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EP&M. . Physical chemistry. .

2. Electrochemistry.

A phenomenon of electric current in solid conductors (class I conductors) is a flow of electrons caused by the electric field applied. For this reason, such conductors are also known as electronic conductors (metals and some forms of conducting non-metals, e.g., graphite). The same phenomenon in conductors (class II conductors) is a flow of caused by the electric field applied. For this reason, such conductors are also known as ionic conductors (solutions of dissociated species and molten salts, known also as ). The question arises, what is the phenomenon causing the current flow across a boundary between the two types of conductors (known as an interface), i.e., when the circuit looks as in Fig.1.

DC source Fig. 1. A schematic diagram of an electric circuit consis- flow of electrons ting of both electronic and ionic conductors.

The phenomenon occuring at the interface must involve both the electrons and ions to ensure the continuity of the circuit. metal metal

flow of ions ions containing solution

When a conventional reaction occurs in a solution, the charge transfer (electron transfer) happens when the two reacting species get in touch with each other. For example:

Ce4+ + Fe 2+→ Ce 3+ + Fe 3+ (2.1)

In this reaction cerium (IV) draws an electron from iron (II) that leads to formation of cerium (III) and iron (III) ions. Cerium (IV), having a strong affinity for electrons, and therefore tending to extract them from other species, is called an oxidizing agent or an oxidant. Iron (II), that readily donates electrons, is called a reducing agent or a reductant. Iron (II) reduces cerium (IV) and at the same is being oxidized. Cerium (IV) oxidizes iron (II) at the same time being reduced in the process.

It is possible to separate the processes of oxidation and reduction in space. The two species do not come into contact with each other and the electron transfer is achieved via an external electrical circuit. Such an arrangement is called an . An electrochemical cell in which reaction (2.1) occurs is shown in Fig. 2.

2.1. Electrochemical cells. electrons Fig. 2. In the cell shown at left, the overall reac- µA tion is the same as reaction (2.1). How- ever, oxidation of iron (II) occurs in the electrolytic bridge left beaker (half-cell) and reduction of cerium (IV) in the right one. Electrons are drawn by cerium from the platinum sheet and delivered by iron to a twin one (Pt terminals are inert in the process).

2+ 3+ 4+ 3+ - - One can write: Fe Fe +e Ce +e Ce 4+ - 3+ Pt Pt Ce + e→ Ce (2.1a) Fe2+→ Fe 3+ + e - (2.1b) 2+ 4+ Fe solution Ce solution

Reactions written in the caption to Fig. 2 are called half-reactions. They can be handled in the usual way as any other chemical reactions (normal rules apply). If we add both half-reactions, we will get back the overall reaction (2.1). The reaction is the phenomenon we asked at the beginning, connecting the

15 EP&M. Chemistry. Physical chemistry. Electrochemistry. flow of electrons in the external circuit with flow of ions in the . To permit flow and close the circuit, solutions in both half-cells are connected by a reversed U-tube (with porous tips) filled with solution of salt, e.g., KCl or KNO3, known as an electrolytic bridge. Some cells are built of that share a common electrolyte. They are known as cells withou liquid junction.

2.1.1. Electrodes (half-cells). and .

Half cells are also called electrodes, though, the term is frequently confused with the terminals of the external circuit (in the example shown - Pt plates). Sometimes, the material of the terminal is actually involved in the half reaction, for example: Cu2+ + Zn 0→ Zn 2+ + Cu 0 (2.2) Cu2+ + 2e -→ Cu 0 (2.2a) Zn02+-→ Zn + 2e (2.2b) when the two terminals are made of copper and zinc. The name of the half cell or the electrode should rather be associated with the half-reaction than with the material of the terminal. One can also say, with a redox couple involved. It is quite obvious, that when both constituents of the couple are ions, or one is in gaseous state, then the terminal is not involved and is made of a neutral material, like platinum or carbon (graphite, glassy carbon).

By definition, the electrode at which an oxidation reaction occurs is called anode and that one at which a reduction takes place is cathode.

An important electrode involving in the redox couple is shown in Fig.3. We will frequently comment on this example in future considerations. R ∞ Fig. 3. The half-reaction occurring in the left mV half-cell is: +- electrolytic bridge H2 (g)→ 2H (aq) + 2e (2.3a) H Note, that the reaction is written in a way 2 p=1 Atm indicating an equilibrium.

Pt 4+ 3+ - Ce +e Ce Pt Standard a=1Ce4+ Hydrogen a=1Ce3+ Electrode 4+ 3+

a=1H+ Ce /Ce solution

In general there are two types of electrochemical cells. Galvanic or voltaic cells are these which are capable to deliver electricity at the expense of a spontaneous reaction occurring in the cell. On the contrary, electrolytic cells are those in which we perform desired chemical changes (reactions) at the expense of an external source of electric (non-spontaneous, externally driven redox reactions). Both kinds have numerous practical applications. Voltaic cells are batteries (storage of chemical energy to be retrieved in a form of electric energy) and electrolytic cells are used in manufacturing processes (e.g., those of metals).

2.1.2. Conventional cell notation. A half-cell, or a redox couple is usually written as a following scheme: Oxidized form/Reduced form (2.4)

+ 2+ 4+ 3+ or Ox/Red for short, e.g., H /H2, Cu /Cu, Ce /Ce . A cell notation, by convention (Stockholm convention) is written in such a way, that anode is written at the left side and cathode - at the right. For example, the cell shown in Fig. 2 and in reaction (2.1) may be written as follows: 2+ 3+ 4+ 3+ Pt Fe (aq,a1 ),Fe (aq,a2 ) Ce (aq,a3 ),Ce (aq,a4 ) Pt (2.5)

16 EP&M. Chemistry. Physical chemistry. Electrochemistry. where symbol | indicates a boundary (interface) and symbol || indicates the liquid junction. Activities are indicated (we will simplify it using molar concentrations). Some other examples: cell corresponding to reaction (2.2) 2+ 2+ Zn Zn (aq,a1 ) Cu (aq,a2 ) Cu (2.6) cell shown in Fig.3 +4+3+ Pt H2 (g,p = 1 Atm)H (aq,a = 1) Ce (aq,a = 1),Ce (aq,a = 1)Pt (2.7)

2.1.3. Electrode reaction. Overall cell reaction.

The basics of this subject have been discussed above. One of common problems, that will become important in future considerations is how to design a cell to perform a desired overall reaction.

Example 2.1: Write a suitable diagram of a cell in which the following overall reaction occurs (in aqueous solution): -2++ 2+4+ 2MnO4 + 5Sn + 16H→ 2Mn + 5Sn + 8H2 O (2.8) Solution: The reaction (as written) suggests the permanganate is the oxidant (undergoes reduction) and tin (II) is the reductant (undergoes oxidation). Hence, the Sn4+/Sn2+ redox couple should be placed at the left side (oxidation → anode).

4+ 2+ - 2+ + Pt Sn (aq,a1 ),Sn (aq,a24 ) MnO (aq,a3 ),Mn (aq,a4 ),H (aq,a5 ) Pt (2.9) Example 2.2: Design a cell diagram in which the following overall reaction occurs in aqueous solution: K AgCl(s)←→s Ag+- (aq) + Cl (aq) (2.10) Solution: The reaction can be "decomposed" into two reactions (each one temporarily written as reduction): Ag+- (aq) + e→ Ag(s) (2.11) AgCl(s) + e--→ Ag(s) + Cl (aq) (2.12)

One can handle the electrochemical reactions in a way analogous to that utilized when one treats thermochemical reactions using the Hess's law. However, one doesn't multiply the potentials when the reaction (stoichiometric coefficients) is multiplied. In the above case, one can reverse the first reaction and add both of them. The outcome is exactly reaction (2.10), as desired. The scheme suggests then, reaction (2.11) to be oxidation (anodic) and reaction (2.12) to be reduction. Therefore, the following scheme should be the solution to the problem: Ag|AgCl(s)|Cl-+ (aq)||Ag (aq)|Ag (2.13)

We will see later, whether the above scheme represents an actual . In addition to the galvanic/electrolytic classification of the cells, there is another, important one. Namely, one can distinguish the reversible and irreversible cells. If, after reversing the direction of the electron flow, the overall reaction doesn't change and simply runs backward, the cell is reversible. In the irreversible cell, changing the direction of current causes an entirely different half-reaction to undergo at one or even both electrodes.

Fuel cells. Let's consider the cell shown schematically below:

+ - Pt H21 (g,p ) H (aq,a1 ) OH (aq,a22 ) O (g,p 2 ) Pt (2.14) We see, that at the anode gaseous hydrogen is oxidized and at the cathode oxygen is reduced. This is so called fuel cell. The hydrogen/oxygen fuel cell is the source of the most clean, environment friendly energy. If hydrogen is produced and stored, then fuel cells can deliver its energy and the sole

17 EP&M. Chemistry. Physical chemistry. Electrochemistry. (as one can easily notice) is water. Moreover, this way of the conversion of energy from chemical to electric one: chemical energy → electric energy fuel cell bypasses the thermodynamic limitations (Carnot's cycle) of the more conventional way: chemical energy → → mechanical energy → electric energy combustion steam generator boiler turbine hence, the overall efficiency of the process is much higher. Fuel cells are already technologically possible but they are not yet economically viable. Considerable R&D efforts are being done all over the world to achieve this goal.

2.2. Electrode potential. Electromotive (EMF).

If the cell behaves as a galvanic cell, it means there is certain ability for pushing and pulling the electrons, which depends on the nature of both the redox couples involved (we will later see, also on their composition). This is measured as the cell potential or, more correctly, its . Electromotive force (EMF) is the cell measured with no current flow between the electrodes (see Fig. 3, where a resistance approaching infinity is included in the circuit. It is *the maximum voltage measured between them. When the resistance is small, and the current drawn, the voltage is smaller due to different limitations (transport of the reacting species in the electrolyte, kinetic effects in electrode reactions, ohmic drop across the electrolyte). Gradually, the voltage falls to zero, when compositions in both half-cells reach equilibrium values for the overall reaction (at equilibrium there is no net reaction, hence no charge transfer or electron flow or current). Electromotive force is the voltage measured between the cathode and the anode. It is a difference of potentials, hence:

EMF = Ecell = E cathode - E anode (2.15) Units of both EMF and electrode potentials (E) are . One is a difference of potentials when moving one Coulomb of electric charge delivers (when reaction is spontaneous) or requires (when forced) energy equal to one Joule: 1[J] 1[V] = (2.16) 1[C] EMF characterizes the cell (overall redox reaction) ability to deliver energy. The higher the EMF (which is always positive for galvanic cells), the larger energy () can be delivered. Similar definition was used to describe ("that part of the energy of a system that can be converted into work", subchapter 1.4.1). Hence, the Gibbs free energy for the overall reaction in the cell is proportional to the EMF:

∆G∝ Ecell or ∆ G = -nFE cell (2.17) where: n - is number of electrons exchanged in the overall reaction, F - is Faraday's constant, equal to 96 485 [C/mol] (approximate but acceptable value is 96500). In other words it is the charge carried by one mole of electrons. One should remember, that there is usually certain contribution of the so called liquid junction potential to the EMF measured. We will neglect it in our considerations here. Nevertheless, it must be accounted for or avoided in practical studies. 2.2.1. Standard potential.

The standard cell potential E0 of a galvanic cell is the cell potential measured when the activities (concentrations) of all the ions involved in the overall cell reaction are equal to 1 mol/dm3 and all the are at 1 Atm .

0 0 ∆G=-nFEcell (2.18)

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Example 2.3: Calculate the standard cell potential of the Daniell cell, in which reaction (2.2) occurs (scheme (2.6)), if the standard free energy of this reaction is equal to -212 kJ. Solution: ∆G0 -212000[J] E=-0 =- = 1.10[V] cell nF 2⋅ 96500[C]

One cannot measure the potential of an individual electrode (individual contriburion to the EMF). It is always necessary, to close the electric circuit, to immerse another terminal in the electrolyte, where immediately another reaction occurs. Hence, the relative scale was established. One can talk about the potential of any electrode only when it is measured vs. another one, named a of choice. The standard hydrogen electrode (SHE) (see Fig.3) has been chosen as the basic (primary) reference electrode or redox couple. It is not very practical in use, thus other reference electrodes has been developed and their potentials determined vs. the SHE.

Standard potentials of many half-reactions have been measured vs. SHE and their values tabulated (usually at 25°C, standard potentials, like thermodynamic functions to which they are related, are temperature dependent). Note, that when the EMF is measured, there is no net current, hence, each of half-reactions remains at equilibrium between the Ox form, Red form and the charge (but the overall reaction is not at equilibrium).

Let's consider the following cells: 2+ + Cu Cu (aq,a = 1) H (aq,a = 1) H2 (g,p = 1 Atm) Pt (2.19)

2+ + Zn Zn (aq,a = 1) H (aq,a = 1) H2 (g,p = 1 Atm) Pt (2.20) i.e., the standard copper electrode vs. SHE and standard zinc electrode vs. SHE. Their measured EMF's are -0.34 V and 0.76 V, respectively. It means, that the second cell diagram has been written correctly, i.e., zinc is actually oxidized (anode at left), or rather would be oxidized by the hydrogen ions if the cell was short circuited. This is apparently true, hydrogen gas evolves when pieces of zinc are added to an acidic solution + and zinc is getting dissolved. It means that H /H2 redox couple reveals a greater tendency to undergo reduction than Zn2+/Zn couple does. What about the first cell? The negative sign simply means that we have written the cell diagram in the wrong way (at least as a galvanic cell as the EMF of such a cell should always be positive). It means that copper oxidation by the hydrogen ions is not spontaneous under standard 2+ + conditions. Also, that Cu /Cu redox couple reveals greater tendency to undergo reduction than H /H2 redox couple and as such, the former should be placed at the right side of the diagram. By convention, all standard potentials of half-reactions are tabulated as reduction potentials, i.e., for cells with the SHE at the left side and the reaction in question at the right side (the way opposite to that applied in diagrams (2.19) and (2.20). Using this convention we would read: for half reaction: Zn2+ (aq,a = 1) + 2e - → Zn E0 = -0.76 V for half reaction: Cu2+ (aq,a = 1) + 2e - → Cu E0 = +0.34 V The higher the standard reduction potential, the more noble, more difficult to get oxidized is the + metal. Those metals with positive (above hydrogen or, more precisely H /H2 redox couple) standard reduction potential cannot be dissolved in acids with accompanying hydrogen evolution. The list of ordered standard reduction potentials constitute so called electrochemical series. On the basis of the series one can predict reaction spontaneity and calculate thermodynamic functions. Example 2.4: Calculate the standard cell potential of the Daniell cell (scheme (2.6)), in which reaction (2.2) occurs, from the standard potentials of the half-reactions. Solution: One can utilize equation (2.15): 0 0 0 Ecell = Ecathode - Eanode = 0.34 - (-0.76) = 1.10[V]

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2.2.2. . Dependence of electrode potential on the electrolyte composition.

Bearing in mind equation (2.18) and, from the preceeding chapter, equation (1.25), one can write: 0 RT⋅ ln(K) = nFEcell (2.21) RT or: E=0 ln(K) (2.22) cell nF Again, taking into account equation (2.17) and equation (1.24), one can: - rearrange equation (1.24) and combine it with (1.25) ∆∆G= G0 -RT⋅ ln(Q) (2.23) - compare right sides of the above equation (2.23) and equation (2.17) 0 ∆G - RT⋅ ln(Q) = nFEcell (2.24) - reinsert equation (1.25) into the above equation (2.24) to eliminate ∆G0 0 nFEcell - RT⋅ ln(Q) = nFE cell (2.25) - and, finally, rearrange the above equation (2.25) RT E=E-0 ln(Q) (2.26) cell cell nF For an overall redox equation of the type: t=const. bRed12 + aOx←→ bOx12 + aRed (2.27) resulted from balancing the two half reactions: - Red11→ Ox + ae (2.27a) - Ox2 + be→ Red2 (2.27b) where n in the overall reaction is a⋅b, equation (2.26) assumes the following form: RT [Ox ]b [Red ] a E=E-0 ln 1 2 (2.28) cell cell b a abF [Red1 ] [Ox2 ] After introducing the constant values R, F, and T (usually 298 K) and a conversion factor from the natural to common logarithm, one gets: 0.0592 [Ox ]b [Red ] a E=E-0 log 1 2 (2.29) cell cell b a ab [Red1 ] [Ox2 ] where the value of 0.0592 is expressed in Volts. The same formula is true for a half-reaction (written as reduction): 0.0592 [Red] E=E0 - log (2.30) n [Ox] Equations (2.21), (2.26), (2.28), (2.29), and (2.30) are variants of the Nernst equation.

Example 2.5: Calculate the equilibrium constant (solubility product) for reaction (2.10); scheme (2.13), if standard reduction potentials for reactions (2.11) and (2.12) are equal to +0.80 and +0.22 V at 298 K, respectively.

Solution:

Step A. The standard cell (2.13) potential, corresponding to reaction (2.10) is equal to: E0 = 0.22-0.80=-0.58[V] equation (2.15). Step B. We utilize equation (2.21)

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 nF   96500  K = exp E0  = exp (-0.58) = 1.55× 10-10  RT cell   8.314⋅ 298  which value is quite satisfactory. From the value of E0 = -0.58 [V] one can come to conclusion that under standard conditions the reaction is not spontaneous (the cell will not be a galvanic one at standard conditions).

We will use data from Appendix 2B in the textbook to solve problems involving standard half-cell potentials.

Example 2.6: (Exercise 17.41 from the textbook) To calculate the reaction quotient, according to equation (2.26) we need: - properly balanced overall redox reaction to get number of electrons exchanged, - a value of the standard cell potential, - temperature (we must assume 298K,as the standard potentials in the tables are given at this temperature). Solution: Part a) Step A. It's rather simple to write the overall reaction corresponding to the scheme given: Cathode (reduction): Pb4+ + 2e- → Pb2+ E0 = +1.67 [V] Anode (oxidation): Sn2+ → Sn4+ + 2e- E0 = +0.15 [V] (standard reduction potential) One can get the overall reaction by simple addition of the 2 half-reactions (hence n=2): Pb4+ + Sn2+ → Sn4+ + Pb2+ Step B. Using equation (2.15) one can get the value of the standard cell potential,: E0 = 1.67-0.80 = 0.87 [V] Step C. Final calculation of the Q value from equation (2.26):  nF   2⋅ 96500  Q = exp E0 - E = exp (0.87 - 1.33) = 2.74× 10-16  RT ()cell cell   8.314⋅ 298  A very small value of the reaction quotient leads us to conclude that the actual composition of the system favors the reactants to a significant degree. Part b) Step A. It's rather complicated to write the overall reaction corresponding to the diagram given. Let's have a look. There are no problems at the cathodic side: 2- + - 3+ 0 Cathode (reduction): Cr2O7 + 14 H + 6e → 2Cr + 7H2O E = +1.33 [V]

At the anodic side, from the scheme given, one cannot identify the redox couple easily. Only O2 and + H are mentioned. If the latter is an Ox, then H2 would be a corresponding Red (and it is not indicated). If the former is an Ox, then its Red counterpart is not given, either. Actually, the missing counterpart, obviously present in the system but missing in the scheme (it's presence is too obvious) is H2O: + - 0 Anode (oxidation): 2H2O → O2 + 4H + 4e E = +1.23 [V] (standard reduction potential) One can get the overall reaction by: i) multiplying the cathodic reaction by 2, ii) multiplying the anodic reaction by 3 (hence, after balancing, n=12), iii) simple addition of the multiplied half-reactions: 2- + - 3+ 2Cr2O7 + 28 H + 12e → 4Cr + 14H2O + - 6H2O → 3O2 + 12H + 12e 2- + 3+ 2Cr2O7 + 16H → 4Cr + 3O2 + 8H2O Step B. Using equation (2.15) one can get the value of the standard cell potential: E0 = 1.33-1.23 = 0.10 [V]

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Step C. We san notice that the given value of E is equal to that of E0. It simply means that the actual conditions are, most probably, standard conditions. Hence, the value of the reaction quotient, Q, will be equal to 1 (by definition at standard conditions). The same can be found by virtue of equation (2.26). Final calculation:  nF   12⋅ 96500  Q = exp E-E=exp0   (0.10 - 0.10) =1  RT ()cell cell   8.314⋅ 298  A very large value of the equilibrium constant leads us to conclude that at equilibrium composition of the system, the products are favored to a significant degree:  nF   12⋅ 96500  K = exp E0  = exp (0.10) = 2.0× 1020  RT cell   8.314⋅ 298  More comments on reaction scheme (remarks in part b), step A): + For any practical purposes, O2 and H do not constitute the redox couple, while O2 and H2O do (O2/H2O). In neither of the reactions in part b) does hydrogen change its oxidation number, while oxygen does (accepts electrons). This is a peculiar situation here, as water plays several roles in the overall reaction: i) it is the solvent, ii) it is a product of H+ reacting with oxide anions, like in the cathodic reaction (with no charge transfer), iii) it participates in the actual redox (involving charge transfer) reaction (in oxidation, it is a reactant). Example 2.7: (Exercise 17.35 from the textbook) To write the expression for the equilibrium constant, we must deduce, from the cell notation, a properly balanced overall redox reaction. Solution: Part a) Step A. The scheme suggests the following half-cells reactions: Cathode (reduction): AgCl(s) + e- → Ag(s) + Cl-(aq) + - Anode (oxidation): H2(g) → 2H (aq) + 2e One can get the overall reaction by: i) multiplying the cathodic reaction by 2, ii) addition of the two half-reactions: 2AgCl(s) + 2e- → 2Ag(s) + 2Cl-(aq) + - H2(g) → 2H (aq) + 2e + - 2AgCl(s) + H2(g) → 2Ag(s) + 2H (aq)+ 2Cl (aq) [H+2 ] [Cl -2 ] Step B. Writing the expression for the equilibrium constant: K= p H2 Part b) Step A. The scheme suggests the following half-cells reactions: - + - Cathode (reduction): NO3 (aq) +4H (aq) + 3e → NO(g) + 2H2O Anode (oxidation): Fe2+(aq) → Fe3+(aq) + e- One can get the overall reaction by: i) multiplying the anodic reaction by 3, ii) addition of the two half-reactions: - + - NO3 (aq) +4H (aq) + 3e → NO(g) + 2H2O 3Fe2+(aq) → 3Fe3+(aq) + 3e-

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- + 2+ 3+ NO3 (aq) +4H (aq) + 3Fe (aq) → NO(g) + 3Fe (aq) + 2H2O Step B. Writing the expression for the equilibrium constant: p[Fe]3+ 3 K= NO 2+ 3 + 4 - [Fe ] [H ] [NO3 ]

Example 2.8: (Exercise 17.56 from the textbook) Solution: One should notice that the charge in question is equal to the Faraday constant, F. To calculate the desired amounts of products, one must write the suitable half reaction: Part a) Metallic silver is produced in the cathodic reduction process: Ag+(aq) + e- → Ag(s) Hence, 1 mole of electrons (1 F) is required to produce 1 mole of silver metal. This is equivalent to 107.9 g. Part b) Gaseous chlorine is produced in the anodic oxidation process: - - 2Cl (aq) → Cl2(g) + 2e Hence, 2 moles of electrons (2 F) are required to produce 1 mole of chlorine gas. This is equivalent to 22.4 dm3 at STP. If 1F is passed, one will get half of this amount, i.e., 11.2 dm3 at STP. Part c) Metallic copper is produced in the cathodic reduction process: Cu2+(aq) + 2e- → Cu(s) Hence, 2 moles of electrons (2 F) are required to produce 1 mole of silver metal. If 1F is passed, one will get half of this amount, i.e., 63.54/2 g = 31.77 g. From the above problem, one can deduce the law of electrolysis, known as Faraday's law: M M m= Q= i(t)dt (2.31) nF nF ∫ where: m - mass of a product of electrolysis in grams, M - molar mass of a product of electrolysis in grams per mole, n - number of electrons required to produce 1 of a product, F - Faraday constant in Coulombs (see above), note that nF is the charge in coulombs needed to produce one mole of the product, Q - charge, in Coulombs, actually passed through the circuit (thus across the interface, causing the reaction to occur). This charge is an integral of current i (variable in time) over time. If i(t)=const. then Q=it.

Example 2.9: (Exercise 17.57 from the textbook) Answers: a) 7.87 seconds, b) 1.295 milligram.

Example 2.10: (Exercise 17.43 from the textbook) Solution: To solve the problem, usually one can utilize suitably modified equation (2.26), i.e., an appropria- te expression for the reaction quotient should be introduced: Part a) First, we must calculate potentials of both half-cells from equation (2.30): That of the right half-cell is equal to zero, as the right half cell is simply an SHE, and that of the left: 1/ 2 0.0592 [Red] pH E=E0 - log = -0.0592⋅⋅ log 2 = +0.0592 log(0.0010) = -0.1776[V] n [Ox] [H+ ] Second, we must calculate the cell EMF, according to equation (2.15):

Ecell = Ecathode - Eanode = 0 - (-0.1776) = 0.1776 [V]

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Final conclusion will be that while on the right side reduction would take place and at the left one - oxidation of hydrogen (concentration of acid would gradually rise in the left compartment). This is an example of a . Part b) First, we must calculate the standard cell potential from equation (2.15): 0 0 0 0 0 E cell = E cathode - E anode = E Ni(II)/Ni(0) - E Zn(II)/Zn(0) = -0.23 - (-0.76) = 0.53 [V] Second, we can find the unknown cell potential (EMF) from equation (2.26):

2+ 0 RT 0.0592 [Zn ] 0.10 E=E-cell cell ln(Q) = 0.53 - log = 0.53 - 0.0296⋅ log = 0.4708[V] nF 2 [Ni2+ ] 0.0010 Part c) We can solve it exactly the same way as part b). Standard cell potential: 0 0 0 0 0 E cell = E cathode - E anode = E H+/H2 - E Cl2/Cl- = 0 - (1.36) = -1.36 [V] The overall reaction is: - + 2Cl + 2H → Cl2 + H2 And the unknown cell potential: 100 450 × ppCl H 0.0592 2 2 760 760 E=-1.36-cell log = -1.36 - 0.0296× log = 2 [Cl-2 ] [H +2 ] 1.0 2× 0.01 2 450 ... = -1.36 - 0.0296× log = -1.36 - 0.0296× 2.892 = -1.446[V] 762-4× 10 It seems that the reaction cannot go the way the cell diagram is written, but rather in the opposite direction. Example 2.11: (Exercise 17.44 from the textbook) Answers: a) 0.482 Volt, b) 0.0046 Volt (The reaction, for all practical purposes is at equilibrium. Well, almost.), c) 0.6068 Volt.

Example 2.12: (Exercise 17.45 from the textbook) Solution: To solve the problem, usually one can utilize suitably modified equation (2.26), i.e., an appropria- te expression for the reaction quotient should be introduced: Our unknown is: X = [Fe2+]/[Fe3+]. Standard cell potential: 0 0 0 0 0 E cell = E cathode - E anode = E Hg2+2/Hg(0) - E Fe3+/Fe2+ = 0.79 - 0.77 = -0.02 [V] 2+ 2+ 3+ The overall reaction is: 2Fe + Hg2 → 2Fe + 2Hg n=2 And the actual cell potential can be expressed by modified equation (2.26):

0 2+ Ecell = E cell + 0.0592 logX as [Hg2 ]=1.0, then:  E-E0   cell cell   0.0592  X = 10 = 100.676 = 4.74 Example 2.13: (Exercise 17.47 from the textbook) Solution: We use the same method as in the previous problem, although with concentrations of individual ions as the unknowns: Answers: a) pH = 1.01, b) 0.10 mol/dm3.

© Wojciech Chrzanowski

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