. Calcium & Magnesium ions form complexes with EDTA at pH 9-10. H To maintain the p 9-10 NH4Cl, NH4OH buffer solution is used.

4. An Alcoholic solution of Eriochrome Black-T (EBT) is used as an indicator

5. The disodium salt of EDTA under the trade name Triplex-III is used for complexation

29 UNIT: III WATER & Its Treatment EDTA

Disodium Salt Of EDTA

M-DETA COMPLEX

30 UNIT: III WATER & Its Treatment BASIC PRINCIPLE:

• When hardwater comes in contact with EDTA, at pH 9-10, the Ca+2 & Mg+2 forms stable, colourless complex with EDTA.

Ca+2 + + EBT pH 9-10  Ca-EBT Mg+2 Mg-EBT (Complex)

(from hard water) (unstable, wine red colour)

• To the hard water sample the blue coloured indicator EBT is added

along with the NH4Cl, NH4OH buffer solution. EBT forms an unstable, winered complex with Ca+2 & Mg+2

Ca+2 + + EDTA pH 9-10  Ca-EDTA Mg+2 Mg-EDTA (Complex)

(from hard water) (stable, colourless) 31 UNIT: III WATER & Its Treatment • The winered coloured [Ca-EBT, Mg-EBT] complex is titrated with EDTA replaces EBT from Ca-EBT complex and form stable colourless Mg-EBT

[Ca-EDTA] [Mg-EDTA] complex releasing the blue coloured

indicator EBT into H2O

• Hence the colour change at the end point is winered to blue colour

• The titration is carried out in the following steps

1. PREPARATION OF STANDARD HARD WATER:

Dissolve 1gm of pure, dry CaCO3 in minimum quantity of dilute HCl and evaporate the solution to dryness on a waterbath.

32 UNIT: III WATER & Its Treatment • Dissolve the residue in distilled water to make 1 litre in a standard flask and shake well.

Molarity of standard hard water solution = wt. of CaCO3 m.wt. of CaCO3

= 1 100 = 0.01 M

(2) PREPARATION OF EDTA SOLUTION:

Dissolve 4 gms of pure EDTA crystals along with 0.1 gm of MgCl2 in one litre of distilled water.

33 UNIT: III WATER & Its Treatment (3) PREPARATION OF INDICATOR (EBT):

Dissolve 0.5 gms of Erichome Black-T in 100 ml of alcohol.

(4) PREPARATION OF BUFFER SOLUTION:

Add 67.5 gm of NH4Cl to 570 ml of concentrated ammonia solution and dilute with distilled water to one litre

(5) STANDARDISATION OF EDTA SOLUTION:

Pipette out 20 ml of standard hard water solution into a conical flask. Add 2-3 ml of buffer solution and 2-3 drops of EBT indicator.

Titrate the wine red coloured complex with EDTA taken in a burette after rinsing it with EDTA solution till the wine red colour changes to clear blue.

34 UNIT: III WATER & Its Treatment Not the burette reading and let the volume be “x”-ml. Repeat the titration to get concurrent values.

(6) STANDARDISATION OF HARD WATER SAMPLE:

Pipette out 20 ml of the water sample into a 250ml conical flask, add 2-3 ml of buffer solution and 2-3 drops of EBT indicator.

• Titrate the wine red coloured solution with EDTA taken in the burette till a clear blue coloured endpoint is obtained

• Let the volume of EDTA be “y” ml. Repeat the titration to get concurrent values

35 UNIT: III WATER & Its Treatment (7) STANDARDISATION FOR PERMANENT HARDNESS:

Pipette out 100 ml of hard water sample in a beaker and boil till the volume reduces to 20 ml. All the bicarbonates of Ca++ and Mg++ decomposes to

CaCO3 and Mg(OH)2

• Cool the solution and filter the water into a flask, wash the beaker and precipitate with distilled water and add the washing to conical flask

• Add 2-3 ml of buffer solution and 2-3 drops of EBT indicator and titrate with EDTA solution taken in the burette till a clear blue colour end point is obtained.

• Note the burette reading. Let the volume be “z” ml

36 UNIT: III WATER & Its Treatment CALCULATIONS:

Molarity of standard hard water solution = 0.01 M. (Calculated in the preparation of standard hard water)

• Molarity of EDTA solution (M2):

V1M1 = V2M2 n1 n2

+2 • n1 & n2 are no. of moles of Ca and EDTA = 1 each

i.e., n1=1, n2=1

37 UNIT: III WATER & Its Treatment V1 = volume of standard hard water M1= Molarity of standard hard water V2= volume of EDTA M2= molarity of EDTA

M2 = V1M1 = 20x 0.01 V2 titre value (xml)

38 UNIT: III WATER & Its Treatment • Molarity of hard water sample (M3):

V2M2 = V3M3 n2 n3

V2= volume of EDTA M2= molarity of EDTA V3= volume of standard hard water M3= Molarity of standard hard water

M3 = V2M2 = titre value (4 ml) x M2 V3 20

39 UNIT: III WATER & Its Treatment • Total Hardness Of Water = M3 x 100 gms/1 litre = M3 X 100 X1000 mg/L or ppm

• Permanent Hardness Of Water:

M3 = V2M2 = V4 M4 n2 n4

V2= volume of EDTA M2= molarity of EDTA V4 = volume of water sample containing permanent hardness (100 ml) M4= Molarity of water sample containing permanent hardness

40 UNIT: III WATER & Its Treatment • Permanent Hardness Of The Water Sample = M4 x 100 x 1000 ppm

• Temporary Hardness Of The Water Sample = (Total Hardness – Permanent Hardness) = (M3 x 100 x1000 –M4 x 100 x1000) ppm

41 UNIT: III WATER & Its Treatment • 1 gm of CaCO3 was dissolved in HCl and the solution was made upto one litre with distilled water. 50 ml of the above solution required 30 ml of EDTA solution on titration. 50 ml of hard water sample required 40 ml of the same solution of EDTA for titration. 50 ml of the hard water after boiling, filtering etc. required 30 ml of the same EDTA solution for titration. Calculate the temporary hardness of the water

Soln: Molarity of CaCO3 solution (M3) = 1/100 = 0.01 M Molarity of EDTA solution (M2) = V1 M1 V2

V1= volume of CaCO3 solution = 50 ml M1= Molarity of CaCO3 solution = 0.01 M V2= volume of EDTA = 30 ml M2= V4 M4 = 50 x 0.01 = 0.016 M n2 30

42 UNIT: III WATER & Its Treatment • Molarity of Hard Water Solution (M3) = V2 M2 V3

V2= volume of EDTA = 40 ml M2= Molarity of EDTA = 0.016 M V3= volume of hard water = 50 ml M3= 40 x 0.016 = 0.0128 M 50

Total Hardness of water = 0.0128 x 100 x 1000 = 1280 ppm

Permanent hardness of water: = V4 M4 = V2 M2 n4 n2

M4 = V2 M2 V4

43 UNIT: III WATER & Its Treatment n4=1; V2= volume of EDTA = 30 ml n2=1; M2= molarity of EDTA = 0.016 V4= volume of permanent hardness containing water = 50

M4 = 30 x 0.016 = 0.0096 M 50

Permanent hardness of water = 0.0096 x 100 x 1000 = 930 ppm

Temporary hardness = Total hardness – Permanent hardness = 1280 – 960 = 320 ppm

44 UNIT: III WATER & Its Treatment . 0.5 g of CaCO3 was dissolved in dil. HCl & diluted to 1000 ml. 50 ml of this solution required 48 ml of EDTA solution for titration. 50 ml of hard water sample required 15 ml of EDTA solution for titration. 50 ml of same water sample on boiling, filtering etc required 10 ml of EDTA solution.

Calculate the different kind of hardness in ppm

Soln:

Volume of EDTA consumed for 50 ml of standard hardwater (V1) = 48 ml, Volume of EDTA consumed for 50 ml of given sample (V2) = 15 ml Volume of EDTA consumed for 50 ml of boiled water (V3) = 10 ml

45 UNIT: III WATER & Its Treatment • Total Hardness = V2 x 1000 mg/L V1 = 15 x 1000 = 312.5 mg/L 48

• Permanent Hardness = V3 x 1000 ml V1 = 10 x 1000 48 = 208.3 mg/L

• Temporary Hardness = Total Hardness – Permanent Hardness = 312.5 – 208.3 = 104.2 mg/L

46 UNIT: III WATER & Its Treatment • BOILER TROUBLES:

Water finds a great use in various industries for generation of steam in boilers. When water is continuosly evaporated to generate steam, the concentration of the dissolved salts increase progressively causing bad effects for steam boilers.

• The following are the boiler troubles that arise:

(1) Priming and foaming (2) Caustic embrittlement (3) Boiler corrosion (4) Scale and sludge formation

47 UNIT: III WATER & Its Treatment • PRIMING & FOAMING:

Priming: When a boiler is steaming rapidly, some particles of the liquid water are carried along with the steam. This process of “WET STEAM” formation is called ‘PRIMING’

• Priming is caused by:

(1) Presence of large amount of dissolved solids (2) High steam velocities (3) Sudden boiling (4) Improper boiler design (5) Sudden increase in steam production rate

48 UNIT: III WATER & Its Treatment • It can be avoided by: (Preventions)

(1) Maintainig low water level (2) Using softened water (3) Fitting mechanical steam purifiers (4) Using a well-designed boiler (5) Avoiding rapid change in steam rate (6) Blowdown of the boiler

49 UNIT: III WATER & Its Treatment • FOAMING: Foaming is phenomenon of formation of foam or bubbles on the surface of water inside the boiler with the result that the foam may pass along with the steam.

• Causes: The presence of large quantity of suspended impurities and oils lowers the surface tension producing foam.

• Preventions: Foaming can be avoided by

(1) Adding anti foaming chemicals like castor oil

(2) Removing oil foam boiler water by adding compounds like “NaAlO2” (3) Blow down of the boiler can prevent the foaming

50 UNIT: III WATER & Its Treatment • Disadvantages Of Priming & Foaming:

Priming & Foaming may cause the following boiler troubles:-

(1) The actual height of the water in boiler is not judged (2) Wastage of heat with the result that it becomes difficult to keep up steam pressure and efficiency of the boiler is lowered

• CAUSTIC EMBRILLEMENT:

Caustic embrillement is a term used for the appearance of cracks inside the boiler particularly at those places which are under stress such as rivetted joints due to the high concentration of alkali leading to the failure of the boiler. The cracks have appearance of brittle fracture. Hence, the failure is called “Caustic Embrillement”

51 UNIT: III WATER & Its Treatment • Reasons for the formation of Caustic Embrillement:

 The boiler feed water containing carbonates and bicarbonates of alkali metals, sodium hydroxide (NaOH) and a small quantity of silica or sodium silicate is purified by Lime-Soda Process.

 During the softening process by lime soda process, free Na2Co3 is usually present in small portion in the soft water which decomposes to give NaOH and Co2 at high pressure of the boilers

Na2CO3 + H2O  2NaOH + CO2

 The precipitation of NaOH makes the boiler water “Caustic”  The NaOH containing water flows into the small pits and minute hair- cracks present on the inner walls of the boiler  As the water evaporates, the concentration of caustic soda (NaOH) increases progessively, creating a “Concentration Cell”

52 UNIT: III WATER & Its Treatment • Thus, dissolving the iron of the boiler as Sodium Ferrate

Fe + 2NaOH  Na2FeO2 + H2

 The cracking of the boiler occurs particularly at stressed parts like bends, joints, rivets etc causing the failure of the boiler  The iron at plane surfaces surrounded by dilute NaOH becomes Cathodic; while the Iron at bends, rivets and joints is surrounded by highly concentrated NaOH becomes Anodic which consequently decayed or corroded

• Prevention Of Caustic Embrittlement:

(1) By adding Na2SO4, tannin etc to the boiler water which blocks hair cracks. There by preventing infiltration of caustic soda solution (2) By using sodium phosphate as the softening agent instead of sodium carbonate

53 UNIT: III WATER & Its Treatment • DISADVANTAGES OF CAUSTIC EMBRITLLEMENT:

 The cracking or weaking of the boiler metal causes failure of the boiler

(3) BOILER CORROSION: Boiler corrosion is a decay of boiler material by chemical/electro chemical attack by its environment is called “Boiler Corrosion”

Reasons for boiler corrosion are:

(a) Dissolved oxygen

(b) Dissolved SO2 (c) Acids from dissolved salts

54 UNIT: III WATER & Its Treatment (a) Dissolved Oxygen: Water usually contains about 8 mg/L of dissolved oxygen at room temperature.

Dissolved O2 at high temperature attacks boiler material

2Fe + 2H2O + O2  2Fe(OH)2 ↓ (ferrous hydroxide)

4Fe(OH)2 ↓ + O2  2[Fe2O3.2H2O] ↓ (ferric oxide cruit)

Removal of dissolved O2: By adding calculated quantity of sodium sulphate (or) hydrazine (or) sodium sulphide

2Na2SO3 + O2  2Na2SO4 Sodium Sulphite

N2H4 + O2  N2 + 2H2O Hydrazine

Na2S + 2O2  Na2SO4

55 UNIT: III WATER & Its Treatment (a) Dissolved CO2 : Dissolved CO2 has slow corrosive effect on the materials of boiler plate.

Source of CO2 into water is the boiler feed water which contains bicarbonates

Under the high temperature and pressure, maintained in the boiler the

bicarbonates decompose to produce CO2

Ca(HCO3)2  CaCO3 + CO2 ↑ + H2O Mg(HCO3)2  Mg(OH)2 + 2CO2 ↑

The disadvantage of the CO2 is slow corrosive effect on boiler plates by producing carbonic acid

CO2 + H2O  H2CO3

56 UNIT: III WATER & Its Treatment Removal of CO2:

By the addition of calculated quantity of ammonia.

2NH4OH + CO2  (NH4)2CO3 + H2O

(c) Acids from dissolved salts: Water containing dissolved Mg-salts liberate acids on hydrolysis

MgCl2 + 2H2O  Mg(OH)2↓ + 2HCl

Disadvantages of the acid production is that the acids react with Iron of the boiler plate in a chain reaction to produce decay of the metal.

Fe + 2HCl  FeCl2 + H2 FeCl2 + H2O  Fe(OH)2 + 2HCl Fe(OH)2 + H2O + 1O2  Fe2O3.3H2O (ferric oxide) 2

57 UNIT: III WATER & Its Treatment Consequently even a small amount of MgCl2 can cause corrosion to a large extent.

Preventions:

(1) Softening of boiler water to remove MgCl2 from the water

(2) Additon of corrosion inhibitors like sodium silicates, sodium phosphate & sodium chromate

(3) By frequent blowdown operation i.e., removal of water, concentrated with dissolved salts and feeding the boiler with fresh soft water.

(4) Sludges and Scales formation: In boiler, water evaporate continously and the concentration reaches saturation point, they form precipitates (scale or sludge) on the inner wall of the boiler

58 UNIT: III WATER & Its Treatment SLUDGE: “Sludge is a soft, loose and slimy precipitate formed within the boiler”. Sludge are formed by substances which have greater solubilities in hot water than in cold water.

Salts like MgCO3, MgSO4, MgCl2, CaCl2 etc., are responsible for sludge formation in boilers.

Disadvantages:

(a) Sludge is a bad conductor of heat, hence it wastes a portion of heat generated.

(b) Excessive sludge formation reduces the effeciency of the boiler.

59 UNIT: III WATER & Its Treatment Prevention:

(a) Frequent blow-down operation should be carried out

(b) By using well-softened water.

SCALES: “Scales are the hard, adhering ppt formed on the inner wall of the boiler”

Very difficult to remove once they are deposited on Inner wall of the boiler

They formed due to decomposition of Calcium bicarbonate, Calcium sulphate etc..,

Ca(HCO3)2  CaCO3 ↓ + H2O + CO2 ↑

60 UNIT: III WATER & Its Treatment REMOVAL OF SCALES:

(1) Frequent blowdown operation can remove the scales which are loosely adhering

(2) By chemical treatment

eg: CaCO3 scale removed by washing with 5-10% of HCl

(3) By giving thermal shocks

Prevention:

By using softening water which is discussed separately in treatment of water.

61 UNIT: III WATER & Its Treatment TREATMENT OF BOILER FEED WATER:

Water used for industrial purpose especially for generation of steam should be sufficiently pure. The treatment of water includes the removal of hardness causing salts either by precipitation or by complex formation. Hence two types of treatments are given as below.

62 UNIT: III WATER & Its Treatment 63 UNIT: III WATER & Its Treatment EXTERNAL TREATMENT (OR) SOFTENING OF WATER:

The removal of hardness causing salts from water is called “Softening of water”

The 3 Industrial methods employed for softening of water are: (1) Lime-Soda Process (2) Zeolite (or) Permutite Process (3) Ion-Exchange (or) Demineralization process

(1) LIME-SODA PROCESS: This process is based on converting the soluble calcium and magnesium salts into Insoluble calcium carbonate and magnesium hydroxide precipitates by addition of calculated amount of lime (Ca(OH)2) and Soda (Na2CO3). The precipitate are removed by filtration. Any free dissolved CO2 and acids are also removed by this process. The various chemical reactions involved in this process are:

64 UNIT: III WATER & Its Treatment (a) For Calcium and Magnesium bicarbonates, only lime is required

(i) Ca(HCO3)2 + Ca(OH)2  2CaCO3↓ + 2H2O

(ii) Mg(HCO3)2 + 2Ca(OH)2  2CaCO3↓ + Mg(OH)2↓ + 2H2O

(b) For MgSO4 & MgCl2, both lime & soda are required

(iii) MgSO4 + Na2CO3 + Ca(OH)2  Mg(OH)2 ↓ + CaCO3↓ + Na2SO4

(iv) MgCl2 + Na2CO3 + Ca(OH)2  Mg(OH2) ↓ + CaCO3 + 2NaCl

(c) For CaSO4 & CaCl2, only Soda is required

(v) CaSO4 + Na2CO3  CaCO3↓ + Na2SO4

(vi) CaCl2 + Na2CO3  CaCO3↓ + 2NaCl

65 UNIT: III WATER & Its Treatment (a) Other Reactions: Free acids, CO2, H2S dissolved iron and aluminium salts etc are also removed in this process

2HCl + Na2CO3  2NaCl + H2O + CO2↑

H2SO4 + Na2CO3  Na2SO4 + H2O + CO2↑

CO2 + Ca(OH)2  CaCO3↓ + H2O

H2S + Ca(OH)2  CaS↓ + 2H2O

FeSO + Ca(OH)2  Fe(OH)2↓ + CaSO4

Al2(SO4)3 + 3Ca(OH)2  2Al(OH)3↓ + 3CaSO4 + H2O

66 UNIT: III WATER & Its Treatment Calculation: 100 parts by mass of CaCO3 are equivalent to 74 parts of Ca(OH)2 and 106 parts of Na2CO3

(a) Amount of lime required for softening = 74 100 2+ 2+ +2 +2 3+ + (Temp Ca + 2 x Temp. Mg + Perm. (Mg + Fe + Al ) + CO2 + H (HCl/H2SO4) + HCO3- all in terms of CaCO3 equivalent)

(b) Amount of lime required for softening = 106 100 2+ 2+ +2 3+ + (Perm. (Ca + Mg + Fe + Al ) + H (HCl/H2SO4) + HCO3- all in terms of CaCO3 eq.)

67 UNIT: III WATER & Its Treatment . COLD-LIME-SODA PROCESS:

68 UNIT: III WATER & Its Treatment . In this method the lime & soda are mixed with hard water at room temperature with constant stirring

. Generally the precipitates formed by this process are finely divided and in order to settle the precipitates, coagulants like alum, ferrous sulphate etc are added

. The hard water to be softened is mixed with calculated quantity of chemicals (Lime + Soda + Coagulant) from the top into the inner chamberon vigorous stirring. The chemical reactions takes place and the hardness producing salts get converted into insoluble precipitates

. The sludge is removed from the bottom of the outer chamber while the softened water passes through a wood fibre filter to ensure the complete removal of any residual sludge particles

. The clear softened water is withdrawn from the top of the outer chamber. The softened water from this process contains a residual hardness of 50-60ppm

69 UNIT: III WATER & Its Treatment . HOT-LIME-SODA PROCESS:

70 UNIT: III WATER & Its Treatment . This process is similar to the cold lime-soda process, but no coagulant is needed.

. Here the process is carried at a temperature of 80o to 150o c. Since the reaction carried out at high temperature.

(a) The reaction takes place faster (b) The sludge settles rapidly (c) Viscosity of soft water is lower, hence filtered easily (d) The dissolved gases such as CO2, air etc driven out of the water (e) The residual hardness is low, compared to cold lime-soda process. Hot lime soda process consists of three parts

. “REACTION TANK” in which complete mixing of water, chemicals and steam takes place and water gets softened.

. “Conical Sedimentation Vessel” where the sludge settle down.

71 UNIT: III WATER & Its Treatment . “SAND FILTER” where sludge is completely removed

. The softened water from this process contains a residual hardness of 15-30 ppm

ADVANTAGES OF LIME-SODA PROCESS:

I. This process is economical II. Mineral content of the water is reduced III. The process increases the pH value of water, which reduces the content of pathogenic bacteria IV. Manganese and Iron salts are also removed by this process V. The process improves the corrosion resistance of the water

72 UNIT: III WATER & Its Treatment DISADVANTAGES OF LIME-SODA PROCESS:

1) Due to residual hardness, water is not useful for high pressure boilers 2) Large amount of sludge is formed which create disposal problem

73 UNIT: III WATER & Its Treatment PROBLEMS: 1) Calculate the quantities of Lime & Soda required to soften 5000 litres of water containing the following salts:

MgCl2 = 15.5 ppm; Ca(HCO3)2 = 32.5 ppm, CaSO4 = 22.4 ppm; Mg(HCO3)2 = 14.6 ppm NaCl = 50 ppm

Soln: Calculation of Calcium Carbonate Equivalents:

Hard Salt Weight (ppm) M.Wt CaCO3 eq. = W x 100 MW

MgCl2 15.5 95 15.5x100 = 16.31 95

Ca(HCO3)2 32.5 162 32.5 x 100 = 20.06 162

CaSO4 22.4 136 22.4 x 100 = 16.47 136

Mg(HCO3)2 14.6 146 14.6 x 100 = 10.00 146

74 UNIT: III WATER & Its Treatment Lime required for litre of water:

= 74 (Ca(HCO ) + 2 x Mg(HCO ) + MgCl as CaCO eq. 100 3 2 3 2 2 3

= 74 (20.06 + 2x10 + 16.31) 100

= 74 (56.37) 100 = 41.71 mg

Lime req’d for 5000 litres of water = 41.71 x 5000 = 208.55 g 1000

75 UNIT: III WATER & Its Treatment Soda req’d for litre of water = 106 [MgCl2 + CaSO4] as CaCO2 eq. 100

= 106 [16.31 + 16.47] 100

= 106 [32.78] 100

= 34.74 mg

Soda req’d for 5000 litres of water:

= 34.74 x 5000 1000

= 0.173 kg

76 UNIT: III WATER & Its Treatment . Zeolite or Permutit Process:

Zeolite is “Hydrated sodium alumino silicate”

Its general formula is: Na2O.Al2O3.xSiO2.yH2o Here: x=2-10 y=2-6

Eg: Natrolite: Na2O.Al2O3.3SiO2.2H2o

. Natural zeolites are generally non-porous

. The artificial zeolite is called Permutit. These are prepared by heating together with chain clay, feldspar and soda ash.

. These are porous and have greater softening capacity than natural zeolite.

77 UNIT: III WATER & Its Treatment . They exchange Na+ ions with the hardness, producing ions (Ca2+, Mg2+, etc) in water.

. Sodium Zeolite is denoted as Na2Ze

. PROCESS: In this process hard water is passed through a bed of zeolite at ordinary temperature. The hard water percolates (filtered), Ca+2, Mg2+ present in hard water are exchangeed with Na+ ions

. The following reactions taking place:

MgCl2 + Na2Ze  MgZe + 2NaCl

MgSO4 + Na2Ze  MgZe + Na2SO4

CaCl2 + Na2Ze  CaZe + 2NaCl

78 UNIT: III WATER & Its Treatment CaSO4 + Na2Ze  CaZe + Na2SO4

Mg(HCO3)2 + Na2Ze  MgZe + 2NaHCO3

Ca(HCO3)2 + Na2Ze  CaZe + 2NaHCO3 Regeneration Of Zeolite: On continuous passing of hard water through sodium zeolite bed it is conveted to calcium and magnesium zeolite which is known as Exhausted Bed. Hence, it must be regenerated.

This can be done by washing zeolite bed with 10% sodium chloride sol’n.

CaZe + 2NaCl  Na2Ze + CaCl2

MgZe + 2NaCl  Na2Ze + MgCl2 (Regenerated Zeolite)

79 UNIT: III WATER & Its Treatment 80 UNIT: III WATER & Its Treatment ADVANTAGES:

1) The equipment is small and easy to handle

2) It requires less time for softening

3) Water obtained from this process contains a residual hardness upto 10 ppm

4) Easy to regenerate

5) No sludge is formed in this process

81 UNIT: III WATER & Its Treatment DISADVANTAGES:

1) Highly turbid water cannot be treated by this process.

2) The process exchanges only Ca+2 & Mg2+ ions by sodium ions and hence the softened water contains more sodium salts.

3) All the acidic ions like HCO3-, CO32- etc are not removed by this process. Sodium bicarbonate decomposes in the boiler releasing CO2 which leads to corrosion. While Na2CO3 is hydrolysed to NaOH which creates castic embrittlement of boiler.

82 UNIT: III WATER & Its Treatment ION EXCHANGE PROCESS/DETERMINERALISATION PROCESS:

Ion exchange resins are insoluble, cross-linked, long chain organic polymers. The functional groups attached to the chains can exchange hardness producing cat-ions and an-ions present in the water

PROCESS: The process involves the following steps:

1) The first chamber is packed with cat-ion exchange resin (RH+). When the hard water is passed through a bed of cation exchange resin it exchanges H+ with Ca+, Mg+2, K+, Na+ etc of hard water.

+ 2+ 2+ + - 2RH + Mg Cl2 R2Mg + 2H Cl

+ 2+ 2+ + - 2RH + Ca Cl2  R2Ca + 2H Cl

+ 2+ +2 + - 2RH + CaSO4  R2Ca + 2H + SO4

83 UNIT: III WATER & Its Treatment Thus, the hardness producing cations (Ca2+, Mg2+ etc) are removed 1) The second chamber is packed with anion exchange resin. The water + - 2 - coming out of the first chamber contains H , Cl , SO4 - and HCO3 ions. It is now passed through anion exchange resin bed which can exchange - 2- - OH ions with anions like Cl-, SO4 and HCO3 RlOH + Cl-  RlCl + OH-

l 2- l 2- - 2R OH + SO4  R 2SO4 + 2OH

l - l - R OH + HCO3  R HCO3 + OH

- 2- - Thus, hardness producing anions like Cl , SO4 and HCO3 are removed.

84 UNIT: III WATER & Its Treatment 3) Thus, H+ ions produced from first chamber combine with OH- ions produced from second chamber to form water.

+ - H + OH  H2O

Hence, the water produced from ion-exchange process is completely free from all cations and anions of salts.

85 UNIT: III WATER & Its Treatment 86 UNIT: III WATER & Its Treatment REGENERATION OF RESINS:

Regeneration of Resin the resin bed gets exhausted, when used for a long period and can be regenerated:

(a) The exhausted cation exchange resin can be regenerated by passing dil. HCl (H+)

2+ + + 2+ R2Mg + 2H  2RH + Mg 2+ + + 2+ R2Ca + 2H  2RH + Ca (b) The exhausted an ion exchange resin can be regenerated by passing dil. NaOH (OH-)

RlCl + OH-  RlOH- + Cl- 2- - l - 2- R2SO4 + 2OH  2R OH + SO4 - - l - - R2HCO3 + OH  R OH + HCO3

87 UNIT: III WATER & Its Treatment ADVANTAGES: 1) The softened water by this method is completely free from all salts and fit for use in boilers

2) It produces very low hardness nearly 2 ppm

3) Highly acidic or alkaline water can be treated by this process

DISADVANTAGES:

1) The equipment is costly

2) More expensive chemicals are required for regeneration

3) Turbid water cannot be treated by this method

88 UNIT: III WATER & Its Treatment INTERNAL TREATEMENT (OR) CONDITIONING OF WATER:

The softening of water carried out inside the boiler is called Conditioning/Internal treatment of water

In this process, the hardness causing dissolved salts were prohibited

1) By complexing the hardness causing soluble salts by adding appropriate reagents

2) By precipitating the scale forming impurities in the form of sludges which can be removed by blowdown operation

3) By converting the scale forming salts into compounds which stay in dissolved form and donot cause any trouble to the boilers

4) All internal treatment methods must be followed by blowdown operation so that accumulated sludges are removed

89 UNIT: III WATER & Its Treatment IMPORTANT INTERNAL CONDITIONING METHODS ARE:

1) Colloidal Conditioning: The scale formation in low pressure boilers is prevented by the addition of Kerosene, tarnnin, agaragar etc. which get coated over the scale forming precipitates. These form loose, non-sticky deposites that can be removed by blowdown.

2) Phosphate Conditioning: The scale formation due to permanent hardness causing salts is avoided by complexation with sodium phosphate in high pressure boilers. The complex formed is soft, non-adherent and easily removable

3CaCl2 + 2Na3PO4  Ca3(PO4)2 + 6NaCl 3MgCl2 + 2Na3PO4  Mg3(PO4)2 + 6NaCl

The Calcium Phosphate and Magnesium Phosphate complexes were removed by blowdown operation.

90 UNIT: III WATER & Its Treatment The 3- Phosphate employed for conditions are:

(1) NaH2PO4 – Sodium dihydrogen phosphate (Acidic) (2) Na2HPO4 – Disodium hydrogen phosphate (weakly alkaline) (3) Na3PO4 – Trisodium phosphate (Alkaline)

Na3PO4 is the most preffered reagent because it not only forms complex with Ca+2 & Mg+2 ions, but also maintains the pH of water between 9-10, where the calcium and magnesium ion undergo complexation.

3) Carbonate Conditioning: The hard and strong adherant scales formed

due to CaSO4 are avoided by the addition of Sodium Carbonate to boiler water and this is called Carbonate Conditioning.

The CaSO4 is converted to CaCO3 which is loose sludge and it can be removed by blowdown.

Na2CO3 + CaSO4  CaCO3 ↓ + Na2SO4

91 UNIT: III WATER & Its Treatment 4) Calgon Conditioning: Sodium hexamet phosphate (NaPO3)6 is called as Calgon. This forms soluble complex compounds with CaSO4 which causes no boiler troubles. The treatment of boiler water with calgon is called calgon conditioning.

5) Treatment With Sodium Aluminate (NaAlO2): Sodium aluminate is hydrolised at the high temperature of boiler to Al(OH)3 and NaOH. The NaOH so formed precipitates some of magnesium salts as Mg(OH)2 ↓

NaAlO2 + H2O  Al(OH)3 + 2NaOH MgCl2 + 2NaOH  Mg(OH)2 + 2NaCl

The flocculent precipitates of Mg(OH)2 + Al(OH)3 produced inside the boiler entraps the finely suspended and colloidal impurities including oil drops and silica which can be removed by blow-down operation. This type of conditioning is called Sodium Aluminate Conditioning.

92 UNIT: III WATER & Its Treatment 6) Electrical Conditioning:Sealed glass bulbs containing mercury connected to a battery are set rotating in the boiler. When water boils mercury bulbs emit electrical dischanges which prevent scale forming particles to adhere/stick together to form scale.

7) Radio-Active Conditioning: In this type of conditioning radio-active salts in the forming tables are placed inside the boiler water at a few points. The radiation emitted by the tablets of radio-active material prevents scale formation.

93 UNIT: III WATER & Its Treatment DESALINATION OF BRACKISH WATER:

Water containing high concentrations of dissolved solids with a peculiar salty or brackish taste is called brackish water

Sea water is an example of brackish water containing about 3.5% of dissolved salts. This water cannot be used for domestic and industrial applications unless the dissolved salts are removed by desalination.

Commonly used methods are:

1) Electrodialysis

2) Reverse Osmosis

94 UNIT: III WATER & Its Treatment 1) Electrodialysis: Electrodialysis is based on the principle that the ions present in saline water migrate towards their respective electrodes through ion selective membranes. Under the influence of applied e.m.f.

2) The unit consists of a chamber two electrodes, the cathode and anode

3) The chamber is divided to 3-compartments with the help of thin, rigid, ion-selective membranes which are permeable to either cation or anion.

4) The anode is placed near anion selective membrane while the cathode placed near cation selective membrane

5) The anion selective membrane is containing positively charged + functional groups such as R4N and is permeable to anions only

6) The cation selective membrane consists of negatively charged functional - groups such as RSO3 and is permeable to cations only.

95 UNIT: III WATER & Its Treatment 96 UNIT: III WATER & Its Treatment . Under the influence of applied e.m.f. across the electrodes the cations move towards cathode through the membrane and the anions move towards anode through the membrane.

. The net result is depletion of ions in the central compartment, while it increases in the cathodic and anodic compartments.

. Desalinated water is periodically drawn from the central compartment while concentrated brackish water is replaced with fresh sample.

ADVANTAGES OF ELECTRODIALYSIS:

. The unit is compact.

. The process is economical as for as capital cost and operational expenses are concerned.

97 UNIT: III WATER & Its Treatment REVERSE OSMOSIS:

. When two solutions of unequal concentration are separated by a semi-permeable membrane which does not permit the passage of dissolved solute particles, i.e., molecules and ions.

. Flow of solvent takes place from the dilute solution to concentrated solution this is called as “OSMOSIS”.

. If a hydrostatic pressure in excess of osmotic pressure is applied on the concentrated side the solvent is forced to move from higher concentration to lower concentrated side across. Thus, solvent flow is reversed hence this method is called “Reverse Osmosis”

. Thus, in reverse osmosis pure water is separated from the contaminated water. This membrane filtration is also called “Super Filtration” or “Hyper-Filtration”.

98 UNIT: III WATER & Its Treatment METHOD OF PURIFICATION:

The reverse osmosis cell consists of a chamber fitted with a semi-permeable membrane, above which sea water/impure water is taken and a pressure of 15 to 40 kg/cm2 is applied on the sea water/impure water. The pure water is forced through the semi permeable membrane which is made of very thin films of cellulose acelate. However superior membrane made of Polymethacrylate and Polyamide polymers have come to use

99 UNIT: III WATER & Its Treatment ADVANTAGES:

. Both ionic and non-ionic colloidal and high molecule weight organic matter is removed from the water sample

. Cost of purification of water is less and maintenance cost is less

. This water can be used for high pressure boilers.

100 UNIT: III WATER & Its Treatment WATER FOR DOMESTIC USE & TREATMENT OF WATER FOR MUNICIPAL SUPPLY:

The following are the specification of water drinking purpose:

. This water should be clear, colourless and odourless.

. The water must be free from pathogenic bacteria and dissolved gases

like H2S.

. The optimum hardness of water must be 125 ppm and pH must be 7.0 to 8.5

. The turbidity in drinking water should not exceed 25ppm

. The recommended maximum concentration of total dissolved solids in potable water must not exceed 500ppm

101 UNIT: III WATER & Its Treatment TREATMENT OF WATER FOR MUNICIPAL SUPPLY:

. The treatment of water for drinking purposes mainly includes the removal of suspended impurities, colloidal impurities and harmful pathogenic bacteria.

. Various stages involved in purification of water for Municipal Supply:

Source Of Water Screening Aeration

Sedimentation

Storage Sterilisation and (or) Filtration Distribution Disinfectation

102 UNIT: III WATER & Its Treatment 1) SCREENING: The process of removing floating matter from water is known as “Screening”

In this process, water is passed through a screen. The floating matter is arrested by the screen and the water is free from the folating matter.

2) AERATION: The water is then subjected to aeration which

i) Helps in exchange of gases between water and air.

ii) Increases the oxygen content of water

iii) Removes the impurities like ‘Fe’ and ‘Mn’ by precipitating as their hydroxides

103 UNIT: III WATER & Its Treatment 3) SEDIMENTATION:

i) Plain Sedimentation: The process of removing big sized suspended solid particles from water is called ‘Plain Sedimentation’. In this process, water is stored in big tanks for several hours.

70% of solid particles settle down due to the force of gravity

ii) Sedimentation By Coagulation:

. This is the process of removing fine suspended and colloidal impurities by adding coagulants like alum, ferrous sulpate and sodium aluminate. When coagulant is added to water, “Floc Formation” takes place due to hydroxide formation which can gather tiny particles together to form bigger particles and settle down quickly

104 UNIT: III WATER & Its Treatment 4) FILTRATION: This process of passing a liquid containing suspended impurities through a suitable porous materials so as to effectively remove suspended impurities and some micro-organisms is called “Filtration”.

It is mechanical process. When water flows through a filter bed, many suspended particles are unable to pass through the gaps and settle in the bed.

105 UNIT: III WATER & Its Treatment 5) DISINFECTION OR STERILISATION: The process of killing pathogenic bacteria and other micro-organisms is called ‘Disinfection or Sterilisation’. The water which is free from pathogenic bacteria and safe for drinking is called Potable water. The chemicals used for killing bacteria are called ‘DISINFECTANTS’.

a) By adding Bleaching Powder: Water is mixed with required amount of bleaching powder, and the mixture is allowed o stand for several hours

CaOCl2 + H2O  Ca(OH)2 + Cl2

Cl2 + H2O  HOCl + HCl (Hypo Chlorous Acid)

Germs + HOCl  Germs are killed

The disinfection action of bleaching powder is due to available chlorine in it. It forms hypochlorous acid which act as a powerful germicide (disinfectant)

106 UNIT: III WATER & Its Treatment b) CHLORINATION: Chlorine is mixed with water in a chlorinator, which is a high tower having a number of baffle plates. Water and required quantity of concentrated chlorine solutin are introduced from its top during their passage through the tower. They get thoroughly mixed and then sterilised water is taken out from the bottom.

ADVANTAGES:

i) Storage required less space ii) Effective and economical iii) Stable and does not deteriorate iv) Produces no salts v) Ideal disinfectant

107 UNIT: III WATER & Its Treatment DISADVANTAGES:

i) Excess of chlorine causes unpleasant taste and odour. ii) More effective at below pH 6.5 and less effective at higher pH values.

c) OZONATION: Ozone (O3) is an excellent, disinfectant which can be prepared by passing silent electric discharge through pure and dry oxygen. Ozone is highly unstable and breaks down, liberating nascent oxygen.

3O2  2O3

O3  O2 + (O) Nascent Oxygen

. This nascent oxygen kills bacteria as well as oxidises the organic matter present in water

108 UNIT: III WATER & Its Treatment ADVANTAGES:

. Removes colour, odour and taste

DISADVANTAGES:

. The method is costly.

109 UNIT: III WATER & Its Treatment BCH101 ENGINEERING CHEMISTRY- I

UNIT – I WATER TECHNOLOGY

INTRODUCTION

Water is the nature’s most wonderful, abundant and useful compound. Of the many essential elements for the existence of human beings, animals and plants, water is rated to be of greatest importance. Without food human being can survive for a number of days, but water is such an essential thing without it one cannot survive. Water is not only essential for the lives of animals and plants but also occupies unique position in industries. Probably its most important use as an engineering material is in the ‘steam generation’. Water is also used as a coolant, in power, and chemical plants. In addition to it, water can also be used in the production of steel, rayon, paper, textiles, chemicals, irrigation, drinking fire fighting, etc.

OCCURRENCE:

Water is the only substance that occurs at ordinary temperatures in all three states of matter: Solid, Liquid and Gas. As a so33.3.33lid, ice, it forms glaciers, frozen lakes and rivers, snow, hail and frost. It is liquid as rain and dew, and it covers three- quarters of the earth’s surface in swamps, lakes, rivers and oceans. Water also occurs in the soil and beneath the earth’s surface asa vast groundwater reservoir. As gas, or water vapour, it occurs as fog, steam and clouds.

WATER PURIFICATION:

Impurities are removed from water by seeming, sedimentation, filtration, chlorination or irradiation. Aeration removes odours and tastes caused by decomposing organic matter, industrial wastes and some gases. Various salts and metals cause hardness in water. Hardness may be removed by boiling, by adding sodium carbonate and lime or by filtering through natural or artificial zeolites. Water is also purified by processes such as desalination, reverse osmosis, electrolysis etc.,

CHARACTERISTICS OF WATER

As per the suggestion given by World Health Organisation (WHO) and by Indian Council of Medical Research (ICMR), the following are the important characteristics of potable water.

1. It should be clear, colourless and odourless. 2. It should be cool and pleasant to taste. 3. It should be free from harmful bacteria and suspended impurities. 4. It should be free from dissolved gases like CO2, H2S, NH3, etc., and poisonous minerals like lead, arsenic, manganese, etc., 5. Hardness should be less than 500 ppm. 6. Chloride ion content should be less than 250 ppm. 7. Fluoride ion content should be less than 1.5 ppm. 8. Total Dissolved Solids (TDS) content should be less than 500 ppm. 9. pH of the potable water should be 6.5 – 8.5.

Acidic Neutral Alkaline pH=0-7 7 pH=7-14

CHEMICAL CHARACTERISTICS OF WATER

The most important chemical characteristics of water are its acidity, alkalinity, hardness and corrosiveness. Chemical impurities can be either natural, man made (Industrial) or be deployed in raw water sources by enemy forces. Some chemical impurities cause water to behave as either an acid or a base. Since either condition has an important bearing on the water treatment process, the pH value must be determined. Generally the pH influences the corrosiveness of the water, chemical dosages necessary for proper disinfection and the ability to detect contaminants.

HARDNESS Hardness is caused by the soluble salts of calcium, magnesium, iron, manganese, sodium, sulphates, chlorides and nitrates. The degree of hardness depends on the type and amount of impurities present in the water. Hardness also depends on the amount of carbon-di-oxide in solution. Carbon-di-oxide influences the solubility of the impurities that cause hardness. The hardness caused by carbonates and bicarbonates is called carbonate hardness. The hardness caused by all others (chlorides, sulphates, nitrates) is called non-carbonated hardness.

HARD WATER Water which does not produce lather with soap solution, but produces whiteprecipitate (scum) is called hard water.In other words, water that contains mineral salts (an calcium and magnesium ions)that limit the formation of lather with soap. This is due to the presence of dissolved Ca and Mg salts.

SOFT WATER Water, which produces lather, readily with soap solution is called soft water. This is due to the absence of Ca and Mg salts.Water that is not hard (ie., does not contain mineral salts that interfere with the formation of lather with soap).

HARDNESS OF WATER How to detect hardness? Hardness of water can be detected in two ways. · When the water is treated with soap solution, if it prevents lathering and forms white scum, the water contains hardness. · Water containing hardness, gives wine red colour with Eriochrome Black –T indicator. The total water hardness (including both Ca2+ and Mg2+ ions) is read as parts per million (ppm) or weight / volume (mg/L) of Calcium Carbonate (CaCO3) in the water. Although water hardness usually measures only the total concentrations of calcium and magnesium (the two most prevalent, divalent metal ions), iron, aluminum and manganese may also be present at elevated levels in some geographical locations. The predominant source of magnesium is dolomite (CaMg (CO3)2).

TYPES OF HARDNESS Depending upon the types of dissolved salts present in water, hardness of water can be classified into two types: · Temporary Hardness · Permanent Hardness

Temporary Hardness (or) Carbonate Hardness (CH) (or) Alkaline Hardness Temporary hardness is caused by a combination of calcium and magnesium bicarbonate ions in the water. It can be removed by · boiling water

· by the addition of lime (Ca(OH)2) Boiling promotes the formation of carbonate from the bicarbonate and precipitates calcium carbonate out of solution, leaving water that is softer upon coving.

Ca (HCO3)2→ CaCO3 ↓+ H2O + CO2

Mg (HCO3)2 + 2Ca(OH)2→ Mg (OH)2↓+ 2CaCO3↓ +2H2O

Permanent Hardness (or) Non – Carbonate Hardness (NCH) (or) Non – alkaline Hardness Permanent hardness is hardness (mineral content) that cannot be removed by boiling. It is usually caused by the presence of calcium and magnesium sulphates and /or chlorides which become more soluble as the temperature rises. Despite the name, permanent hardness can be removed using water – softener or ion-exchange column, where the calcium and magnesium ions are exchanged with the sodium ions in the column. It can be removed by · Lime – Soda process · Zeolite process

CaCl2 + Na2 CO3 → CaCO3↓ +2Nacl (Soda) CaSO4 + Na2Ze →CaZe + Na2SO4

Zeolite =(Na2 Al2 Si2 O8. X H2O) Hard water causes scaling, which is the left- over mineral deposits that are formed after the hard water had evaporated. This is also known as lime scale.

Total Hardness The sum of temporary hardness and permanent hardness.

Table 1 :1 Molecular weights of some hardness producing salts.

Hardness Molecular Molecular producing salt weight Hardness producing salt weight

Ca(HCO3)2 162 MgSO4 120

Mg(HCO3)2 146 MgCO3 84

Mg(NO3)2 148 MgCl2 95

Ca(NO3)2 164 CaCl2 111

2+ CaCO3 100 Ca 40

2+ CaSO4 136 Mg 24

Expression of hardness in terms of equivalents of CaCO3 The concentration of hardness producing salts are usually expressed in terms of an equivalent amount of CaCO3. CaCO3 is chosen as a standard because, i) Its molecular weight (100) and equivalent weight (50) is a whole number, so the Calculations in water analysis can be simplified.

ii) It is the most insoluble salt, that can be precipitated in water treatment. If the concentration of hardness producing salt is x mgs/lit. then

Example

If the concentration ( or) weight of CaSO4 is 43mgs/lit, then weight equivalent to

UNITS OF HARDNESS 1. Parts per million (ppm)

It is defined as the number of parts of CaCO3 equivalent hardness per 106 parts of water. 2. Milligrams per litre (mg/lit)

It is defined as the number of milligrams of CaCO3 equivalent hardness per 1 litre of water. 3. Clarke’s degree (oCl)

It is defined as the number of parts of CaCO3 equivalent hardness per 70,000 parts of water. 4. French degree (oFr)

It is defined as the number of parts of CaCO3 equivalent hardness per 105 parts of water.

Relationship between various units 1ppm = 1 mg/lit = 0.10 Fr = 0.070 Cl

ESTIMATION OF TOTAL HARDNESS OF WATER BY EDTA METHOD The hardness of water is estimated by EDTA method using Eriochrome Black –T [EBT]. Principle: The calcium ion in the water is capable of forming complex with Indicator EBT and also with the EDTA in the pH range 8- 10.To keep the solution at this pH range , a buffer [ mixture of ammonium chloride and ammonium hydroxide ] is used . The complex between EDTA and indicator is more stable that of between the metal ion and indicator

Experiment : 1.Preparation of Standard hard water : Hard water is prepared in such a way that 100 ml of containing 100 mg of Calcium carbonate; So , 1 ml of Std. hard water = 1mg 2.Standadisation of EDTA : EDTA is taken in the burette; 20 ml of Std. hard water is pipette out in beaker; 5ml of buffer solution and 2 a few drops of indicator are added ; now the solution becomes wine –red colour because the colour of the complex between calcium and indicator [ M- In ] is wine –red. Then it is titrated against EDTA; at the end point the colour changes to Pale blue, which is colour of the free indicator. Since the complex between metal ion -EDTA is more stable that of between the metal ion and indicator , the Metal moves from [M-In] towards EDTA and forms complex with that , which is colourless; now the indicator is freed and the solutions attains steel blue colour , which is the colour of the free indicator. Let the titre value be V1.

PREPARATION OF SOLUTIONS;

1. Standard hard water: 1 gm of dry CaCO3 is dissolved in minimum quantity of HCl and evaporate the solution to dryness on a water bath, and then diluted to 1 lit with water. Each ml of this solution then contains 1 mg of CaCO3 hardness.

2. EDTA solution: 4 gm of EDTA crystals + 0.1 gm MgCl2 in 1lit

3. Indicator: 0.5 gm of EBT in100 ml of alcohol.

4. Buffer solution: 67.5 gm NH4Cl + 570 ml of Con. Ammonia solution diluted with distilled water to 1 lit.

5. Titration of permanent hardness of water: Take 250 ml of the water sample in a large beaker .Boil till the volume is reduced to 50 ml. Filter, wash the precipitate with distilled water collecting filtrate and . Finally make the volume to 250 ml with distilled water. Then titrate 50 ml of the boiled water sample just as in step (5).

Let volume used by V3 ml Calculations: 50 ml of standard hard water = V1 ml of EDTA :.

50 x1 mg of CaCO3 = V1 ml of EDTA :.

1 ml of EDTA = 50/V1 mg of CaCO3 eq.

Now 50 ml. of given hard water = V2 ml EDTA = V2 x 50/V1 mg of CaCO3

eq. :. 1 L (1,000 mL) of given hard water = 1000 V2/V1 mg of CaCO3 eq. :.

Total hardness of water = 1000 V2/V1 mg/L = 1000 V2/V1 ppm Now 50 ml of boiled water = V3 ml of EDTA . . . V3 x 50 /V1 mg of CaCO3 eq

. . . V3 x 50/ V1 mg of CaCO3 eq 1000 ml (= 1 L) of boiled water = 1000 V3 /V1 mg of CaCO3 eq 1000 V3 Permenent hardness = /V1 ppm = . . .

And Temporary hardness = Total hardness – Permanent hardness= ppm =

3.1 Advantages of EDTA method:

This method is definitely preferable to the other methods, because of the

(i) Larger accuracy; (ii) Convenience; (iii) Rapid procedure

ALKALINITY

Alkalinity is classified as Depending up on the anions that are responsible for the alkalinity of water, there are three types of alkalinity: 1. Hydroxide alkalinity – due to hydroxide ions 2. Carbonate alkalinity - due to carbonate ions 3. Bicarbonate alkalinity - due to bicarbonate ions The alkalinity due hydroxide and carbonate can be detected by Phenolphthalein indicator and so they are collectively called as Phenolphthalein Alkalinity , represented by P. The alkalinity due hydroxide, carbonate and bicarbonate can be detected by Methyl orange indicator and so it is called as in Methyl orange Alkalinity, represented by M. 1.Determination of Phenolphthalein Alkalinity ,P : 100 ml of given water sample is taken in the conical flask , a few drops of Phenolphthalein indicated are added and titrated against N/50 H2SO4 ; let the titre value when the solution becomes colourless, be V1. 2.Determination of Methylorange Alkalinity ,M : The in the same solution a few drops of Methylorange indicator are added and titrated against the same acid until the colour changes from yellow to red ; let the titre value be X .

So, M = V1 + X = V2

(a)Calculation of P:

Volume of the acid = V1 cc

Normality of the acid , N1 = 1/50

Volume of water , V2 = 100 cc

Normality of water N2 = V1 X 1/50 X 1/ 100

P in terms of CaCO3 = N2 x equivalent of CaCO3 X 1000 mg of CaCO3

= V1 X 1/50 X 1/ 100 X 50 X 1000 mg of CaCO3

= 10 V1

P = 10 V1 ppm

(a)Calculation of M:

Volume of the acid = V2 cc

Normality of the acid, N2 = 1/50

Volume of water, V1 = 100 cc

Normality of water N1 = V2 X 1/50 X 1/ 100

M in terms of CaCO3 = N1 x equivalent of CaCO3 X 1000 mg of CaCO3

= V2 X 1/50 X 1/ 100 X 50 X 1000 mg of CaCO3

= 10 V2

M = 10 V2 ppm

ALKALINITY HYDROXIDE CARBONATE BICARBONATE P= 0 0 0 M P = 1/2 M 0 2P 0 P = M M or P 0 0 P < 1/2 M 0 2P M - 2P P > 1/2 M 2P - M 2(M - P ) 0

- - 2- 1.When OH only present ; [ H CO3 = 0 & CO3 = 0 ] P = M

OH P M

1/2CO3

H CO3

H CO3

2- - - 2. when CO3 only present ; [ H CO3 = 0 & OH = 0 ] ; 2P = M or P = M/2

OH

1/2CO3 P M - HCO3 P

- HCO3

- 2- - 3. when HCO3 only present [ CO3 = 0 & OH = 0 ]

- HCO3 =M ; P = 0

OH

1/2CO3

- HCO3

- HCO3 M

4. When OH & CO3 are present ; [ H CO3 = 0 ]

X OH P – (M-P) P 1/2CO3 M - P M

HCO3 M - P

HCO3

X +M – P = P ; X + M = 2P ; or 2P > M ; or P > M/2 X = OH = P – (M - P) = (2P – M)

CO3 = (M – P )+ (M – P) = (2M – 2P) = 2( M – P )

5. When CO3& HCO3 are present ; [ OH = 0 ]

CO3 = 2P ; HCO3 = ( M – 2P )

OH

1/2 CO3 P

HCO3 P M

HCO3 M - 2P

SCALE AND SLUDGE FORMATION IN BOILERS

In boilers, water evaporates continuously and the concentration of the dissolved salts increases progressively. When their concentrations reaches saturation point, they are thrown out of water in the form of precipitates which stick to the inner walls of the boiler. If the precipitation takes place in theform of loose or slimy precipitate it is called sludge. On the other hand, if the precipitated matter forms a hard adhering crust/ coating on the inner walls of the boiler, it is a scale. Sludge is a soft, loose and slimy precipitate formed within the boiler. Sludge can be easily scrapped off with a wire brush. It is formed at comparatively colder portions of the boiler and collects in areas of the system, where the flow rate is slow at bends. Sludges are formed by substances which have greater solubilities in hot water than in cold water. Examples are MgCO3, MgCl2, CaCl2, MgSO4 etc.

Differences between scale and Sludge.

S.No. Scale Sludge 1. Scale is hard and adherent . Sludge is loose , slimy and non – adherent. 2. formed by the salts like Calcium formed by the salts like bicarbonate , Calcium sulphate , etc. magnesium Sulphate , magnesium carbonate , etc,., 3. formation can be prevented by formation can be prevented by i. dissolving scale using dilute acids like periodically removing the

HCl , H2SO4 . concentrated water by fresh water ii. taking soft water

4.1 Disadvantages of sludge formation: 1. Sludges are poor conductors of heat, so they tend to waste a portion of heat used. 2. If sludges are formed along with scales, the former get entrapped in the later and both get deposited as scales. 3. Excessive sludge formation disturbs the working of the boiler. It settles in the regions of poor water circulation such as pipe connection, plug opening, gauge glass connection thereby causing even chocking of the pipes. 4.2 Prevention of sludge formation: 1. By using well softened water. 2. By a frequent blow down operation, i.e., drawing off a portion of the concentrated water Scales are hard deposits, which stick very firmly to the inner surface of the boiler. Scales are very difficult to remove even with the help of hammer and chisel. Scales are the main source of boiler troubles. Formation of scales may be due to 1. Decomposition of calcium bicarbonate Ca(HCO3)2 → CaCO3 + H2O + CO2 However, scale composed chiefly of calcium carbonate is soft and is the main cause of scale formation in low pressure boilers. But in high pressure boilers CaCO3 is soluble. CaCO3 + H2O→Ca(OH)2 + CO2 2. Decomposition of calcium sulphate: The solubility of calcium sulphate in water decreases with increase in temperature. Thus, solubility of calcium sulphate is 3,200 ppm at 15 oC and it reduces to 55 ppm at 230 oC and 27 ppm at 320 oC. In other words, calcium sulphate is soluble in cold water, but almost completely insoluble in superheated water. Consequently calcium sulphate gets precipitated as hard scale on the heated portions of the boiler. This is the main cause in the high pressure boilers. 3. Hydrolysis of magnesium salts: Dissolved magnesium salts undergo hydrolysis at prevailing high temperatures in the boiler forming magnesium hydroxide precipitate, which forms a soft type of scale. MgCl2 + 2H2O →Mg(OH)2 + 2HCl 4. Presence of silica: (SiO2), even if present in small quantities, deposits as calcium silicate (CaSiO3) and/ or magnesium silicate (MgSiO3). These deposits stick very firmly to the inner walls of the boiler surface and are very difficult for removal. One important source of silica in water is the sand filter used. 4.3 Disadvantages of scale formation: 1. Wastage of fuels: Scales have a low thermal conductivity, so the rate of transfer of heat from boiler to inside water is largely decreased. In order to provide a steady supply of heat to water, excessive or over heating is done which causes unnecessary increase in fuel consumption. Thickness of the scale (mm) 0.325 0.625 1.25 2.5 12 Wastage of fuel 10% 15% 50% 80% 150% 2. Lowering of boiler safety: Due to scale formation, over-heating of the boiler has to be done in order to maintain a constant supply of steam. The over-heating of the boiler tube makes the boiler material softer and weaker and this causes distortion of the boiler tube and makes the boiler tube unsafe to bear the pressure of the steam especially in high-pressure boilers. 3. Decrease in efficiency: Scales may sometimes get deposited in the valves and condensers of the boiler and choke them partially or totally. This results in decrease the efficiency of the boiler. 4. Danger of explosion: When thick scales crack due to uneven expansion, the water comes in contact with the overheated iron plates. This causes a release of a large amount of steam suddenly, developing a high pressure, which may cause explosion in the boiler.

4.4 Removal of scales: 1. With the help of scraper or piece of wood or wire brush, if they are loosely adhering. 2. By giving thermal shocks like heating the boiler and suddenly cooling it with cold water. 3. Dissolving scales by adding suitable chemicals, if they are adherent and hard. Thus calcium carbonate scales can be dissolved by the addition of 5% HCl. Calcium sulphate scales can be dissolved by the addition of EDTA ( ethylene diamine tetra acetic acid), with which they form complexes. 4. By frequent blow down operation, if the scales are loosely adhering.

5. CAUSTIC EMBRITTELMENT Caustic embrittlement is a type of boiler corrosion, caused by using highly alkaline water in the boiler. During softening process by lime- soda process, free sodium carbonate is usually present in small proportion in the softened water. In high pressure boilers, sodium carbonate decomposes to give sodium hydroxide and carbon dioxide, and their presence makes the boiler water caustic. Na2CO3 + H2O → NaOH + CO2 The water containing sodium hydroxide flows into the minute hair cracks always present, by capillary action in to the inner sides of the boiler. Here as water evaporates the dissolved caustic soda concentration increases progressively. This concentrated caustic soda attacks the surrounding area dissolving inner iron side of the boiler by forming sodium ferroate. This causes the embrittlement of the boiler parts, particularly stressed parts such as bends, joints, rivets etc., causing even failure of the boiler operations. Caustic cracking can be explained by the following concentration cell Iron at Bends, rivets and joints The iron surrounded by the dilute NaOH becomes the cathodic surface and the iron present with the high concentration of NaOH becomes anodic which is consequently dissolved or corroded.

5.1 Caustic embrittlement can be avoided by 1. By using sodium phosphate as a softening agent instead of sodium carbonate. 2. By adding tannin or lignin to the boiler water, since these substances block the hair cracks, thereby preventing the infiltration of the caustic soda solution in to these. 3. By adding sodium sulphate to boiler water: Sodium sulphate blocks the hair cracks preventing the infiltration of caustic soda solution in to these. It has been observed that caustic cracking can be prevented, if sodium sulphate is added to the boiler in the ratio of Na2SO4 : NaOH as 1:1; 2:1; 3:1 in boilers working respectively at pressures up to 10, 20 and above 20 atmospheres.

6. BOILER CORROSION

Boiler corrosion is the decay of boiler material (iron) either by chemical or electro chemical attack of its environment. Main reasons for the boiler corrosion are:

6.1 Dissolved oxygen:

Water usually contains 8 mg of dissolved oxygen per liter at room temperature. Dissolved oxygen in water in the presence of prevailing high temperature of the boiler, attacks the boiler material as

2Fe + 2 H2O + O2→ 2 Fe(OH)2

4 Fe(OH)2 + O2→2 [Fe2O3.2 H2O]

Removal of the dissolved oxygen: a. By adding calculated amount of sodium sulphite or hydrazine or sodium sulphide.

2Na2SO3 + O2 →2 Na2SO4

N2H4 + O2 → N2 + 2 H2O

Na2S + O2 → Na2SO4 b. Mechanical de-aeration:

In this process water is sprayed in to a tower fitted with perforated plates (Fig), heated from sides and connected to vacuum pump. High temperature, low pressure and large exposed surface area reduce the dissolved oxygen in water.

6.2 Dissolved carbon dioxide:

Carbon dioxide dissolved in water forming carbonic acid, has a slow corrosive effect on the boiler material. Carbon dioxide is also released inside the boiler, if water, containing bicarbonates is used for steam generation

CO2 + H2O → H2CO3

Mg(HCO3)2 → MgCO3 + CO2 + H2O

Removal of dissolved carbon dioxide: a. By adding calculated amount of ammonia

2NH4OH + CO2 → (NH4)2CO3 b. By mechanical de-aeration process along with oxygen (described above)

6.3 Acids from dissolved salts:

Water containing dissolved salts of magnesium liberates acids on hydrolysis.

MgCl2 + H2O → Mg(OH)2 + 2 HCl

The liberated acid reacts with the iron material of the boiler in chain like processes, producing HCl again and again.

Fe + 2HCl→FeCl2+ H2

FeCl2 + 2H2O → Fe(OH)2 + 2 HCl

Consequently, presence of even small amount of magnesium chloride will cause corrosion to a large extent and may cause damage to the boiler material. Removal of acids: a) Softening boiler water to remove magnesium chloride, if any. b) By frequent blow down operation of removal of concentrated water with fresh soft water. c) Addition of inhibitors as sodium silicate/sodium phosphate/sodium chromate, which protect the boiler material against acid attack.

7. PRIMING AND FOAMING

When a boiler is producing steam rapidly, some particles of the condensed liquid water are carried along with the steam. The process of wet steam formation is called priming. Priming is causedby

1. The presence of large amounts of dissolved solids

2. High steam velocities

3. Sudden boiling

4. Improper boiler design

5. Sudden increase in the steam production rate.

Foaming is the production of persistent foam or bubbles in boilers, which do not break easily. Foaming is due to the presence of substances like oils in water, which reduce the surface tension of water.Priming and foaming usually occur together. They have to be eliminated because a. Dissolved salts in boiler water are carried by the wet steam to super heater and turbine blade, where they get deposited as water evaporates. This deposit reduces the efficiency of the boiler. b. Dissolved salts may enter the other parts of the machinery, where steam is being used, thereby decreasing the life of the machinery c. Actual height of the water column cannot be judged properly making the maintenance of the boiler pressure difficult.

Priming can be avoided by fitting mechanical steam purifiers, avoiding the rapid change in steaming rate, maintaining low water levels in boilers, efficient softening and filtration of the boiler feed water. Foaming can be avoided by adding anti foaming chemicals like castor oil, or removing oil from boiler water by adding compounds like sodium aluminate.

8. SOFTENING METHODS

Water used for industrial purposes (such as for steam generation) should be sufficiently pure. it should, therefore, be freed from hardness- producing salts before it is being put to use. The process of removing hardness-producing salts from water is known as softening of water. In industry three methods are mainly employed for softening of water.

8.1 Lime soda process:

In this method, the soluble calcium and magnesium salts in water are chemically converted into insoluble compounds, by adding calculated amounts of lime [Ca(OH)2] and soda [Na2CO3].

Calcium carbonate [CaCO3] and magnesium hydroxide [Mg(OH)2] are precipitated and removed.

8.2 Ion exchange or de-ionization or de-mineralization process:

Ion-exchange resins are insoluble, cross-linked, long chain organic polymers with a microporous structure, and the “functional groups’ attached to the chains are responsible for the ion- exchanging properties. Resins containing acidic functional groups (-COOH, -SO3H etc.) are capable of exchanging their H+ ions with other cations, which come into their contact; whereas those containing basic functional groups (-NH2=NH as hydrochloric acid) are capable of exchanging their anions with other anions, which come into their contact.

The ion-exchange resins may be classified as: (i) Cation exchange resins (RH+ ) are mainly styrene-divinyl benzene copolymers, which on sulphonation or carboxylation become capable to exchange their hydrogen ions with the cations present in the raw water

(ii) Anion exchange resins (R1OH- ) are styrene-divinyl benzene amineformaldehyde copolymers, which contains amino or quaternary ammonium or quaternary phosphonium or tertiary sulphonium groups as an integral part of the resin matrixs. These, after treatment with dil. NaOH solution, become capable to exchange their OHanions with anions present in the raw water.

Process: The hard water is passed first through cation exchange column, which removes all the cations like Ca2+.Mg2+, etc., from it and an equivalent amount of H+ ions are released from this column to water. Thus: 2

2RH + + Ca2+ →R2Ca 2+ + 2H +

2RH + + Mg 2+ →R2Mg 2+ + 2H The water which is now free from cations, is passed through anion exchange column, which removes all the anions like SO4 2- , Cletc., present in the water and equivalent amount OHions are released from this column to water. Thus:

R'OH - + Cl-→R'Cl + OH –

- 2- 2- – 2R'OH + SO4 →R'2 SO4 + 2OH

- 2- 2- - 2R'OH + CO3 →R'2 CO3 + 2OH

H + and OHions (released from cation exchange and anion exchange columns respectively) combine to produce water.

+ - H + OH → H2O

Thus the water coming out from the exchanger is free from all cations as well as anions. Ion-free water is known as deionised or demineralised water.

Regeneration: When capacities of cation and anion exchangers to exchange H+ and OHions respectively are lost, they are then said to be exhausted

The exhausted cation exchange column is regenerated by passing a solution of dil. HCl or H2SO4. The regeneration can be represented as:

2+ + + 2+ R2Ca + 2H → 2RH + Ca (Washing)

The column is washed with deionized water and such washing (which containing Ca2+, Mg2+, 2- etc. and cation SO4 ) is passed into sink or drain. The exhausted anion exchange column is regenerated by passing a solution of dil. NaOH. The regeneration can be represented as:

2- - - 2- R' 2 SO4 + 2OH → 2R'OH + SO4 (Washing)

2- The column is washed with deionized water and such washing (which contains Na+ and SO4 or Clions) is passed into sink or drain.

Advantages

(1) The process can be used to soften highly acidic or alkaline waters. (2) It produces water of very low hardness (2 ppm) .

Disadvantages

(1) The equipment is costly and expensive chemicals are needed.

(2) If water contains turbidity, then the output of the process is reduced.

(3) The turbidity must be below 10 ppm. If it is more, it has to be removed first by coagulation followed by filtration.

9. Potable water

The water which is fit for human consumption is known as potable water

Municipalities have to supply potable water, i.e., water which is safe to human consumption should satisfy the following essential requirements

1. It should be sparkling clear and odourless.

2. It should be pleasant in taste

3. It should be perfectly cool

4. Its turbidity should not exceed 10 ppm

5. It should be free from objectionable dissolved gases like hydrogen sulphide.

6. It should be free from objectionable minerals such as lead, arsenic, chromium and manganese salts.

7. Its alkalinity should not be high. Its pH should not be above 8.0

8. It should be reasonably soft

9. Its total dissolved solids should be less than 500 ppm 10. It should be free from disease- producing micro- organisms.

Purification of domestic water for domestic use:

For removing various types of impurities in the natural water from various sources, the following treatment process is employed;

9.1 Removal of suspended impurities :

The treatment water for municipal supply involves the following steps: 1.Screening :

It is a process of removing the floating materials like , leaves , wood pieces ,etc., from water. Here water is passed through a screen having a number of holes . 2.Aeration :

The process of mixing air with water is called aeration; here the gases like CO2,

H2S and other volatile impurities responsible for the bad taste and odour, are removed; further ferrous and manganeous salts are converted into insoluble ferric and manganic salts .

3.Sedimentation :

In this process suspended impurities are removed by keeping the water undisturbed for 2 – 6 hours in a tank. This removes only 75 % of the suspended impurities.

4.Coagulation :

In this method by adding coagulants like aluminium sulphate, the colloidal impurities like finely divided clay, silica, etc., are also removed. The aluminium sulphatehydrolysed to give gelatinous precipitate Al(OH)3 ; The suspended impurities adhere to the precipitate and settle down at the bottom.

Al2(SO4)3 + H2O Al(OH)3 + H2 SO4

5.Filtration : The bacterias, colour, taste ,odour and suspended impurities are removed by passing the water through the layers of fine sand , coarse sand and fine gravel and coarse gravel successively placed in a filter tank.

After a long time the rate of filtration is ceased as the holes in the filter are blocked by the impurities; so filtration is stopped and the top concentrated sand layer is scrapped off and replaced by fresh sand .

9.2 Removal of micro-organisms:

The process of destroying /killing the disease producing bacteria, micro-organisms, etc., from the water and making it safe for the use, is called disinfectation. a. Boiling: By boiling water for 10-15 minutes, all the disease producing bacteria is killed and the water becomes safe for use. b. Adding bleaching powder: In small water works, about 1 kg of bleaching powered per 1000 kiloliter of water is mixed and allowed to standing undisturbed for several hours. The chemical action produces hypochlorous acid (a powerful germicide)

CaOCl2+ H2O → Ca(OH)2+Cl2 Cl2+ H2O → HCl+ HOCl

Germs+ HOCl → Germs are killed c. Chlorination: Chlorination (either gas or in concentrated solution from) produces hypochlorous acid, which is a powerful germicide.

Cl2+ H2O → HCl+ HOCl

Bacteria+ HOCl → Bacteria are destroyed

9.3. Break point chlorination (or) or free residual chlorination :

It involves addition of sufficient amount of chlorine to oxidize: (a) organic matter (b)reducing substance and (c) free ammonia in raw water; leaving behind mainly free chlorine, which possesses disinfecting action against disease- producing bacteria The addition of chlorine at the dip or break is called “break point” chlorination. This indicates the point at which free residual chlorine begins to appear..

Advantages:

(1) It oxides completely organic compounds, ammonia and other reducing compounds.

(2) It removes color, odour and taste of water. (3) It removes completely all the disease causing bacteria/micro-organism (4) It prevents the growth of any weeds in water.

9.4. Using Chloramine (ClNH2):

When chlorine and ammonia are mixed in the ratio of 2:1 by volume, chloramine is formed. Cl2+NH3 → ClNH2+ HCl

Chloramine is a better bactericidal than chlorine.

9.5. Disinfection by Ozone:

Ozone gas is an excellent disinfectant, which is produced by passing silent electric discharge through cold and dry oxygen. 3O2 → 2O3

O3 → O2 + [O]

10. Desalination of brackish water

The process of removing common salt (NaCl) from the water is known as desalination. Water containing high concentration of dissolved salts with a peculiar salty taste is called brackish water. Sea water is an example containing 3.5% of dissolved salts. The common methods for the desalination of brackish water are;

10.1 Eletctrodialysis:

It is a method in which the ions are pulled out of the salt water by passing direct current, using electrodes and thin rigid plastic membrane pair.

An Eletctrodialysis cell consists of a large number of paired sets of rigid plastic membranes. Hard water is passed between the membrane pairs and an electric field is applied perpendicular to the direction of water flow. Positively charged membrane and negatively charged membrane repel positively charged ions and negatively charged ions respectively to pass through. So, in one compartment of the cell, the salt concentration decreases while in the adjacent compartment it increases

Thus, we get alternative stream of pure water and concentrated brine. Advantages: 1. It is most compact unit

10.2 Reverse osmosis:

When two solutions of unequal concentrations are separated by a semi permeable membrane, flow of solvent takes place from dilute to concentrate sides, due to osmosis. If, however a hydrostatic pressure in excess to osmotic pressure is applied on the concentrated side, the solvent flow is reversed, i.e, solvent is forced to move from concentrated side to dilute side across the membrane. This is the principle of reverse osmosis.(RO) Thus in reverse osmosis method, pure solvent is separated from its contaminants, rather than removing contaminants from the water. The membrane filtration is sometimes also called super- filtration or hyper filtration.

METHOD:

In this process, pressure is applied to the sea water or impure water to force the pure water content of it out the semi-permeable membrane, leaving behind the dissolve solids. The principle of reverse osmosis as applied for treating saline/sea water The membrane consists of very thin film of cellulose acetate, affixed to either side of a perforated tube. However, more recently superior membranes made of polymethacrylate and polyamide polymers have come into use.

ADVANTAGES

1. Reverse osmosis possesses distinct advantages of removing ionic as well as non-ionic, colloidal and high molecular weight organic matter.

2. It removes colloidal silica, which is not removed by demineralization.

3. The maintenance cost is almost entirely on the replacement of the semi permeable membrane. 4. The life time of membrane is quite high, about 2 years,

5. The membrane can be replaced within a few minutes, thereby providing nearly uninterrupted water supply.

6. Due to low capital cost, simplicity, low operating cost and high reliability, the reverse osmosis is gaining grounds at present for converting sea water into drinking water and for obtaining water for very high –pressure boilers.

BCH101 ENGINEERING CHEMISTRY- I

UNIT – II POLYMERS

A polymer (Greek poly-, "many" + -mer, "parts") is a large molecule, or macromolecule, composed of many repeated subunits. Because of their broad range of properties, both synthetic and natural polymers play an essential and ubiquitous role in everyday life. Polymers range from familiar synthetic plastics such as polystyrene to natural biopolymers such as DNA and proteins that are fundamental to biological structure and function. Polymers, both natural and synthetic, are created via polymerization of many small molecules, known as monomers. Their consequently large molecular mass relative to small molecule compounds produces unique physical properties, including toughness, viscoelasticity, and a tendency to form glasses and semi crystalline structures rather than crystals.

The term "polymer" derives from the ancient Greek word (polus, meaning "many, much") and (meros, meaning "parts"), and refers to a molecule whose structure is composed of multiple repeating units, from which originates a characteristic of high relative molecular mass and attendant properties. The units composing polymers derive, actually or conceptually, from molecules of low relative molecular mass. The term was coined in 1833 by Jöns Jacob Berzelius, though with a definition distinct from the modern IUPAC definition. The modern concept of polymers as covalently bonded macromolecular structures was proposed in 1920 by Hermann Staudinger, who spent the next decade finding experimental evidence for this hypothesis.

Polymers are of two types:

 Natural polymeric materials such as shellac, amber, wool, silk and natural rubber have been used for centuries. A variety of other natural polymers exist, such as cellulose, which is the main constituent of wood and paper.

 The list of synthetic polymers includes synthetic rubber, phenol formaldehyde resin (or Bakelite), neoprene, nylon, polyvinyl chloride (PVC or vinyl), polystyrene, polyethylene, polypropylene, polyacrylonitrile, PVB, silicone, and many more.

Most commonly, the continuously linked backbone of a polymer used for the preparation of plastics consists mainly of carbon atoms. A simple example is polyethylene ('polythene' in British English), whose repeating unit is based on ethylene monomer. However, other structures do exist; for example, elements such as silicon form familiar materials such as silicones, examples being Silly Putty and waterproof plumbing sealant. Oxygen is also commonly present in polymer backbones, such as those of polyethylene glycol, polysaccharides (in glycosidic bonds), and DNA (in phosphodiester bonds).

TYPES OF POLYMERS 1. Oligopolymers 2.Highpolymers 3. Homopolymers 4.Heteropolymers.

MONOMER ARRANGEMENT IN COPOLYMERS

Monomers within a copolymer may be organized along the backbone in a variety of ways.

 Alternating copolymers possess regularly alternating monomer residues [AB...]n

 Periodic copolymers have monomer residue types arranged in a repeating sequence:

[AnBm...] m being different from n.  Statistical copolymers have monomer residues arranged according to a known statistical rule. A statistical copolymer in which the probability of finding a particular type of monomer residue at a particular point in the chain is independent of the types of surrounding monomer residue may be referred to as a truly random copolymer

 Block copolymers have two or more homopolymer subunits linked by covalent bondsPolymers with two or three blocks of two distinct chemical species (e.g., A and B) are called diblock copolymers and triblock copolymers, respectively. Polymers with three blocks, each of a different chemical species (e.g., A, B, and C) are termed triblock terpolymers.

 Graft or grafted copolymers contain side chains that have a different composition or configuration than the main chain.

TACTICITY Tacticity describes the relative stereochemistry of chiral centers in neighboring structural units within a macromolecule. There are three types: isotactic (all substituents on the same side), atactic (random placement of substituents), and syndiotactic (alternating placement of substituents).

POLYMERIZATION.

The repeating unit of the polymer polypropylene

Polymerization is the process of combining many small molecules known as monomers into a covalently bonded chain or network. During the polymerization process, some chemical groups may be lost from each monomer. This is the case, for example, in the polymerization of PET

polyester. The monomers are terephthalic acid (HOOC-C6H4-COOH) and ethylene glycol (HO-

CH2- CH2-OH) but the repeating unit is -OC-C6H4-COO-CH2-CH2-O-, which corresponds to the combination of the two monomers with the loss of two water molecules. The distinct piece of each monomer that is incorporated into the polymer is known as a repeat unit or monomer residue. The process in which a large number of micro molecules, called monomers, are linked to form a giant molecule, called polymer is known as Polymerization.

1.AdditionPolymerisation : In this type polymer is produced from the monomers with multiple bonds, without loss of any material; the product polymer is the integral multiple of the monomer. The reaction is conducted by the application of heat, light , pressure or catalyst.

2.CondensationPolymerisation: In this type , polymer is produced from the monomers with loss of simple molecules such as water, ethanol etc., ex. Nylon – 6,6

nH2N – (CH2)6 – NH2 + n HOOC – (CH2)4 – COOH hexa methylene diaminesAdipic acid

Nylon – 6,6 + 2nH2O

3. CopolymerisationPolymerisation: Here two different type of monomers [ one is aliphatic & the other is aromatic ] combine to give polymers; Ex. Butadiene and styrene gives Buna –S – rubber OR Polybutadiene –co - styrene

DIFFERENCES BETWEEN ADDITION AND CONDENSATION POLYMERIZATION

S.No Addition Polymerization. Condensation Polymerization.

Monomers combine to form Monomers combine to form polymers with elimination.( few 1. polymers without elimination. exceptions )

Monomers must have at least two Monomers must have at least one 2. identical or different functional multiple bond ; ex. CH2 =CH2 groups; ex. CH2 OH– CH2OH H2N(CH2)4-COOH By product such as H S , HCl , etc., No other by product is formed. 2 3. are formed. M.wt. of the polymer need not be an M.wt. of the polymer is an integral 4. integral multiple of M.Wt. of the multiple of M.Wt. of the monomer. monomer. High m.wt. polymer is formed at 5. M.Wt. of polymer increases steadily. once Longer reaction time give higher Longer reaction times are essential 6. yield ; but no effect on M.Wt. to get high M.Wt. Polymer. 7. Homo- chain polymer is obtained. Hetro - chain polymer is obtained. Thermo plastics are produced. ex. Thermosetting plastics are produced. 8. PVC , polyethylene , Ex. Bakelite , Polyester,

FREE – RADICAL MECHANISM OF POLYMERIZATION. The free radical mechanism occurs in three major steps: 1.Initiation 2.Propagation 3.Termination I.Initiation : This involves two reactions : (i)First reaction involves production of free radical by hemolytic dissociation of an initiator ( or catalyst ): (ii ) Second reaction : Addition of the free radical with the first monomer to give chain initiating species :

II Propagation : Here growth of chain initiating species occurs by successive addition of large number of monomers.

The growing chain of the polymer is known as living polymer.

III Termination :

Termination of a growing chain of polymer may occur by

(i) Coupling (or ) combination

(ii) Disproportionation

Coupling or combination

Here coupling of free radical of one chain end to another free radical occurs to form a macro molecule.

Disproportionation : Here transfer of hydrogen from one radical to another radical occurs to form two macromolecules, one is saturated and the other is unsaturated.

The products of addition polymerisation is known as Dead Polymers.

PREPARATION, PROPERTIES AND USES OF THE POLYMERS a) PVC ; b) Teflon ; c ) PET; d ) nylon 6-6

a) PVC

Preparation: This involves two steps :

In the Step – I vinyl chloride is formed from acetylene

Step II : By heating water emulsion of vinyl chloride in the presence of H2O2 or benzoylperoxide under pressure

Properties of PVC : 1.colourless , odourless 2.chemically inert powder 3. insoluble in inorganic & alkalis but soluble in hot chlorinated hydrocarbons like ethyl chloride. 4. Undergoes degradation in the presence of light or heat. uses : 1. production of pipes, cable insulators ,table covers & rain coats 2. for making sheets which are employed for light fittings , 3. Refrigerator components, etc., b)Teflon :

Preparation : It is nothing but polytetrafluoroethylene (PTFE) which is an Engineering polymer. It is prepared by polymerization of water emulsion of tetrafluoroethylene in the presence of benzoylperoxide under pressure

Properties : 1. Extremely tough & flexible

2. has extremely good electrical & mechanical properties

3. Chemically resistant towards all chemicals ( except hot alkali metal & hot

(Fluorine )

4. excellent thermal stability.

Uses of Teflon ;

1.as a very good electrical insulating material in motors,cables , transformers, electrical fittings 2. For making gaskets, pump parts, tank linings , etc., 3. For making non-lubricating bearings, chemical pipes, etc., 4. For making non-sticking stop cocks for burettes. c) PET : Preparation:

PET is nothing but Polyethyleneteterephthlate; it is prepared by the condensation of ethylene glycol and terephthalic acid(benzene-1,4-dicarboxylic acid).

PET Poly (ethyleneterephthalate) Properties 1.goodfibre – forming material 2.Fibre possess high stretch – resistance & wrinkle – resistance . 3.Highly resistant to mineral & organic acids but less resistant to alkali. Uses : 1.for making synthetic fibres like terylene , Dacron , etc., 2.blended with wool to provide better wrinkle – resistance. 3.In safety helmets , aircrafts , battery boxes , etc.,

D) NYLON 6-6 preparation: it is an engineering plastic obtained by the condensation polymerization of 1 , 6 – hexamethylenediamine with adipic acid . 6,6 indicates the number of carbon atoms in each monomer.

Properties: 1. Nylons are transparent, whitest and high melting polymers. 2. Possess high temperature stability and good abrasion – resistance. 3. Insoluble in common organic solvents and soluble in phenol and formic acid. Uses: 1. For making filaments for ropes , bristles for tooth –brushes , etc., 2. Nylon 6 and nylon 11 are mainly used for moulding purposes for gears , bearings, etc., 3. Forfibres, which is used in making socks , dresses , carpets, etc.,

PLASTICS

Plastics are high molecular weight organic or inorganic materials, which can be moulded into any stable form when subjected to heat and pressure in the presence of a catalyst. The name plastics or plastic materials in general is given to organic materials of high molecular mass, which can be moulded into any desired form when subjected to heat and pressure in presence of catalysts. Polymer resin is the basic binding material, which forms the major part of a plastic. In recent years plastics have attained greater importance in every walk of life due to their unique properties. Now, plastics substitute all engineering materials like wood, metal, glass etc because of their special advantages over other conventional materials. Plastics are products of polymers

Plastics are classified into two types: 1. Thermoplastics They are the resins which soften on heating and set on cooling. Therefore, they can be remoulded any number of times and used. Example: Polythene, PVC, Nylon, etc. 2. Thermosetting plastics. They are the resins which set on heating and cannot be resoftened. Hence, their scrap cannot be reused. Examples: Phenol-formaldehyde resin (Bakelite), urea formaldehyde resin, etc. The differences between two types of plastics arise mainly due to the difference in their chemical structure.

Properties of Engineering plastics They possess 1. High abrasion resistance 2. High load bearing properties 3. Fairly good thermal stability 4. Light weight 5. Readilymoldable into complicated shapes 6. Rigidity 7. Dimensional stability 8. High performance properties which permit them to be used in the same manner as metals, alloys and ceramics DISADVANTAGES OF PLASTICS 1. Softness 2. Embrittlement at low temperature

3. Deformation under load 4. Low heat resistant 5. Combustibility 6. Tend to degrade upon exposure to heat and radiation 7. Non bio – degradable

Differences between Thermo Plastics and Thermosetting Plastics

S.No. Thermo Plastics Thermosetting plastics Formed by condensation 1 Formed by addition Polymerization Polymerization Consists of linear long chain 2. Consists of three dimensional networks. polymers. Polymer chains are held together by Polymer chains are held together by 3. weak Vander wall’s forces. strong covalent forces. 4. Weak , soft and less brittle Strong, hard and more brittle. Soften on heating and harden on Harden on heating ;but once hardened 5. cooling. they cannot be softened. 6. Can be remoulded Cannot be remoulded. 7. Low M.Wt. High M.Wt. 8. Soluble in organic solvents. Insoluble in organic solvents.

RUBBER Rubbers are non-crystalline, high polymers (linear polymers) having elastic and other rubber – like properties.Rubber is a natural elastic polymer of isoprene. It is obtained from the milk of rubber called 'Latex'. The structure of natural rubber is as follows. 1. Latex is rubber milk containing about 30 to 45% of rubber.

2. The rubber milk is diluted with water and allowed to stand for sometime.

3. The clear liquid from the top is treated with acetic acid or formic acid to precipitate rubber.

4. The precipitated rubber is collected and passed through rollers to get sheets of rubber.

5. Rubber sheets are finally dried by smoking. This rubber is called 'Smoked rubber'. 6. During the coagulation of rubber milk with acetic or formic acid, retardants like sodium bisulphite (NaHSO) are added to prevent oxidation of rubber. This is called 'Creep rubber'.

The natural rubber obtained from latex cannot be used in industries because it has some defects.

1. It becomes soft and sticky during summer.

2. It became hard and brittle during winter.

3. It swells up in oils.

4. It flows plastically due to prolonged stress.

5. Chemicals easily affect rubber.

NATURAL AND SYNTHETIC RUBBER. Natural rubber: Natural rubber is obtained from the tree as a latex , which is a dispersion of Isoprene. During the treatment of latex isoprene molecules undergo polymerization to form long coiled chain of polyisoprene.

Isoprene

Polyisoprene Properties : 1.Plastic in nature – soften at high temperature and too brittle at low temperature. 2.Has large absorbing capacity. 3.Tensile strength is very less . 4.non – resistant to non-polar solvents like benzene .

5.attacked by oxidizing agents like nitric acid and con.H2SO4 6.becomes brittle on long exposure to air. To improve the properties of rubber, it is mixed with some chemicals like sulphur , hydrogen sulphide etc. the properties of rubber improved using sulphur is called vulcanisation.

SYNTHETIC RUBBER : Synthetic rubber is nothing but any vulcanisable , manmade rubber like polymer ; it is superior to natural rubber in certain properties. Ex. Styrene Rubber (Buna – S rubber) or SBR; It is prepared by copolymerization of Butadiene and Styrene

DRAWBACKS OF NATURAL RUBBER 1.plastic in nature – soften at high temperature and too brittle at low temperature. 2.Has large absorbing capacity. 3.Tensile strength is very less . 4.non – resistant to non-polar solvents like benzene .

5.attacked by oxidizing agents like nitric acid and con. H2SO4 VULCANISATION OF RUBBER : Vulcanization is a process of heating the raw rubber with Sulphur at about 100 – 140 0 c. to improve its properties. The added Sulphur combines at the double bonds of different long chain rubber springs. Thus vulcanization prevents intermolecular movement of rubber springs. The extent of stiffness of vulcanized rubber depends on the amount of sulphur added.

Ex. 1.Tyre rubber contains 3 – 5 % sulphur 2.Battery case rubber contains 30 % sulphur

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BCH101 ENGINEERING CHEMISTRY- I

UNIT – III ELECTRO CHEMISTRY

Electrochemistry is the branch of physical chemistry that studies the relationship between electricity, as a measurable and quantitative phenomenon, and identifiable chemical change, with either electricity considered an outcome of a particular chemical change or vice versa. These reactions involve electric charges moving between electrodes and an electrolyte (or ionic species in a solution). Thus electrochemistry deals with the interaction between electrical energy and chemical change.

When a chemical reaction is caused by an externally supplied current, as in electrolysis, or if an electric current is produced by a spontaneous chemical reaction as in a battery, it is called an electrochemical reaction. Chemical reactions where electrons are transferred directly between molecules and/or atoms are called oxidation-reduction or (redox) reactions. In general, electrochemistry describes the overall reactions when individual redox reactions are separate but connected by an external electric circuit and an intervening electrolyte.

The term "redox" stands for reduction-oxidation. It refers to electrochemical processes involving electron transfer to or from a molecule or ion changing its oxidation state. This reaction can occur through the application of an external voltage or through the release of chemical energy. Oxidation and reduction describe the change of oxidation state that takes place in the atoms, ions or molecules involved in an electrochemical reaction. Formally, oxidation state is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. An atom or ion that gives up an electron to another atom or ion has its oxidation state increase, and the recipient of the negatively charged electron has its oxidation state decrease.

Electrochemical reactions in water are better understood by balancing redox + − reactions using the ion-electron method where H , OH ion, H2O and electrons (to compensate the oxidation changes) are added to cell's half-reactions for oxidation and reduction.

Acidic medium

In acid medium H+ ions and water are added to half-reactions to balance the overall reaction. For example, when manganese reacts with sodium bismuthate.

Basic medium

In basic medium OH− ions and water are added to half reactions to balance the overall reaction. For example, on reaction between potassium permanganate and sodium sulfite

Neutral medium

The same procedure as used on acid medium is applied, for example on balancing using electron ion method to complete combustion of propane

Electrochemical cells

An electrochemical cell is a device that produces an electric current from energy released by a spontaneous redox reaction. This kind of cell includes the Galvanic cell or Voltaic cell, named after Luigi Galvani and Alessandro Volta, both scientists who conducted several experiments on chemical reactions and electric current during the late 18th century.

Electrochemical cells have two conductive electrodes (the anode and the cathode). The anode is defined as the electrode where oxidation occurs and the cathode is the electrode where the reduction takes place. Electrodes can be made from any sufficiently conductive materials, such as metals, semiconductors, graphite, and even conductive polymers. In between these electrodes is the electrolyte, which contains ions that can freely move.

The galvanic cell uses two different metal electrodes, each in an electrolyte where the positively charged ions are the oxidized form of the electrode metal. One electrode will undergo oxidation (the anode) and the other will undergo reduction (the cathode). The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state and become an ion. At the cathode, the metal ion in solution will accept one or more electrons from the cathode and the ion's oxidation state is reduced to 0. This forms a solid metal that electrodeposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electric current that can be used to do work, such as turn a motor or power a light.

A galvanic cell whose electrodes are zinc and copper submerged in zinc sulfate and copper sulfate, respectively, is known as a Daniell cell.

Half reactions for a Daniell cell are these

2+ − Zinc electrode (anode): Zn(s) → Zn (aq) + 2 e 2+ − Copper electrode (cathode): Cu (aq) + 2 e → Cu(s)

In this example, the anode is the zinc metal which is oxidized (loses electrons) to form zinc ions in solution, and copper ions accept electrons from the copper metal electrode and the ions deposit at the copper cathode as an electrodeposit. This cell forms a simple battery as it will spontaneously generate a flow of electric current from the anode to the cathode through the external connection. This reaction can be driven in reverse by applying a voltage, resulting in the deposition of zinc metal at the anode and formation of copper ions at the cathode. To provide a complete electric circuit, there must also be an ionic conduction path between the anode and cathode electrolytes in addition to the electron conduction path. The simplest ionic conduction path is to provide a liquid junction. To avoid mixing between the two electrolytes, the liquid junction can be provided through a porous plug that allows ion flow while reducing electrolyte mixing. To further minimize mixing of the electrolytes, a salt bridge can be used which consists of an electrolyte saturated gel in an inverted U-tube. As the negatively charged electrons flow in one direction around this circuit, the positively charged metal ions flow in the opposite direction in the electrolyte.

A voltmeter is capable of measuring the change of electrical potential between the anode and the cathode.

Electrochemical cell voltage is also referred to as electromotive force or emf.

A cell diagram can be used to trace the path of the electrons in the electrochemical cell. For example, here is a cell diagram of a Daniell cell:

2+ 2+ Zn(s) | Zn (1M) || Cu (1M) | Cu(s)

First, the reduced form of the metal to be oxidized at the anode (Zn) is written. This is separated from its oxidized form by a vertical line, which represents the limit between the phases (oxidation changes). The double vertical lines represent the saline bridge on the cell. Finally, the oxidized form of the metal to be reduced at the cathode, is written, separated from its reduced form by the vertical line. The electrolyte concentration is given as it is an important variable in determining the cell potential.

Reversible cell : Here , the cell reactions will take place in the reverse direction , when the applied voltage is greater than the cell voltage ; The voltage of the Daniel cell = 1.1V; Cell reaction ; Cu + Zn 2+Cu 2+ + Zn when the applied voltage is greater than the cell voltage , the cell reaction is Cu 2+ + Zn Cu + Zn 2+ Ex. Daniel cell Irreversible cell : Here , the cell reactions cannot not be reversed by applying higher voltage greater than the cell voltage. Ex., Dry cell

Standard electrode potential

To allow prediction of the cell potential, tabulations of standard electrode potential are available. Such tabulations are referenced to the standard hydrogen electrode (SHE). The standard hydrogen electrode undergoes the reaction

+ − 2 H (aq) + 2 e → H2 which is shown as reduction but, in fact, the SHE can act as either the anode or the cathode, depending on the relative oxidation/reduction potential of the other electrode/electrolyte combination. The term standard in SHE requires a supply of hydrogen gas bubbled through the electrolyte at a pressure of 1 atm and an acidic electrolyte with H+ activity equal to 1 (usually assumed to be [H+] = 1 mol/liter).

The SHE electrode can be connected to any other electrode by a salt bridge to form a cell. If the second electrode is also at standard conditions, then the measured cell potential is called the standard electrode potential for the electrode. The standard electrode potential for the SHE is zero, by definition. The polarity of the standard electrode potential provides information about the relative reduction potential of the electrode compared to the SHE. If the electrode has a positive potential with respect to the SHE, then that means it is a strongly reducing electrode which forces the SHE to be the anode (an example is Cu in aqueous CuSO4 with a standard electrode potential of 0.337 V). Conversely, if the measured potential is negative, the electrode is more oxidizing than the SHE (such as Zn in ZnSO4 where the standard electrode potential is −0.76 V).

Standard electrode potentials are usually tabulated as reduction potentials. However, the reactions are reversible and the role of a particular electrode in a cell depends on the relative oxidation/reduction potential of both electrodes. The oxidation potential for a particular electrode is just the negative of the reduction potential. A standard cell potential can be determined by looking up the standard electrode potentials for both electrodes (sometimes called half cell potentials). The one that is smaller will be the anode and will undergo oxidation. The cell potential is then calculated as the sum of the reduction potential for the cathode and the oxidation potential for the anode.

E°cell = E°red (cathode) – E°red (anode) = E°red (cathode) + E°oxi (anode)

For example, the standard electrode potential for a copper electrode is:

Cell diagram + 2+ Pt(s) | H2 (1 atm) | H (1 M) || Cu (1 M) | Cu(s)

E°cell = E°red (cathode) – E°red (anode)

At standard temperature, pressure and concentration conditions, the cell's emf (measured by a multimeter) is 0.34 V. By definition, the electrode potential for the SHE is zero. Thus, the Cu is the cathode and the SHE is the anode giving

2+ + Ecell = E°(Cu /Cu) – E°(H /H2)

Or,

E°(Cu2+/Cu) = 0.34 V

Changes in the stoichiometric coefficients of a balanced cell equation will not change E°red value because the standard electrode potential is an intensive property

Spontaneity of redox reaction

During operation of electrochemical cells, chemical energy is transformed into electrical energy and is expressed mathematically as the product of the cell's emf and the electric charge transferred through the external circuit.

Electrical energy = EcellCtrans where Ecell is the cell potential measured in volts (V) and Ctrans is the cell current integrated over time and measured in coulombs (C); Ctrans can also be determined by multiplying the total number of electrons transferred (measured in moles) times Faraday's constant (F).

The emf of the cell at zero current is the maximum possible emf. It is used to calculate the maximum possible electrical energy that could be obtained from a chemical reaction. This energy is referred to as electrical work and is expressed by the following equation:

Wmax = Welectrical = -nFEcell where work is defined as positive into the system.

Since the free energy is the maximum amount of work that can be extracted from a system, one can write:[23]

∆G = -nFEcell A positive cell potential gives a negative change in Gibbs free energy. This is consistent with the cell production of an electric current from the cathode to the anode through the external circuit. If the current is driven in the opposite direction by imposing an external potential, then work is done on the cell to drive electrolysis.

A spontaneous electrochemical reaction (change in Gibbs free energy less than zero) can be used to generate an electric current in electrochemical cells. This is the basis of all batteries and fuel cells. For example, gaseous oxygen (O2) and hydrogen (H2) can be combined in a fuel cell to form water and energy, typically a combination of heat and electrical energy.

Conversely, non-spontaneous electrochemical reactions can be driven forward by the application of a current at sufficient voltage. The electrolysis of water into gaseous oxygen and hydrogen is a typical example. The relation between the equilibrium constant, K, and the Gibbs free energy for an electrochemical cell is expressed as follows:

O O ∆G =-RT ln K=-nFE cell

Rearranging to express the relation between standard potential and equilibrium constant yields

O E cell =RTln K nF

The previous equation can use Briggsian logarithm as shown below:

EO cell =0.0591V log K n

Cell emf dependency on changes in concentration

Nernst’s equation

APPLICATIONS OF NERNST’S EQUATION The applications of Nernst equation are 1. It can be used to calculate potential of an electrode; 2. Whether a reaction is feasible or not, can be decided;

3. Whether a metal can displace H2 from the acid solution can be determined ;

4. Equilibrium constant Keq. can be calculated.

GLASS ELECTRODE CONSTRUCTION

Construction :

A glass electrode consists of thin – walled bulb (the glass is a special type having low melting point and high electrical conductivity) containing a Pt wire in 0.1 M HCl. The glass electrode is represented as Pt, 0.1 M HCl / glass The thin walled glass bulb called glass membrane functions as an ion – exchange resin, and equilibrium is established between the Na+ ions of glass and H+ ions in solution and its emf is given by the expression 0 + EG= E G - 0.0595 log [H ] (or) 0 + EG= E G + 0.0595 p[H ]

ADVANTAGES AND DISADVANTAGES OF GLASS ELECTRODE Advantages : 1.It can be easily constructed 2.results are accurate 3.can be used in any solutions. 4.a small solution is sufficient for the determination of Ph. Disadvantages : 1.it cannot be used in strong alkaline solution as it affects the glass. 2.special electronic potentiometer should be used for measurements.

GLASS ELECTRODE USED TO DETERMINE THE PH OF A SOLUTION The glass electrode is placed in the solution under test and is coupled with saturated calomel electrode as shown in the fig.

Electrode 1 is glass electrode and electrode 2 is calomel electrode; they are connected through a potentiometer as shown in the diagram.

It is an internal reference electrode. The glass electrode & calomel electrode are in touch with the test solution and connected to a potentiometer; the emf of the cell is determined; then

0 E G value is taken from the tables ; now ,in the equation [2] ,

0 EG and E G are known ; so pH value can be determined.

Thus the pH of a given solution is determined using glass electrode.

STANDARD HYDROGEN ELECTRODE Construction : Hydrogen electrode consists of platinum foil connected to Pt wire and sealed in a glass tube. Hydrogen gas is passed through the side arm of the glass tube. The electrode is dipped in a 1N HCl and hydrogen gas at 1 atm. Pressure is passed at a temp. of 250c . the electrode potential of SHE is zero at all temp.

It is represented as,

0 Pt, H2 (1atm.) / HCl(1M) ; E = 0 V

In a cell , when this acts as anode , electrode reaction can be written as,

+ H2 (g) 2H + 2e when this acts as cathode , electrode reaction can be written as,

+ 2H + 2e H2 (g)

CALOMEL ELECTRODE WITH A NEAT DIAGRAM.

Construction:

It consists of a glass tube containing Hg at the bottom ; above that Hg2Cl2 (calomel) paste is placed; above this saturated solution of KCl is taken.

The Pt wire is sealed in a glass tube and its bottom is in contact with Hg as shown in the figure. The side tube is used for making electrical contact as salt bridge. The electrode potential of saturated calomel electrode = 0.2422 V

It is represented as: Pt, Hg, Hg2Cl2(s)/ KCl (saturated)

If this electrode acts as anode, reaction is

2+ 2Hg (l) Hg2 + 2e

If this electrode acts as cathode , reaction is

2+ - Hg2 + 2e 2Hg(s)

The potential of this electrode depends on the concentration of

Cl-& it decreases as the concentration of Cl– increases.

Its reduction potential is given by E(0.1N.) = +0.3338 V ; E( 1N.) = +0.280 V ;

E( sat.) = +0.2422 V ;

EMF OF A GALVANIC CELL MEASURED BY POGGENDORFF’S COMPENSATIONMETHOD Principle : The emf of a cell can is measured on the basis of Poggendorff’s compensation method. Here the emf of the test cell is opposed and compensated by an external emf; measurements are are made when there is no flow of current in the circuit ;ie emf of the test cell is balanced ( ie., compensated ) by the external emf. Circuit Diagram

Procedure :

Standard cell C1 and the test cell C2 are connected to a potentiometer through a three way plug and also to a galvanometer G and a sliding contact J as shown the circuit diagram. At first the standard cell is connected & J is moved here and there until there is no current flow in G; now the length l1α C1

Then test cell is connected & J is moved here and there until there is no current flow in G; now the length l2 α C2

Hence , C2 /C1 = l2 /l1

C2 = l2 /l1 x C1 Since current C is directly proportional to emf E ,

So , E2 = E1 . l2 /l1 thus the emf of a cell is determined on the basis of Poggendorff’s compensation method.

THE POTENTIOMETRIC TITRATION WITH A SPECIFIC EXAMPLE The titration in which the end point is noted by the change in the potential values of the solution .let us consider the titration between ferrous Ammonium Sulphate

(FAS) and K2Cr2O7 .

Principle:

The potential value of a solution depends on its concentration. During titration its concentration changes gradually and so its potential value also changes gradually; at the end point the change will be very sharp; after the end point the change will not be appreciable.

Let us consider the titration of FAS Vs K2Cr2O7 In the titration at the beginning there are only F2+ ions and no Fe3+ ions; as the dichromate is added , the F2+ ions are oxidized in to Fe3+ ; the concentration of to Fe3+ ions goes on increases ; so the potential value also goes on increases & at the end point the emf value will be sharp; further addition will result no appreciable change in emf value.

In the graph plotted drawn between EMF and volume of K2Cr2O7 , the emf goes on increases gradually as shown ; at the end point all the ions are completely oxidized in to Fe3+ ; so sharp increase results ; further the graph is almost horizontal .

The line which is almost perpendicular, gives the value of dichromate necessary for the end point; from this the strength and hence amount of FAS in the whole of the given solution can be calculated

CONDUCTOMETRIC TITRATION WITH SUITABLE EXAMPLE.

Principle :

A titration, in which concentration changes are followed by the changes in the conductance values, is called Conductometric titration. The basic principle involved in this titration is that for a given electrolyte at a particular temperature, the conductivity value depends on its concentration alone. The variation in conductance of the solution during titration is utilized to locate the end point. By plotting C values against the volume of titrant, a graph obtained consists of two straight lines; the point of intersection of these two lines gives the end point.

Let us consider the titration of Strong acid (HCl) verses Strong Base (NaOH) NaOH is taken in the burette and HCl is taken in a beaker; conductivity cell is dipped in it and connected to a Conductometer. Conductance value ,C is noted at the beginning , after each addition( 1cc) of NaOH ; C value goes on decreases as the H+ ions in the HCl is replaced by the bigger Na+ ions ; when all the H+ ions are replaced by Na+ ions , further addition of NaOH increases the value of C as it NaOH is a strong electrolyte. Then C values are plotted against volume of NaOH added; two straight lines are obtained as shown in the fig. ; the point of intersection gives the end point which give the volume of NaOH required to neutralize the HCl taken in the beaker. From these values , the strength of HCl and the amount of HCl present in the whole volume of the given solution can be calculated.

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BCH101 ENGINEERING CHEMISTRY- I

UNIT – IV CORROSION AND CORROSION CONTROL

Corrosion can be defined as the gradual destruction of materials, (usually metals), by the chemical or electrochemical reaction with its environment. Corrosion is the breakdown of materials due to reactions. It is usually oxidation with air molecules and often in the presence of water. Corrosion also occurs when an acidic or basic material touches another material. When a material corrodes, it changes and becomes weaker. Materials that corrode include metal, plastic, and wood. Also, corrosion is a form of erosion.

One form of high temperature corrosion can lead to the formation of compacted oxide layer glazes, which under certain circumstances reduces wear.Iron corrosion is called rusting.

Corrosion is a natural process, which converts a refined metal to a more chemically-stable form, such as its oxide, hydroxide, or sulfide. It is the gradual destruction of materials (usually metals) by chemical and/or electrochemical reaction with their environment. Corrosion engineering is the field dedicated to controlling and stopping corrosion.

In the most common use of the word, this means electrochemical oxidation of metal in reaction with an oxidant such as oxygen or sulfur. Rusting, the formation of iron oxides, is a well-known example of electrochemical corrosion. This type of damage typically produces oxide(s) or salt(s) of the original metal, and results in a distinctive orange colouration. Corrosion can also occur in materials other than metals, such as ceramics or polymers, although in this context, the term "degradation" is more common. Corrosion degrades the useful properties of materials and structures including strength, and permeability to liquids and gases.

PILLING –BEDWORTH RULE According to this rule if the volume of the oxide film formed is at least as large as the volume of the metal consumed , the oxide layer is protective or nonporous ; on the other hand , if the volume of the oxide is less than the volume of the metal consumed , the layer will be porous or non-protective

The Pilling–Bedworth ratio (P–B ratio), in corrosion of metals, is the ratio of the volume of the elementary cell of a metal oxide to the volume of the elementary cell of the corresponding metal (from which the oxide is created).On the basis of the P–B ratio, it can be judged if the metal is likely to passivate in dry air by creation of a protective oxide layer.

MECHANISM OF CORROSION

[1] Chemical Corrosion

1. Chemical corrosion simply involves reaction between chemicals and materials ; it occurs due to the attack of the metal surfaces by atmospheric gases such as oxygen , nitrogen , hydrogen sulphide ,sulphurdioxide , etc.,

iii) M2+ andO2- react to form metal oxide film on the surface .

M + 1/2 O2 MO Once the metal surface is converted to a monolayer of metal oxide , the metal ion diffuses outward through metal oxide film for further corrosion to occur. Thus the growth of oxide film commences perpendicular to the the metal. Surface . If the oxide layer formed is non-porous &non-volatile ,it stops further O2 attack through diffusion. Such a film behaves as a protective coating; but if it is porous and volatile , further corrosion occurs by diffusion or direct of oxygen in its ionic form with metal ion.

[2] Mechanism of Electro chemical or wet Corrosion:

This type of corrosion occurs under the following conditions :

1. When two dissimilar metals are in contact with each other in the presence of moisture or an aqueous solution.

2. When a metal is exposed to different concentration of O2 or electrolyte. Under the above conditions, one part of the metal becomes anode and the another part becomes cathode; i) At anode : Metal oxidizes to M2+ M M2+ ii) At cathode : Here reduction occurs depending on the nature of corrosive environment:

GALVANIC CORROSION.

This is one of the Two types of electrochemical corrosion. This type of corrosion occurs when two different metals are in contact with each other in the presence of an aqueous solution or moisture.

Her , more active metal (with more positive oxidation potential) acts as anode and less active metal (with less positive oxidation potential) acts as cathode.

Zn Fe Fe Cu

We know ,

Zn/Zn2+ = +0.76V ; Fe/Fe2+ = -0.44 V ; Cu/Cu2+ = -0.34 V

So , in the Zn – Fe couple . since Zn is more active than Fe , Zn acts as anode and undergoes corrosion and Fe acts as cathode; But in the Fe – Cu couple , since Fe is more active than Cu , Fe acts as anode and undergoes corrosion and Cu acts as cathode. Ex. Steel screw in a brass material Here due to galvanic corrosion , iron having more oxidation potential than brass undergoes corrosion ; Galvanic corrosion can be avoided by having bolt and nut made by the same metal. Prevention : it can be prevented by providing insulating material between the two metals.

DIFFERENTIAL AERATION CORROSION.

This is one of the Two types of electrochemical corrosion. This type of corrosion occurs when the metal is exposed to varying concentration of oxygen or any electrolyte on the surface of the metal.

Ex. Metals partially immersed in water Examples for differential aeration corrosion:

1. Pitting or localized corrosion 2. Crevice corrosion 3. Pipiline corrosion 4. Corrosion on wire fence

Let us discuss one in detail; 1.Pitting or localized corrosion;

Pitting is the localized attack , resulting in the formation of a hole around which the metal is relatively unattacked. Let us consider a drop of water resting on a metal surface as shown in the fig. the area covered by the drop of water acts as anode due to less oxygen concentration and suffers corrosion. The uncovered area freely exposed to air, acts as cathode due to high oxygen concentration.

Rate of corrosion will be more , when the area of the cathode is larger and the area of anode is smaller. Therefore , more and more material is removed from the same spot. Thus a small hole or pit I formed on the surface of the metal.

This type of intense corrosion is called pitting.

CREVICE CORROSION

Crevice corrosion is a localized form of corrosion occurring in confined spaces (crevices), to which the access of the working fluid from the environment is limited. Formation of a differential aeration cell leads to corrosion inside the crevices. Examples of crevices are gaps and contact areas between parts, under gaskets or seals, inside cracks and seams, spaces filled with deposits and under sludge piles.

Crevice corrosion is influenced by the crevice type (metal-metal, metal-nonmetal), crevice geometry (size, surface finish), and metallurgical and environmental factors. The susceptibility to crevice corrosion can be evaluated with ASTM standard procedures. A critical crevice corrosion temperature is commonly used to rank a material's resistance to crevice corrosion.

FACTORS INFLUENCING THE CORROSION.

1.Nature of the metal ; 2.Nature of environment ;

1.Nature of the metal : a)position in emf series : when two metals or alloys are in electrical contact , in presence of an electrolyte the more active metal suffers corrosion. b)relative areas of the anodic and cathodic parts ; Corrosion is more rapid and severe , if the anodic area is relatively small than the cathodic part. c) purity of the metal ; the rate of corrosion increases with increase of the impurities. d)nature of the corroded products ; if the corrosion products are porous or volatile , rapid and continuous corrosion takes place.

II. Nature of the environment ; a) Temperature: The rate of corrosion increases with increase of temperature. b) Humidity of air: The rate of corrosion increases with increase of humidity. c) Presence of impurities in atmosphere: The rate of corrosion increases with increase of impurities , gases likeCo2,H2S,SO2 , and fumes of HCl ,H2SO4. c) influence of pH : The rate of corrosion can be reduced by increasing the pH solution. e) Concentration of O2 ; when oxygen supply is greater , corrosion rate also increases.

THE ROLE OF THE NATURE OF FILM FORMED ON METAL SURFACE IN THE OXIDATION

Corrosion. The role of the nature of film formed on metal surface in the oxidation Corrosion can be discussed by “Pilling - Bedworth” rule ; According to that rule , If the volume of the oxide film formed is less than the volume of the metal corroded , the oxide layer is porous and non- protective ; Ex. The volumes of the oxides of alkali and alkaline earth metals , such as Na , Mg , ca , etc., is less than the volume of the metal consumed. Hence the oxide layer formed is porous and non- protective. If the volume of the oxide film formed is greater than the volume of the metal corroded , the oxide layer is non-porous and protective ; Ex. The volumes of the oxides of heavy metals , such as Pb , Sn , etc., is greater than the volume of the metal consumed. Hence the oxide layer formed is non- porous and protective.

CORROSION CONTROL BY a) Sacrificial anodic protection method b) Impressed current cathodic protection method

a) Sacrificial anodic protection method

In this method the metal to be protected is connected to more anodic ( active ) metal by a wire . the more active metal itself gets corroded slowly , while the parent structure is protected. The more active metal used , is called Sacrificial anode. Metals commonly employed as Sacrificial anodes are Mg , Zn , Al ,and their alloys.

Application :

1.ships hull ; 2.marine equipments ; 3. Water tanks

b)Impressed current cathodic protection method.

In this method , an impressed (direct) current is applied in the opposite direction of the corrosion current to nullify it .

In this method , the metal to be protected converted to cathode by connecting to the negative terminal of the battery. The positive terminal of the battery is connected to an inert anode. The inert anodes used for this purpose are graphite ,platinised titanium. The anode is buried in a “back fill ”It is a mixture of gypsum , coke , breeze , sodium sulphate; it provides good electrical contact to anode.

ELECTROPLATING Electroplating is a process by which the coating metal is deposited on the base metal by passing a direct current through an electrolyte solution containing the soluble salt of the coating metal.

ELECTROLESS PLATING The surface of the plate is first degreased using organic solvents or alkali followed by acid treatment; then they are dipped in different solutions to activate the surface.

DIFFERENCES BETWEEN ELECTROPLATING AND ELECTROLESS PLATING

S.N ELECTRO PLATING ELECTROLESS PLATING O. It is carried out By auto catalytic 1. It is carried out by passing current redox reactio 2. Separate anode is employed Catalytic surface acts as anode Anode reaction Anode reaction 3. M ------Mn+ + ne R ------O + ne Cathodic reaction Cathodic reaction 4. Mn+ + ne ------M Mn+ + ne ------M Not satisfactory for objects with irregular satisfactory for all objects even 5. shape. with irregular shape. It is carried out on conducting , 6. It is carried out on conducting materials. semiconducting (plastics)materials. Thickness of the plating is 1 – 100 7. Thickness of the plating is 1 – 100 µm µm

A) ELECTROPLATING OF GOLD OVER COPPER OBJECT B) ELECTROLESS PLATING OF NICKEL a) Electroplating of gold over copper object .

The electroplating is the process by which the coating metal is deposited on the base metal by passing a direct current through an electrolytic solution containing the soluble salt of the coating metal.

The base metal to be plated is made cathode of an electrolytic cell and anode is either made of coating metal itself or an inert metal of good electrical conductivity.

THEORY :

If the anode is made of coating metal itself , the concentration of the electrolyte remains unaffected during electrolysis; since metal ions deposited from solution on cathode , are replaced continuously by the reaction of free negative ions with anode.

PROCESS : The copper object to be coated is cleaned using dil HCl or H2SO4; then it is made cathode; Au foil is anode ; AuCl3 is the electrolyte; current is passed from battery. Au dissolves in the electrolyte and deposit uniformly on Cu object.

In order to get strong , smooth &adherent deposit , certain additives like gelatin , glue , etc., are added to the bath.

The conditions for a good deposit are : 1. the current density : 1 – 10 mA / cm2 2. optimum temperature : 600 C 3. low metal ion concentration. b) Electroless plating of Nickel

It is the process by the coating metal is deposited on the base metal by passing direct current through an electrolytic solution containing a soluble salt of the coating metal.

It is a technique of depositing a noble metal on an catalytically active surface of the metal to be protected , by using a suitable reducing agent without using electrical energy.

Metal ions + reducing agents ------metal + oxidized products.

( deposited ) Step.I ; Pre treatment and activation of the surface : the surface to be plated is first degreased by using organic solvents or alkali , followed by acid treatment.

Step II ; Plating bath : this consists of 1. coating solution – NiCl2 2. reducing agent - sodium hypophosphite 3. complexing agent – sodium succinate 4. buffer – sodium acetate 5. optimum ph – 4.5 6. optimum temperature - 930 C Step III ;Procedure ; The pre - treated object is immersed in the plating bath for the required time. During which the following reduction reaction will occur and the Ni gets coated over the object. The reducing agent provides electrons for reduction ,

- - + H2PO2 + H2O H2PO3 - + 2 H + 2e

At cathode : Ni+2 + 2e Ni

THE CONSTITUENTS OF PAINTS AND THEIRFUNCTIONS.

Paint can be defined as a mixture consisting of a solid pigment or pigments suspended in a medium (mixture of vehicle and thinner) , that when applied to a surface dries to form a hard coating.

THE FUNCTION OF PIGMENT IN A PAINT Pigments are solid and colour producing substances in the paint. Its functions are 1.it gives colour and opacity to the film 2.also provides strength to the film 3.protects the film by reflecting the destructive UV rays.

THE CONSTITUENTS OF PAINTS AND THEIR FUNCTIONS:

1. Pigment : these are solids and colour producing substances in the paint. Function : 1. It gives colour and opacity to the film. 2. It also provides the strength to the film. 2.Vehicle or drying oil ; This is the non- volatile portion of the medium. This is the film forming constituent of the paint. Function : 1. they form a protective film by the oxidation and polymerization of the oil. 2. They hold the pigment particle together on the metal surface. 3. Thinner or solvent This is a volatile portion of the medium; it easily evaporates after application of a paint. Function : 1. Reduces the viscosity of the paint 2. Increases the elasticity of the film 3. Increases the penetrating power of the vehicle. 4. Extender or Filler They are white or colourless pigments. Function : 1. Reduces the cost of the paint. 2. Modifies the shades of the paint 3. Prevents the shrinkage and cracking 5.Driers : They are the substances , used to accelerate the process of drying. Function ; 1. They act as oxygen – carrier or catalysts 2. Provide oxygen , which is essential for oxidation 6.Plasticisers : They are chemicals added to the paints to provide elasticity to the film and to prevent the cracking of the film .

7.anti-skinning agents :

These are chemicals added to the paints to prevent gelling and skinning of the paint.

********** BCH101 ENGINEERING CHEMISTRY- I

UNIT – V

NON – CONVENTIONAL ENERGY SOURCES

NUCLEAR FISSION It can be defined as “the process of splitting of heavier nuclei nucleus into two or more smaller nuclei with simultaneous liberation of large amount of energy.” It is always accompanied by two or more neutrons.

NUCLEAR FUSION It can be defined as “the process of combination of lighter nuclei into heavier nuclei, with simultaneous liberation of large amount of energy. ”

ENERGY LOSS IN A NUCLEAR CHAIN REACTION Some of the neutrons released in the nuclear chain reaction may escape from the surface to the surrounding o may be absorbed by U238 present as impurity. This will result in breaking of the chain and the amount of energy released will be less than expected; this less in energy is called Energy loss in a nuclear chain reaction. (A)NUCLEAR ENERGY CRITICAL MASS .(B) SUPER CRITICAL MASS .

(a)Nuclear Energy Critical mass .

The minimum amount of fissionable material required to continue the nuclear chain reaction is called Critical Mass. (b) Super Critical mass . If the amount of fissionable material is more than the Critical Mass , it is called Super Critical mass. (c) Sub -critical mass: If the amount of fissionable material is less than the Critical Mass , it is called Sub - Critical mass.

DIFFERENCE BETWEEN NUCLEAR FISSION AND FUSION REACTIONS

S.No Nuclear fission Nuclear fusion . It is a process of breaking a heavier It is a process of combination of lighter 1. nucleus. nuclei. 2. Does emits radioactive rays. Does not emit radioactive rays. 3. Occurs at ordinary temperature. Occurs at high temperature; >106 k The mass no.& the atomic no. of new The mass no.& the atomic no. of new 4. elements are lower than that of parent elements are lower than that of parent nucleus. nucleus. 5. It gives rise to chain reaction. It doesn’t give rise to chain reaction. 6. Emits neutrons . Emits positrons . 7. Can be controlled. Can’t be controlled.

LIGHT WATER NUCLEAR POWER PLANT

235 LWNP Plant is the one in which U fuel rods are submerged in H2O; here water act as coolant and moderator.

CONTROL ROD

Control rods are usually combined into control rod assemblies — typically 20 rods for a commercial pressurized water reactor assembly — and inserted into guide tubes within a fuel element. A control rod is removed from or inserted into the central core of a nuclear reactor in order to control the number of neutrons which will split further uranium atoms. This in turn affects the thermal power of the reactor, the amount of steam generated, and hence the electricity produced. The control rods are partially removed from the core to allow a chain reaction to occur. The number of control rods inserted and the distance by which they are inserted can be varied to control the reactivity of the reactor.

COOLANT Nuclear reactor coolant The light-water reactor also uses ordinary water to keep the reactor cooled. The cooling source, light water, is circulated past the reactor core to absorb the heat that it generates. The heat is carried away from the reactor and is then used to generate steam. Most reactor systems employ a cooling system that is physically separate from the water that will be boiled to produce pressurized steam for the turbines, like the pressurized-water reactor. But in some reactors the water for the steam turbines is boiled directly by the reactor core, for example the boiling-water reactor.

Many other reactors are also light-water cooled, notably the RBMK and some military plutonium-production reactors. These are not regarded as LWRs, as they are moderated by graphite, and as a result their nuclear characteristics are very different. Although the coolant flow rate in commercial PWRs is constant, it is not in nuclear reactors used on U.S. Navyships.

FUEL Nuclear fuel

The use of ordinary water makes it necessary to do a certain amount of enrichment of the uranium fuel before the necessary criticality of the reactor can be maintained. The light- water reactor uses uranium 235 as a fuel, enriched to approximately 3 percent. Although this is its major fuel, the uranium 238 atoms also contribute to the fission process by converting to plutonium 239; about one-half of which is consumed in the reactor. Light- water reactors are generally refueled every 12 to 18 months, at which time, about 25 percent of the fuel is replaced.

MODERATOR Neutron moderator

A neutron moderator is a medium which reduces the velocity of fast neutrons, thereby turning them into thermal neutrons capable of sustaining a nuclear chain reaction involving uranium-235. A good neutron moderator is a material full of atoms with light nuclei which do not easily absorb neutrons. The neutrons strike the nuclei and bounce off. After sufficient impacts, the velocity of the neutron will be comparable to the thermal velocities of the nuclei; this neutron is then called a thermal neutron.

The light-water reactor uses ordinary water, also called light water, as its neutron moderator. The light water absorbs too many neutrons to be used with enriched natural uranium, and therefore uranium enrichment or nuclear reprocessing becomes necessary to operate such reactors, increasing overall costs. This differentiates it from a heavy water reactor, which uses heavy water as a neutron moderator. While ordinary water has some heavy water molecules in it, it is not enough to be important in most applications. In pressurized water reactors the coolant water is used as a moderator by letting the neutrons undergo multiple collisions with light hydrogen atoms in the water, losing speed in the process. This moderating of neutrons will happen more often when the water is denser, because more collisions will occur.

WORKING:- The fission reaction is controlled by inserting or removing the control rods B10 or Cd113from the space between fuel rods. The heat emitted by fission of U 235 is absorbed by the coolant; the heated coolant [at 3000c ] then goes to the heat-exchanger containing sea water . The coolant here, transfers heat to sea water, which is converted it into steam; the steam then drives the turbines, generating electricity.

SOLAR ENERGY CONVERSION Solar energy conversion is the process of conversion of direct sun light into more useful forms. This conversion occurs by two mechanisms: 1. Thermal conversion 2. Photo conversion 1. Thermal conversion:- It involves absorption of thermal energy in the form of infra-red radiation. Solar energy is an important source for low heat [ T< 100 c ]which is useful for heating water , building , refrigeration, etc., 2. Photo – conversion: - It involves conversion of light energy directly into electrical energy. Solar cells:- Photo conversion can be done by Solar cells. Solar cell converts sunlight energy directly into electrical energy; it does not utilize chemical or nuclear reactions to produce electric power. Principle;- The basic principle involved in the solar cells is based on the photovoltaic effect. When solar rays fall on a two layer of semi – conductor devices, a potential difference between the two layers is produced. This potential difference causes flow of electrons & produces electricity.

Construction: Solar cell consists of a p-type semiconductor[ such as Si doped with B ] & n-type semiconductor [ such as Si doped with P ].They are in close contact with each other.

Working:- When solar rays fall on the top layer of p-type semiconductor , the electrons from the valence band get promoted to the conductor band by crossing p-n junction into n-type; thereby potential difference is created between the two layers; this causes flow of electrons [i.e., an electric current. ] If more solar rays fall on the surface of top layer , the potential difference & hence the current also increases. Thus when p and n – layers are connectedto an external circuit, electrons flow from n – layer to p-layer, and hence current is generated.

APPLICATIONS OF SOLAR CELLS; 1. for lighting purpose; 2. in boilers to produce hot water; 3. incalculators, electric watches, etc., 4. Solar energy can be stored in Ni- cd batteries & Pb – acid batteries; 5. To drive vehicles,motors, etc., 6. used as a source of electricity in space craft& satellites. 7. Solar cells can be used to produce hydrogen by electrolysis of water.

WORKING OF HYDROGEN – OXYGEN FUEL CELL. Hydrogen – Oxygen fuel cell It is the simplest and most successful fuel cell. In this , fuel H2 , Oxidizer O2 and the liquid electrolyte are continuously passed through the cell. Construction It consists of two porous carbon electrodes [anode and cathode]. These electrodes are made of compressed carbon containing a small amount of catalyst [ Pt, Pd ,Ag ]. In between the electrodes , an electrolytic solution , such as 25 % KOH or NaOH is filled. Two electrodes are connected through a voltmeter.

LEAD – ACID STORAGE BATTERY. It is a secondary battery which can act as a voltaic cell and as an electrolytic cell. When it acts as a voltaic cell , it supplies electrical energy ; when it is recharged [ ie., current is sent in ] , the cell operates as an electrolytic cell. Construction : It consists of a number of [ 3 to 6 ] voltaic cells, connected in series to get 6 to 12 V battery . anode is made of Pb ; cathode is made of PbO2. A number of Pb & PbO2 plates are connected in parallel separately ; the plates are separated from adjacent ones by insulators rubber or glass fibre. The entire set up is immersed in dil. H2SO4 having a density of 1.3 gm/ml. the cell is represented as :

From these reactions it is clear that PbSO4 is precipitated at both electrodes and

H2SO4 is used up. As a result , the concentration of H2SO4 decreases & hence the

density of H2SO4 falls below 1.2 gm/ml; so battery needs recharging. Recharging: it is done by sending current in the opposite direction. The electrode reactions get

reversed. As a result Pb is deposited on anode & PbO2 on cathode .

NI- CD BATTERY This is also a rechargeable battery. Description :

Nickel – Cadmium cell consists of a Cadmium anode and a metal grid containing a

paste of NiO2 acting as a cathode. The electrolyte is KOH

Cell is represented as : Cd ,Cd(OH)2 // KOH (aq) / NiO2,Ni

Working :

At anode : 2+ - Cadmium is oxidized to Cd and further it combines with OH to form Cd(OH)2 - Cd + 2 OH ------Cd(OH)2 + 2e At cathode : - NiO2 + 2 H2O + 2e ------Ni(OH)2 + 2 OH Net cell reaction :

Cd + NiO2 + 2 H2O + 2e ------Cd(OH)2 + Ni(OH)2 + energy

The products Cd(OH)2 and Ni(OH)2 adhere well to the surfaces of the corresponding electrodes. Recharging: this is similar to lead storage battery. When the current is passed in the opposite direction, the electrode reaction gets reversed. As a result Cd gets deposited on the anode and NiO2 on the cathode.

LITHIUM BATTERY

Lithium battery is a solid state battery because instead of liquid or a paste electrolyte , solid electrolyte is used.

Construction ;

The lithium battery consists of a lithium anode and TiS2 cathode. A solid electrolyte , generally a polymer is packed between the electrolytes. The electrolyte (polymer) permits the passage of ions but not that of electrons

Working (discharging)

When the anode is connected to cathode , lithium ions move from anode to cathode . The anode is elemental lithium, which is the source of the lithium ions and electrons. The cathode is a material capable of receiving the lithium ions and electrons. Recharging the Battery: The lithium battery can be recharged by supplying an external current , which derives the lithium ions back to the anode . the overall reaction is + – Li Ti S2 Li + TiS2

This cell is rechargeable and produces a cell voltage of 3.0 V

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