Reference Book: 1. Principle of Physical Chemistry by Haque and Molla 2. Essentials of Physical Chemistry by B.S. Bahl and Arun Bahl 3. Physical Pharmacy by Alfred Martin, James Swarbrick and Arthue Cammarata Electrochemistry Electrochemistry deals with the phenomenon of, their mechanism of such transformation inter conversion of electrical and chemical energy and other phenomenon associated with this. i. Conversion of electrical energy to chemical energy ii. Conversion of chemical energy to electrical energy
- Electrolytic Cell - Electrochemical Cell Electrolytes: - Electrolytes are electrovalent substances that form ions in solution which conduct an electric current. - Based on the conductivity, the electrolytes can be divided into three group: - Strong electrolytes: Good conductors of electricity e.g. NaCl, KCl etc. - Weak electrolytes: Poor conductors of electricity e.g. acetic acid, aluminium hydroxide etc. - Non-eletrolytes: Not able to conduct electricity e.g. sugar Electrolysis -The phenomenon of decomposition of an electrolyte by passing electric current through its solution is called electrolysis.
- The process of electrolysis is carried out in an apparatus is called Electrolytic cell. - The poles, usually plates or wires, through which electricity is supposed to leave the solution is called anode. It is positive charge. Oxidation takes place at anode. - The poles, usually plates or wires, through which electricity is supposed to enter the solution is called cathode. It is negative charge. Reduction takes place at cathode. Electrolysis of NaCl At normal condition: NaCl Na+ + Cl- Reaction at anode: Cl- - e- Cl (Oxidation) Reaction at cathode: Na+ + e- Na(Reduction) Total reaction: Electrolysis of NaCl
2NaCl 2Na + Cl2
Electrolysis of HCl: ??????????? Electrolyte Electrode Cathodic Anodic reaction reaction 2+ - → - Aqueous acidified Pt Cu + 2e → Cu 2Cl Cl2 +2e CuCl2
2+ - → - Molten PbBr2 Pt Pb + 2e → Pb 2br Br2 + 2e
+ - - - Sodium chloride Hg 2Na + 2e →2Na 2Cl → Cl2 + 2e solution
+ - - - Silver nitrate Pt Ag + e →Ag 2OH →1/2 O2 + H2O + 2e solution
+ - - - Sodium nitrate Pt 2H + 2e → H2 2OH →1/2 O2 + H2O + 2e solution Electrical Unit: Coulomb: - It is a unit of quantity of electricity - It is the amount of electricity which will deposit 0.001118 gm of silver from a 15% solution of silver nitrate in a coulometer. Ampere: -It is a unit of rate of flow of electricity -It is the amount of current which will deposit 0.001118 gm of silver in one second. -An ampere is a current of one coulomb per second. Ohm: - An ohm is a unit of electric resistance. - It is the resistance offered by at 0 0C to a current by a column of mercury 106.3 cm long of about 1 sq mm cross-sectional area and weighing 14.4521 grams. Volt: - A volt is a unit of electromotive force. - It is the difference in electrical potential required to send a current of one ampere through a resistance of one ohm. Faraday’s Law of Electrolysis: The relationship between the amount of a given product liberated at an electrode and the quantity of electricity was established by Scientist Faraday in 1834. This is known as Faraday’s law of electrolysis. First Law: The amount of a given product liberated at an electrode during electrolysis is directly proportional to the quantity of electricity which passes through the electrolytic solution. If the mass (m) of a substance (in grams) deposited on electrode by passing Q coulombs of electricity, then m Q or, m I X t [ I is the strength of current in ampere, t = time in second] So, m = Z X I X t [ Z = constant, electrochemical equivalent] If I =1 and t = 1, then m = Z i.e. The electrochemical equivalent is the amount of substance deposited by 1 ampere current for 1 sec i.e. 1 Coulomb. The quantity of electricity required to liberate one gram equivalent of a substance is 95,600 coulombs. This quantity of electricity is known as Faraday, denoted by F.
Importance of Faraday’s 1st Law: With the help of this law it is possible to calculate - the value of electrochemical equivalent of different electrolytes. - the mass of different substances produced by passing a known quantity of electricity through their solutions.
Problem: 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What is the electrochemical equivalent of copper? Z = m/It Second Law: If the same quantity of electricity is passed through electrolyte solutions the amount of different substances liberated or dissolved at the electrode will be proportional to their chemical equivalent.
O2 H2
Cu Ag Cu Ag
Acidulated water CuSO4 solution AgNO3 solution mass of hydrogen liberated Equivalent weight of hydrogen = mass of copper deposited Equivalent weight of copper
mass of copper deposited Equivalent weight of copper = mass of silver deposited Equivalent weight of silver Importance of Faraday’s Second Law: The second law of electrolysis helps to calculate - The equivalent weight of metals - The unit of electric charge - The Avogadro’s number
Problem: What current strength in amperes will be required to liberate 10 g of iodine from potassium iodide solution in one hour? (1 F = 96,500 coulombs, g. eq. wt. of I is 127)
Problem: An electric current is passed through three cells in series containing respectively solutions of copper sulfate, silver nitrate and potassium iodide. What weight of silver and iodine will be liberated while 1.25 g of copper is being deposited? (equivalent weight of copper, silver and iodine are 31.7, 108 and 127 respectively) Conductance of electrolytes: The power of electrolytes to conduct electric currents is termed conductivity or conductance. Like metallic conductors the electrolytes obey Ohm’s laws: The current (I) flowing through an electrolyte solution is directly proportional to the potential difference (E) at two ends of solution. I E or I = E/R where R is resistance of the electrolytes Again the resistance of a conductor is directly proportional to the length (l) and inversely proportional to its cross-section (A). R l /A R = l/A ------(i) where is constant called specific resistance When, l = 1 cm and A = 1 cm2 then R = i.e the specific resistance is the resistance in ohms between the opposite faces of a cube of the material having an edge 1 cm long. Mechanism of Electrolytic Conduction 1. Grotthus (1805) Chain Theory: - + - In solution, the electrolyte molecule, + - + - + - + - composed of positive and negative part, move in random manner. + - + - + - + - - Solute molecules arrange themselves in chain showing positive end toward + - + - + - anode when electric field is applied. - Attraction and liberation of ions at electrode-----splitting of inner molecules ------rolling of molecules ------same process continues.
Drawbacks: Only small amount of energy is required in electrolysis whereas in Grotthus theory requires large amount of energy. 2. Savante Arrheneus (1887) Theory: - When dissolved in water, neutral electrolyte molecules are split up into two types of charged particles. These particles are called ions and the process is called ionisation. Positive particles are called cation and negative one is called anion. - The ions present in solution constantly reunite to form neutral molecules. AB ↔ A+ + B- - The charged ions are free to move through the solution to the oppositely charged electrode. This movement of ions constitute the electric current. - The electrical conductivity depends upon the number of ions present in the solution. Thus degree of dissociation of an electrolyte determines whether it is a strong electrolyte or weak electrolyte. - Specific conductance The reciprocal of the specific + anode resistance is called specific - Solution Cathode conductance. i.e = 1/ = l/RA + (from equation I) 1 cm 1 cm Specific conductance depends on the number of ions present in unit volume (1 ml ) solution and speed of the ion concerned Unit of specific conductance is ohm-1cm-1 or siemens/cm.
Effect of concentration on specific conductance???? -↓Concentration ------↓SC - specific conductance of NaCl in water at 18 0C. Vol. containing 1 gm mole Sp. Conductance 1,000 0.0744 5,000 0.0176 Equivalent conductance: The conductance of an electrolyte obtained by dissolving one gram equivalent of it in V volume (cc) of solution. It is denoted by (lambda). = specific conductance X volume (V in cc) of solution containing 1 gram equivalent of solute. = X V = 1/R X l/A X V.
In case, if the concentration of the solution is N g-equivalent per litre, then the volume containing 1 g equivalent of the electrolyte will be 1000/N. So equivalent conductance = X 1000/N
Unit= ohm-1cm2eqvt-1 Variation of Equivalent conductance with concentration: - Equivalent conductance of a solution does not vary linearly with concentration - The effect of concentration on equivalent conductance can be studied by plotting values against the square root of concentration.
Weak electrolytes Strong electrolytes
√C √C For strong electrolytes: - The equivalence conductance decrease with the increase on concentration. - At higher concentration, the forces of attraction between the opposite ions increases and consequently it decreases the speed of the ions with which they move towards oppositely charged electrode that ultimately results in decrease equivalence conductance . - As the solution becomes more and more diluted, the equivalence conductance increases till it reaches a limitary value. This value is known as equivalence conductance at infinite dilution. - = X V For Weak electrolytes: - The equivalence conductance decrease with the increase on concentration. - Weak electrolytes have low ionic concentration and hence interionic forces are negligible. - Increase in equivalent conductance in case of weak electrolytes at reduced concentration is due to the increase in the number of ions. Problem: 0.5 N solution of a salt placed between two platinum electrodes 20 cm apart and of cross-section 4 sq cm has a resistance of 25 ohms. Calculate the equivalent conductance of the solution.
= X V = 1/R X l/A X 1000/V. Variation of conductance with temperature: - Generally, temperature ----- conductance of a solution. - The conductance of a given solution increases by 2-3% for one degree (C) rise in temperature. - Conductance of 0.1 M KCl solution At 18 0C, conductance is 1.12 x 10-2 ohm-1 cm-1 At 25 0C, conductance is 1.29 x 10-2 ohm-1 cm-1 - Conductance of an electrolytes depends upon two factors: - Number of ions present in a given volume of solution - Speed at which ions move towards the electrodes - First factor does not change or reduce in a very small number - Increase temperature decreases the viscosity of the liquid that makes the ions to move freely towards the electrodes. Experimental determination of Conductance Basic principle is based on Wheatstone’s bridge
Battery (a-c source) R = variable decade resistance Buzzer C = conductance cell R / M M C M (= thick metal plate)
AB = Bridge wire A l S (100-l) B 0 (= Jockey) 100
T (= telephone) Based on the principle of Wheatstone’s bridge, we can write
R l = Cell resistance (100-l) R(100-l) Cell resistance = l l Cell conductance = (100-l)R
Problems: - Direct current should not be used because eletrolysis will take place - Polarization of the electrode occurs. - Complication may arise due to evolution of gas in many cases. - Secondary reaction at the electrode may occur. - Sometime difficult to determine the exact point at meter bridge Application of Conductance Measurements Conductance depends upon - number of electrolytes. - velocity of the ions/electrolytes 1. Conductance titration (acid-base reaction) 2. Conductance titration (precipitation reaction) 3. Determination of the solubility of sparingly soluble salt
Strong acid (HCl) and strong base (NaOH) reaction
- conductance of H+ is high compared to Na+ - conductance decreases
- conductance of OH- is high Conductance
Vol of NaOH added Weak acid (CH3COOH) and strong base (NaOH) reaction - conductance of H+ is high. Due to weak acid, dissociation of acid is low so number of ion is also low. As a result, conductance of the solution is
low Conductance - Reaction product is salt which is dissociated to ions that ultimately Vol of NaOH added increase the conductance. - After complete neutralization, there is high increase in the number of OH- ions that cause sharp increase of the conductance of the solution.
Strong acid (HCl) and weak base (NH4OH) reaction
?????????????????? Strong acid (HCl) and weak base (NH4OH) reaction - conductance of H+ is high. High conductance. - As the concentration of H+ decreases, the conductance
also decreases. Conductance - After complete neutralization, there is little increase in the Vol of NH4OH added number of OH- ions due to weak base. So there is little increase of the conductance of the solution.
Weak acid (CH3COOH) and weak base (NH4OH) reaction
?????????????????? Weak acid (CH3COOH) and weak base (NH4OH) reaction - conductance of H+ is high. Due to weak acid dissociation of acid is low so number of ion is also low. As a result, conductance of the solution is low
- Reaction product is salt which is Conductance dissociated to ions that ultimately increase the conductance. Vol of NH4OH added - After complete neutralization, there is high increase in the number of OH- ions that cause sharp increase of the conductance of the solution.
Advantages: - No indicator is required for the determination of end point. - Colored solution can be conveniently used by this method - Capable of giving very accurate result. - Also suitable for very dilute solution v- v+ + v- v+ = velocity of cation v- = velocity of anion v+ v+ + v- Determination of Transport Number: Hittorf’s Method The electrodes are connected to a d-c supply, variable resistance, Cu/Ag coulometer * Coulometer is to determine the quantity of electricity passed.
Procedure:
- Filled with AgNO3 solution(0.1- 0.05 N). - Air bubbles are removed - Small current (10-20 mA) is passed through the solution for a known period of time (2 h) - Collect the anode and cathode compartment solution + - Determine the concentration of Ag by using standard NH4SCN solution using ferric alum as indicator. Calculation: After passing electric current: Let the weight of anode solution taken out = a gm Weight of silver nitrate present in it by titration = b gm Weight of water = (a-b) gm
Before passing electric current: Let the weight of silver nitrate in (a-b) gm of water before passing electric current = c gm Fall in concentration =(c-b) gm of silver nitrate =(c-b)/170 gm eqv = d Let the weight of silver deposited in silver coulometer = w1 gm = w1/108 gm eqv = W gm eqv
Transport number of Ag+ (tAg+) = (Fall in conc. around anode)/ (Amount of Ag deposited) = d/W - Transport number of NO3 (tNO3-) = (1-d/w) Problem: A solution of silver nitrate containing 12.14 g of silver in 50 ml of solution was electrolyzed between platinum electrodes. After electrolysis, 50 ml of the anode solution was found to contain 11.55 g of silver while 1.25 g of metallic silver was deposited on + - the cathode. Calculate the transport number of Ag and NO3 ions.