8. ELECTROCHEMICAL CELLS

n Electrode Reactions and Electrode Potentials

8.1. a. H2 → 2H+ + 2e–

Cl2 + 2e– → 2Cl–

H2 + Cl2 → 2H+ + 2Cl–; z = 2

E = E° – RT2F ln ( )a

2H+a

2Cl–

u

b. 2Hg(l)+ 2Cl– → Hg2Cl2 + 2e–

2H+ + 2e– → H2

2Hg + 2H+ + 2Cl– → Hg2Cl2 + H2; z = 2

E = E° + RT2F ln ( )a

2H+a

2Cl–

u

c. Ag + Cl– → AgCl(s)+ e–

2e– + Hg2Cl2(s) → 2Hg + 2Cl–

2Ag(s) + Hg2Cl2(s) → 2AgCl(s) + 2Hg(s)

E = E° (no concentration dependence)

d.12 H2(g) → H+ + e–

AuI(s) + e– → Au(s) + I–

AuI(s) + 12 H2(g) → Au(s) + H+ + I–; z = 1

E = E° – RTF ln (a

H+aI–

)u

158 n CHAPTER 8

e. Ag(s) + Cl–(a1) → AgCl(s) + e–

AgCl(s) + e– → Ag(s) + Cl–(a2)

Cl–(a1) → Cl–(a2); z = 1

E = RTF ln

a1a2

8.2. E°

a. A + 2H+ + 2e– → AH2 –0.60 V

B + 2H+ + 2e– → BH2 –0.16 V

AH2 + B → A + BH2

AH2 is oxidized by B; the half-cell A + 2H+ + 2e– → AH2 is written as a reduction. In theoverall reaction this equation is reversed and therefore represents an oxidation. Because the netreaction has a positive emf (exergonic), the reaction is spontaneous in the forward direction.

b. –0.16 V – (–0.60 V) = 0.44 V

c. [H3O+] does not appear in the equilibrium expression, and the hydrogen-containing entitiescancel in the numerator and denominator. Therefore, there is no effect of pH on the equilibriumratio.

8.3. The ∆G° values for the two reactions are

Cr3+ + 3e– → Cr ∆G°1 = –zE

°1 F = –3 × –0.74 × 96 485 J mol–1 (1)

Cr3+ + e– → Cr2+ ∆G°2 = –zE

°2 F = –1 × –0.41 × 96 485 J mol–1 (2)

The reaction Cr2+ + 2e– → Cr is obtained by subtracting reaction (2) from reaction (1), and the ∆G°

value for Cr2+ + 2e– → Cr is obtained by subtracting ∆G°2 from ∆G

°1 :

∆G° = –3 × –0.74 × 96 500 – (–1 × –0.41 × 96 485) J mol–1

= 1.81 × 96 485 J mol–1

Since Cr2+ + 2e– → Cr involves two electrons, and since ∆G = –zE°F, it follows that

1.81 F = –2(E°/V)F

or E° = –0.90 V

8.4. Left-hand electrode H2 → 2H+(1 m) + 2e–

Right-hand electrode 2e– + 2H+(aq) + F2– → S2–

Overall reaction 2H+(aq) + F2– + H2 → 2H+(1 m) + S2–

The expression for the emf of the cell is

E = E° – RT2F ln

[S2–] [1 m]2

[F2–] c2

ELECTROCHEMICAL CELLS n 159

8.5. a. In writing the representation of the cell, the oxidation reaction occurs at the anode, which isplaced at the left-hand position of the cell. In this case Fe2+ is losing electrons, and theoxidation process is

Fe2+ → Fe3+ + e–

The cathode reaction is written on the right-hand side of the cell and is

Ce4+ + e– → Ce3+

where reduction occurs. The overall reaction is the sum of these two reactions. The cellrepresentation is

Fe3+(aq) Fe2+(aq) MMCe4+(aq) Ce3+(aq)

The symbols for both sides are always written as though they were a reduction process, i.e.,oxidized form reduced form. The voltage of this cell is then the reduction potential of theright-hand electrode minus the reduction potential of the left-hand electrode. Thus

E° = E°Ce4+ Ce3+ – E°

Fe3+ Fe2+ = 1.44 V – 0.771 V = 0.67 V

b. Upon examining the standard reduction potentials in Table 8.1, we see that the following half-cell reactions can be combined to give the cited reaction.

Ag+(aq) + e– → Ag 0.7991 V (1)

AgCl(s) + e– → Ag + Cl– 0.2224 V (2)

Reversal of the second equation and addition gives the desired equation:

Ag+(aq) + Cl–(aq) → AgCl(s)

Equation (1) is the reduction reaction and is placed on the right-hand side. The anode reactionis placed on the left-hand side of the cell representation.

Ag AgCl(s) Cl–(aq) MM Ag+(aq) Ag(s)

The voltage of this cell is the right-hand reduction potential minus the left-hand reductionpotential.

E° = E°Ag+ Ag

– E°AgCl Ag = 0.7991 V – 0.2224 V = 0.5777 V

c. HgO undergoes reduction to Hg and is the cathode. H2 is oxidized and is the anode. Theelectrode potentials are found in the CRC Handbook.

2H2O + 2e– → H2 + 2OH– –0.8277 V (3)

HgO + H2O + 2e– → Hg + 2OH– 0.0977 V (4)

Reversing the sense of equation (3) and adding to (4) gives

HgO(s) + H2(g) → Hg(l) + H2O(l)

The cell is represented by

Pt, H2O(l) H2(g), OH–(aq) MM OH–(aq), H2O HgO(s) Hg(l)

The cell potential is

E° = E°HgO Hg – E°H2O H2 = 0.0977 V – (–0.8277 V) = 0.9254

160 n CHAPTER 8

n Thermodynamics of Electrochemical Cells

8.6. Fe3+/Fe2+ = 0.771 V; I2/I– = 0.5355 V

E° = 0.771 – 0.5355

= 0.2355 V

ln Kuc =

2 × 0.23550.0257

= 18.3268

Kc = 9.10 × 107 dm6 mol–2

8.7. The E° for the process

Sn + Fe2+ →← Sn2+ + Fe

is –0.4402 – (–0.136) = –0.304 V

Since z = 2 we have

–0.304 = 0.0257

2 ln Kuc

ln K = –23.658

Kc = 5.312 × 10–11

8.8. ∆G° = –z × 96 485 × 0.25 J; z = 1

= –24 100 J mol–1 = –24.1 kJ mol–1

8.9. From Table 8.1 the relevant values are

O2 + 4H+ + 4e– → 2H2O E° = 1.23 V (1)

2H+ + 2e– → H2 E° = 0 (2)

Subtraction of (2) from 12(1) gives the required equation

H2 + 12 O2 →← H2O

with E° = 1.23 V and z = 2. Then,

∆G° = –2 × 96 485 × 1.23 J mol–1

= –237 400 J mol–1 = –237.4 kJ mol–1

8.10. From Table 8.1,

Cu2+ + 2e– → Cu E°1 = 0.337 V (1)

Cu2+ + e– → Cu+ E°2 = 0.153 V (2)

ELECTROCHEMICAL CELLS n 161

If we subtract 2 × (2) from (1) we obtain

2Cu+ → Cu2+ + Cu E° = E°1 – E

°2 = 0.184 V; z = 2

∴ ln Ku = 2 × 0.184 F

(8.3145)(298.15) = 2 × 0.1840.0257 = 14.32

K = 1.66 × 106 dm3 mol–1

If Cu2O is dissolved in dilute H2SO4, half will form Cu2+ and half Cu.

8.11. Note that the ∆G° given is for the reaction of 3 moles of H2 to form 2 moles of Sb. The electrodereactions may be written as

3H2 → 6H+ + 6e–,

Sb2O3 + 6H+ + 6e– → 2Sb + 3H2O.

Using Eq. (8.2), we get E° = 83 700 J/(6 × 96 485) = 0.1446 V.

Since this reaction is spontaneous, the electron flow is from the hydrogen electrode (negative) tothe antimony electrode (positive).

n Nernst Equation and Nernst Potentials

8.12. E = RTzF ln

m1m2

= 0.0257 ln 2 = 0.0178 V

8.13. The process is

Pyruvate– + 2H+ + 2e– →← lactate–

and the Nernst equation is

E′ = E° – RT2F ln

[lactate–][pyruvate–]

Then

E′/V = –0.185 – 0.0257

2 ln

10

90

E′ = –0.157 V

8.14. a. From Table 8.1,

Fe3+ + e– → Fe2+ E°1 = 0.771 V (1)

Fe2+ + 2e– → Fe E°2 = –0.4402 V (2)

The corresponding ∆G° values are:

162 n CHAPTER 8

∆G°1 = –0.771 × 96 485 = –74 389.9 J mol–1 (1)

∆G°2 = –2(–0.4402) × 96 485 = 84 945.4 J mol–1 (2)

The required reaction is obtained by adding (1) and (2):

Fe3+ + 3e– → Fe ∆G° = 10.56 kJ mol–1

Then, since z = 3,

E° = – 10 560

3 × 96 485 = –0.0365 V

b. The electrode reactions are

Sn2+ → Sn4+ + 2e– E°1 = –0.15 V (1)

3e– + Fe3+ → Fe E°1 = –0.0365 V (2)

The cell reaction is

3Sn2+ + 2Fe3+ → 3Sn4+ + 2Fe E° = –0.186 V; z = 6

From the Nernst equation, Eq. 8.13,

E = –0.186 – 0.0257

6 ln (0.01)3

(0.1)3 (0.5)2

= –0.186 + 0.024 = –0.16 V

8.15. From Ka = [H+][CH3COO–]

[CH3COOH] , we obtain

[H+] = Ka [CH3COOH]

[CH3COO–] .

The concentration of acetate ions formed from the dissociation of acetic acid will be negligiblecompared to the acetate concentration from the fully dissociated sodium salt. Therefore,

[H+] = 1.81 × 10–5 × 0.01000.0358 = 5.056 × 10–6 M.

The cell reactions are (see Table 8.1):

H2(g) → 2H+ + 2e–; E° = 0.0000 V (by definition)

Hg2Cl2 + 2e– → 2Hg + 2Cl–; E° = 0.2415 V

However, since the cathode reaction (reduction) is accounted for by the standard reductionpotential of the electrode, the Nernst equation contains only the hydrogen ion concentration. Forthis concentration, the Nernst equation gives

E = (0.0000 + 0.2415) – 0.0257

2 ln [5.056 × 10–6]2 = 0.5549 V.

8.16. The reactions taking place during the electrolysis are

ELECTROCHEMICAL CELLS n 163

Cu2+ + 2e → Cu; E° = 0.3419 V

2Br → Br2 + 2e ; E° = –1.0873 V

If the reverse reaction were taking place in a galvanic cell, the initial cell voltage is given by

E = E° – 0.0257

2 ln [Cu2+][Br ]2.

E = (1.0873 – 0.3419) – 0.0257

2 ln (0.0500 × 0.10002)

= 0.843 V.

Therefore, a minimum voltage of 0.843 V will have to be applied at the beginning in order for theelectrolysis reaction to occur.

At the end of the electrolysis, the concentrations are:

[Cu2+] = 0.0500 M – (2.872/63.456)

1.0000 = 0.00474 M.

[Br ] = 0.1000 M – 2 × (2.872/63.456)

1.0000 = 0.0094 M.

Therefore, the voltage required will be

(1.0873 – 0.3419) – 0.0257

2 ln (0.0047 × 0.00942) = 0.934 V.

8.17. For the reaction

I2 + I– → I–3 E° = –0.0010 V and z = 2

Kc = [I

–3]

[I –] =

[I–3]

0.5 mol dm–3

–0.0010 = 0.0257

2 ln [I

–3]

0.5 mol dm–3

ln[I

–3]u

0.5 = –0.0778

[I–3]u

0.5 = 0.925

[I–3 ] = 0.463 M

8.18. ∆Φ/V = 8.3145 × 298.15

96 500 ln 0.180.12

∆ Φ = 0.0104 V = 10.4 mV

8.19. From Table 8.1, we see that Au+ has a much higher reduction potential than Pb2+. Therefore, thegold will be deposited first.

As the Au+ concentration falls, the lead begins to be deposited, i.e., we have

164 n CHAPTER 8

2Au(s) + Pb2+ →← Pb(s) + Au+

for which we write (see Equation 8.7)

E° = (–1.692 – 0.126) V – 0.0257

2 ln [Au+]2

0.0100

or [Au+]2 = 0.0100e(–1.818 × 2/0.0257) = 3.603 × 10–64.

Therefore, [Au+] = 1.898 × 10–32 M.

The conclusion is that only an infinitesimal amount of gold will be left in the solution by the timethe lead starts to deposit at the electrode. Therefore, this is an acceptable way of separating the twometals.

8.20. Reaction at the right-hand electrode:

e– + H+(0.2 m) → 12 H2(10 bar)

At the left-hand electrode:

12 H2(1 bar) → e– + H+(0.1 m)

Overall reaction:

H+(0.2 m) + 12 H2(1 bar) →

12 H2(10 bar) + H+(0.1 m); z = 1

Cell emf:

E = RTF ln

0.2 × (1 bar)1/2

0.1 × (10 bar)1/2

E/V = 0.0257 ln 210

= –0.0118 V

E = –11.8 mV

8.21. The reactions at the two electrodes are

12 H2(1 bar) → H+(0.1 m) + e–

H+(0.2 m) + e– → 12 H2(10 bar)

Every H+ ion produced in the left-hand solution will have to pass through the membrane to preserveelectrical neutrality.

H+(0.1 m) → H+(0.2 m)

The net reaction is 12 H2(1 bar) →

12 H2(10 bar)

E = RTF ln

110

= 0.0257 ln 110

= –0.0296 V = –29.6 mV

ELECTROCHEMICAL CELLS n 165

8.22. The capacitance of the cell membrane (see Eq. 8.20) is

C = 8.854 × 10–12 × 3 × 10–10

10–8 F = 2.66 × 10–13 F

a. With a potential difference of 0.085 V, the net charge on either side of the wall is

Q = CV = 2.66 × 10–13 × 0.085 = 2.26 × 10–14 C

b. The number of K+ ions required to produce this charge is

Qe =

2.26 × 10–14

1.602 × 10–19 = 1.41 × 105

The number of ions inside the cell is

0.155 × 10–12 × 6.022 × 1023 = 9.33 × 1010

The fraction of ions at the surface is therefore

1.41 × 105

9.33 × 1010 = 1.5 × 10–6

8.23. At the left-hand electrode:

12 H2 → H+ + e–

At the right-hand electrode:

CrSO4(s) + 2e– → SO24 – + Cr(s)

Overall reaction:

CrSO4(s) + H2 → 2H+ + SO24 – + Cr(s)

E° = –0.40 V and z = 2

a. E = E° – RT2F ln ([H+]2[SO

24 –])u

E/V = –0.40 – 0.0257

2 ln (0.002)2(0.001)

E = –0.152 V

b. The ionic strength is

I = 12{ }0.002 + (0.001 × 4) = 0.003 M

Historically, common logs have been used in activity coefficient calculations where the valueof B is easily remembered as 0.51.

log10γ± = –2 × 0.51 0.003

= –0.0559

γ± = 0.879

166 n CHAPTER 8

E = E° – RT2F { ln ([H+]2[SO

24–])u + ln γ2

+γ– }

= E° – RT2F { ln ([H+]2[SO

24–])u + ln γ3

± }

E/V = –0.152 – 0.0257

2 ln (0.879)3

E = –0.152 + 0.050 = –0.147

8.24. At the left-hand electrode:

Cu(s) → Cu2+ + 2e–

At the right-hand electrode:

AgCl(s) + e– → Ag(s) + Cl–

Overall reaction:

2AgCl(s) + Cu(s) → 2Ag(s) + Cu2+ + 2Cl–, z = 2

To a good approximation, it can be assumed that the activity coefficients are unity at 10–4 M

(DHLL gives γ± = 0.988). Then

E = E° – RTzF ln ([Cu2+][Cl–]2)u

0.191 = E°/V – 0.0257

2 ln [ ]10–4 (2 × 10–4)2 u

= E°/V + 0.337

E° = –0.146 V

Suppose that at 0.20 M the activity coefficients are γ+ and γ–. Then

–0.074 = –0.146 – 0.0257

2 ln [0.20(0.40)2] – 0.0257

2 ln γ+γ2–

–0.074 = –0.146 + 0.044 – 0.01285 ln γ3±

ln γ3± = –2.179

γ3± = 0.113

γ± = 0.48

8.25. a. The anode reaction:

2Tl(s) + 2Cl–(0.02 m) → 2TlCl(s) + 2e–

The cathode reaction:

Cd2+(0.01 m) + 2e– → Cd(s)

The overall cell reaction:

ELECTROCHEMICAL CELLS n 167

2Tl(s) + Cd2+(0.01 m) + 2Cl–(0.02 m) → 2TlCl(s) + Cd(s); z = 2

b. This reaction can alternatively be written as

(2) 2Tl(s) + Cd2+(0.01 m) → 2Tl+(in 0.01 m CdCl2) + Cd(s); z = 2

for which the Nernst equation is

E = E°Cd2+ Cd

– E°Tl+ Tl

– 0.0257 V

2 ln ([Tl+]2/[Cd2+])u

E° = –0.40 – (–0.34) = –0.06 V

When aCdCl2

= 1, E1 is the value of E, and substituting [Tl+] = Ksp/[Cl–] we have

E°1 /V = –0.06 – 0.01285 ln (K

2sp /[Cd2+][Cl–]2)u

= –0.06 – 0.01285 ln (1.6 × 10–3)2

E°1 = –0.06 – (–0.165) = 0.105 V

When m = 0.01 m,

E/V = –0.06 – 0.01285 ln (1.6 × 10–3)2/(0.01)(.02)2

E = –0.06 – (–0.0057) = –0.054 V

8.26. The diffusible K+ ions are at a higher potential on the right-hand side of the membrane; there is thusa tendency for a few of them to cross to the left-hand side and create a positive potential there. (Thesame conclusion is reached by considering the diffusible Cl– ions; they are at a higher concentrationon the left-hand side, and a few tend to cross to the right-hand side and create a negative potentialthere.)

The Nernst potential is given by Eq. 8.19:

Φ = RTzF ln

c1c2

= 8.3145 × 310.15

96 485 ln 0.160.04

= 0.0370 V

= 37 mV

8.27. The overall reaction is

Lactate– + 2 cytochrome c (Fe3+) → pyruvate– + 2 cytochrome c (Fe2+) + 2H+

with z = 2 and E°′ = 0.254 + 0.185 = 0.439 V. If K′ is the ratio at pH 7,

E°′ = 0.439 = RT2F ln K′ =

0.02572 ln K′

ln K′ = 34.16

K′ = 6.87 × 1014

If K′′ is the ratio at pH 6,

168 n CHAPTER 8

Ktrue = K′ × (10–7)2 = K′′ × (10–6)2

so that

K′′ = K′ × 10–2 = 6.87 × 1012

8.28. The processes are

AgCl(s) + e– → Ag(s) + Cl–(0.1 m)

Ag(s) + Cl–(0.01 m) → AgCl(s) + e–

The electrical neutrality is maintained by the passage of H+ ions from right to left:

H+(0.01 m) → H+(0.10 m)

The overall process is

H+(0.01 m) + Cl–(0.01 m) → H+(0.10 m) + Cl–(0.10 m)

The emf is

E = – RTF ln

0.1 × 0.10.01 × 0.01

E/V = –0.0257 ln 100

E = –0.118 V

8.29. a. The processes at the electrodes are

12 H2 → H+(m1) + e–

H+(m2) + e– → 12 H2

To maintain electrical neutrality of the solutions, for every mole of H+ produced in the left-hand solution, t+ mol of H+ ions will cross the membrane from left to right, and t– mol of Cl–

ions will pass from right to left. In the left-hand solution there is therefore a net gain of

(l – t+) mol = t– mol of H+

and of t– mol of Cl–.

In the right-hand solution, the net loss is

(l – t+) mol = t– mol of H+

and t– mol of Cl–.

The overall process is thus

t–H+(m2) + t–Cl–(m2) → t–H+(m1) + t–Cl–(m1)

The emf is

ELECTROCHEMICAL CELLS n 169

E = – RTF ln

mt–1 m

t–1

mt–2 m

t–2

= 2t–RT

F ln m2m1

b. For m1 = 0.01 m and m2 = 0.10 m,

0.0190 = 2t– × 0.0257 ln 10

t– = 0.161

t+ = 0.839

8.30. M → M+ (0.1 m) + e–

Cell reaction: H+(0.2 m) + e– →

12 H2(1 bar)

H+(0.2 m) + M → M+ (0.1 m) + 12H2(1 bar)

E° = E°H+ H2

– E°M+ M

The Nernst equation is

–0.4 = E°M+ M

/V – 0.0257 ln (0.1)(0.2) = –E°

M+ M/V + 0.018

E°M+ M

= 0.418 V

Upon addition of KCl, almost all of the M+ precipitates, and 0.10 m Cl– is in excess. The value ofM+ in solution is found from the Ksp, namely

Ksp = [M+][Cl–] = (M+) (0.10 M)

and from the Nernst equation

–0.1 = –E°M+ M

– 0.0257 ln Ksp/(0.10)

0.2

–0.1 = –0.418 – 0.0257 ln Kusp + 0.0257 ln 0.02

–0.1 + 0.418 + 0.100 = –0.0257 ln Kusp

ln Kusp = –

0.4180.0257 = –7.08

Ksp = 8.2 × 10–8 mol2 kg–2

8.31. The E° values for the half reactions (Table 8.2) are

pyruvate– + 2H+ + 2e– → lactate– E°′ = –0.19 V (1)

NAD+ + H+ + 2e– → NADH E°′ = –0.34 V (2)

170 n CHAPTER 8

The required reaction is obtained by subtracting (2) from (1). Hence

E°′ = –0.19 – (–0.34) = 0.15 V

Since n = 2, the corresponding ∆G°′ is

∆G°′ = –2 × 96 485 × 0.15 J

= –28 946 J

a. The equilibrium ratio at pH 7 can be calculated from the relationship

∆G°′ = –RT ln K′

= –8.3145 × 298.15 ln K′

ln K′ = 28 946/(8.3145 × 298.15)

= 11.68

K′ = 1.2 × 105 dm3 mol–1

Alternatively, K′ could have been calculated directly from E°′, using Eq. (8.7):

0.15 = 0.02569

2 ln K′

ln K′ = 11.68

K′ = 1.2 × 105 dm3 mol–1

This K′ at pH 7.0 is related to the true (pH-independent) K by the equation

K = [lactate–][NAD +]

[pyruvate–][NADH][H +] =

K′[H+]

= K′

10–7

b. Similarly the K′′ at pH 8.0 is related to K by

K = K′′

10–8

Thus

K′′10–8 =

K′10–7

and

K′′ = 1.2 × 105 × 10–8

10–7

= 1.2 × 104

ELECTROCHEMICAL CELLS n 171

n Temperature Dependence of Cell emfs

8.32. a. We are given that

fumarate2– + 2H+ + 2e– → succinate2– E°′ = 0.031 V (1)

pyruvate– + 2H+ + 2e– → lactate– E°′ = –0.185 V (2)

Subtraction of (2) from (1) gives

fumarate2– + lactate– → succinate2– + pyruvate– E°′ = 0.216 V; z = 2

(Note that this is also E°, the hydrogen ions having canceled out.)

The Gibbs energy change is

∆G° = –zFE° = –2 × 96 485 × 0.216 = –41.7 kJ mol–1

The entropy change is obtained by use of Eq. 8.23:

∆S° = 2 × 96 500 × 2.18 × 10–5 J K–1 mol–1

= 4.21 J K–1 mol–1

b. The enthalpy change can be calculated by use of Eq. 8.25, or more easily from the ∆G° and∆S° values:

∆H° = ∆G° + T∆S°

= –41 680 + (298.15 × 4.207) J mol–1

= –40 430 J mol–1 = –40.4 kJ mol–1

8.33. a. The individual electrode processes are

Cd(Hg) → Cd2+ + 2e–

Hg2+2 + 2e– → 2Hg

and the overall reaction is

Cd(Hg) + Hg2+2 → Cd2+ + 2Hg; z = 2

Since the solution is saturated with Hg2SO4•83H2O the overall reaction can be written as

Cd(Hg) + Hg2SO4(s) + 83 H2O(l) → CdSO4•83H2O(s) + 2Hg(l)

b. ∆G° = –2 × 96 487 × 1.01832 = –196 510 J mol–1

= –196.5 kJ mol–1

∆S° = 2 × 96 487 × (–5.00 × 10–5) = –9.65 J K–1 mol–1

∆H° = –196 510 – (9.65 × 298.15) = –199 390 J mol–1

= –199.39 kJ mol–1

172 n CHAPTER 8

8.34. Since we are only interested in the slope (i.e., not the intercept) of the straight line that best fits theE vs. t data, we need not bother to convert the temperature to Kelvin.

We perform a linear regression analysis using t as the independent variable and E as the dependentvariable. The result is:

E = 0.93046 – 2.8834 × 10–4 t.

Differentiation with respect to t gives

The entropy change for the conversion of one mole of (½)Br2 to Br (z = 1) is

96 485 C mol–1 × 2.8834 × 10–4 V K–1 = –27.820 J K–1 mol–1.

8.35. The cell reaction is Mg(s) + Cl2(g) → Mg2+(aq) + 2Cl(aq); i.e., z = 2. From Eq. (8.23), we get

8.36. E°′ = E° = 0.031 + 0.197 = 0.228 V

∆G° = –2 × 96 485 × 0.228 = 44 000 J mol–1

a. If x kJ mol–1 = ∆G°f (fumarate) ,

–690.44 – 139.08 – x + 181.75 = –44.00

x = –603.8

∆G°f = –603.8 kJ mol–1

b. ∆S° = 2 × 96 485 × (–1.45 × 10–4)

= –28.0 J K–1 mol–1

∆H° = ∆G° + T∆S° = –44 000 – (298.15 × 28.0)

= –52 350 J mol–1

If y kJ mol–1 = ∆H°f (fumarate) ,

–908.68 – 210.66 – y + 287.02 = 52.35

y = –780.0

∆H°f = –780.0 kJ mol–1

8.37. a. The cell reaction can be written

Tl + H+(a = 1) → Tl+(in HBr; a = 1) + 12 H2(1 bar)

E = E°H +H2

– E°Tl+T1 – 0.0257 ln ([T1+]/[H+])u

1414 K V 108834.2 C)( V 108834.2 −−−− ×−=

∂∂=°×−=

∂∂

PP T

E

t

E

. KV 107477.1mol C 485 962

mol KJ 3.337 131

11−−

−

−−×−=

×−=∆=

∂∂

zF

S

T

E

P

ELECTROCHEMICAL CELLS n 173

Since Tl+ = Ksp/[Br–]

E/V = 0.34 – 0.0257 ln (Ksp/[H+][Br–])u

= 0.34 – 0.0257 ln Kusp = 0.34 + 0.236

E = 0.58 V

b. ∆H for the reaction is ∆H° for the half cell

Tl → Tl+ + e–

∆H° = –zF(E – TdE/dT)

= –96 485 [(0.34 – 298.15(–0.003)]

= –96 485 (1.234) J mol–1

= –119 062 rounded to –119 kJ mol–1

n Applications of emf Measurements

8.38. Subtraction of the second reaction from the first gives

AgBr(s) → Ag+ + Br– E° = –0.7278 V; z = 1

Then

–0.7278 = 0.0257 ln ([Ag+][Br–])u

ln ([Ag+][Br–])u = –28.319

Ksp = [Ag+][Br–] = 5.03 × 10–13 mol2 kg–2

Solubility = Ksp = 7.09 × 10–7 mol kg–1

8.39. The standard emf of the AgClAg electrode is 0.2224 V and the cell reaction is

AgCl(s) + 12 H2 → H+ + Cl– + Ag(s); z = 1

E = E° – RTF ln (aH+)u

≈ E° – 2RTF ln (aH+)u

0.517 = 0.2224 – 2 × 0.059 16 log10 (aH+)u

log10(aH+)u = 0.2224 – 0.5172 × 0.059 16

= –2.49

pH = 2.49

8.40. The cell is represented as Ag AgI(s) I–(aq) MM Ag+(aq) Ag(s).

174 n CHAPTER 8

At the two electrodes:

e– + AgI(s) → Ag(s) + I–

Ag(s) → Ag+ + e–

Overall reaction:

AgI(s) → Ag+ + I–; z = 1

E = RTF ln[mAg+ × mCl–]

u = RTF ln Ksp

∴ –0.9509 = 0.0257 ln Kusp

Ksp = 8.53 × 10–17 mol2 kg–2

Solubility = Ksp = 9.24 × 10–9 mol kg–1

8.41. The standard cell potential for the cell shown is E° = (0.2224 – 0.0254) = 0.1970 V. Since themolality is exactly 1.0, Eq. (8.43) simplifies to 0.2053 = 0.1970 – 2 × 0.0257 ln γ± . Solving forthe mean activity coefficient, we calculate

γ± = exp[–(0.2053 – 0.1970)/(2 × 0.0257)] = 0.8509.

8.42. a. The electrical work is –∆G.

∆G° = ∆H° – T∆S°

∆G°/J mol–1 = –2 877 000 – 298.15 × (–432.7)

= –2 784 000 J mol–1 = –2750 kJ mol–1

Electrical work available = 2750 kJ mol–1

b. The total work done is –∆A.

∆G° = ∆A° – ΣνRT

Σν = 4 – 1 – 132 = –3.5

ΣνRT = –3.5 × 298.15 × 8.3145 = –8676 J mol–1

∆A° = –2748 + 8.68 = –2739 kJ mol–1

Total available work = 2740 kJ mol–1

8.43. a. The overall process is

Cd + 2AgCl → 2Ag + CdCl2

and is a two-electron process.

∆G = ∆G° + RT2F ln a

Cd2+a2

Cl–

where a is the activity of the respective ions.

ELECTROCHEMICAL CELLS n 175

E = E° – RT2F ln a

Cd2+a2

Cl–

0.7585 = 0.5732 – RT2F ln (m

Cd2+γCd2+m

2

Cl– γ

2

Cl– )

0.1853 = – RT2F ln[(0.01γ

Cd2+)(0.02)2γ2

Cl– )]

= – 8.3145(298.15)

2(96 485) ln (4 × 10–6)γ3±

ln γ3± = –1.161

γ± = 0.679

b. The ionic strength of a 0.010 m solution of CdCl2 is

I = 12 Σmizi

2 = 12 [(0.010 × 22) + (2 × 0.010 × 12)]

= 0.030 mol kg–1

From the Debye-Hückel limiting law,

log10γ± = –Az+z– I

= –0.51 × 2 × 1 × 0.030

= –0.177

γ± = 0.6653 = 0.67 (to the degree of accuracy allowed by the problem)

The two values are in good agreement.

8.44. The LiOH is required for the hydrogen electrode and the LiCl salt is used to complete the AgClelectrode. Both the Cl– ion and the H+ ion will behave according to their activities in solution.Begin by determining the emf of the cell.

Ecell = EAgCl – EH2 = E°

AgCl – RTF ln a

Cl– –

RTF ln a

H+

This can be rewritten using Kw = aH+a

OH–, and so the desired relationship is established.

Ecell = E°AgCl –

RTF ln a

Cl– –

RTF ln Kw +

RTF ln a

OH–

Combining the two activity terms gives

Ecell = E°AgCl –

RTF ln

aCl–

aOH–

– RTF ln Kw

176 n CHAPTER 8

Rewriting this in terms of activity coefficients and molalities gives, after rearrangement,

(Ecell – E°AgCl )F

RT + ln (mCl–

/mOH–) = –ln Kw – ln

γCl–

γOH–

The value of mOH– is 0.01 mol kg–1. With that substitution, plot the left-hand side of the equation

with E°AgCl = 0.2224 V at 298.15 K against the ionic strength, which varies with the concentration,

and extrapolate to zero ionic strength. At zero ionic strength the γCl–

and γOH– approach unity. Then

the value of the curve is –ln Kw.

In the following data, I is based on m + 0.01 m OH–. The latter is constant. For example,I = 1/2[0.01 × 12 + 0.01 × (–1)2 + 0.01 × 12 + 0.01 × (–1)2].

m/mol kg–1 0.01 0.02 0.05 0.10 0.20

I/mol kg–1 0.02 0.03 0.06 0.11 0.21

(Ecell –E°AgCl )F

RT 32.2086 31.5079 30.566 29.834 29.087

ln m

0.01 0.000 0.693 1.609 2.303 2.996

Sum of last 2 terms 32.209 32.201 32.175 32.137 32.083

From the indicated plot shown, the value of –ln Kw = 32.226. The value of Kw is

1.010 × 10–14.

ELECTROCHEMICAL CELLS n 177

32.24

I/mol kg–1

.22 Plot of

against I

lnRT

(Ecell – E°AgCl)F

.20

.18

.16

.14

.12

.10

32.080 0.10 0.20 0.30

+ ( (0.01 mol kg–1m

lnR

T

(Ece

ll – E

° AgC

l)F+

((

0

.01

mol

kg–1

m