Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap10

10

Equilibrium electrochemistry

Solutions to exercises
Discussion questions
E10.1(b)

The Debye–H¨uckel theory is a theory of the activity coefficients of ions in solution. It is the coulombic
(electrostatic) interaction of the ions in solution with each other and also the interaction of the ions
with the solvent that is responsible for the deviation of their activity coefficients from the ideal value
of 1. The electrostatic ion–ion interaction is the stronger of the two and is fundamentally responsible
for the deviation. Because of this interaction there is a build up of charge of opposite sign around any
given ion in the overall electrically neutral solution. The energy, and hence, the chemical potential
of any given ion is lowered as a result of the existence of this ionic atmosphere. The lowering of the
chemical potential below its ideal value is identified with a non-zero value of RT ln γ± . This non-zero
value implies that γ± will have a value different from unity which is its ideal value. The role of the
solvent is more indirect. The solvent determines the dielectric constant, , of the solution. Looking
at the details of the theory as outlined in Justification 10.2 we see that enters into a number of the
basic equations, in particular, Coulomb’s law, Poisson’s equation, and the equation for the Debye
length. The larger the dielectric constant, the smaller (in magnitude) is ln γ± .

E10.2(b)

The potential difference between the electrodes in a working electrochemical cell is called the cell
potential. The cell potential is not a constant and changes with time as the cell reaction proceeds.
Thus the cell potential is a potential difference measured under non-equilibrium conditions as electric
current is drawn from the cell. Electromotive force is the zero-current cell potential and corresponds
to the potential difference of the cell when the cell (not the cell reaction) is at equilibrium.

E10.3(b)

The pH of an aqueous solution can in principle be measured with any electrode having an emf that is
sensitive to H+ (aq) concentration (activity). In principle, the hydrogen gas electrode is the simplest
and most fundamental. A cell is constructed with the hydrogen electrode being the right-hand electrode
and any reference electrode with known potential as the left-hand electrode. A common choice is
the saturated calomel electrode. The pH can then be obtained from eqn 10.43 by measuring the emf
(zero-current potential difference), E, of the cell. The hydrogen gas electrode is not convenient to
use, so in practice glass electrodes are used because of ease of handling.

Numerical exercises
E10.4(b)

NaCl(aq) + AgNO3 (aq) → AgCl(s) + NaNO3 (aq)
NaCl, AgNO3 and NaNO3 are strong electrolytes; therefore the net ionic equation is
Ag+ (aq) + Cl− (aq) → AgCl(s)
rH

−−

=

fH

−−

(AgCl, s) −

fH

−−

(Ag+ , aq) −

fH

−−

(Cl− , aq)

= (−127.07 kJ mol−1 ) − (105.58 kJ mol−1 ) − (−167.16 kJ mol−1 )
= −65.49 kJ mol−1
E10.5(b)

Pb2+ (aq) + S2− (aq)

PbS(s)

aJνJ

KS =
J

Since the solubility is expected to be low, we may (initially) ignore activity coefficients. Hence
KS =

b(Pb2+ ) b(S2− )
×
b −−
b −−

b(Pb2+ ) = b(S2− ) = S


INSTRUCTOR’S MANUAL

148

KS =

S2
(b −− )2

S = (KS )1/2 b −−
− r G−−
to obtain KS
RT

Use ln KS =
rG

−−

=

fG

−−

(S2− , aq) +

= (+85.8 kJ mol

−1

fG

−−

(Pb2+ , aq) −

) + (−24.43 kJ mol

−1

rG

−−

(PbS, s)

) − (−98.7 kJ mol−1 )

= 160.07 kJ mol−1
ln KS =

−160.07 × 103 J mol−1
(8.314 J K−1 mol−1 ) × (298 K)

= −64.61

KS = e−64.61 = 8.7 × 10−29
KS =
E10.6(b)

S2
b −−

S = (KS )1/2 b −− = (8.735 × 10−29 )1/2 = 9.3 × 10−15 mol kg−1

2

The ratio of hydration Gibbs energies is
−− (NO− )
3
G−− (Cl− )

hyd G
hyd

We have
So
E10.7(b)

hyd G

I = 21

hyd G
−−

i

−−

=

r(Cl− )

r(NO−
3)

=

181 pm
= 0.958
189 pm

(Cl− ) = −379 kJ mol−1 [Exercise 10.6a]

−1
−1
(NO−
3 ) = (0.958) × (−379 kJ mol ) = −363 kJ mol

(bi /b −− )zi2 [10.18]

and for an Mp Xq salt, b+ /b −− = pb/b −− , b− /b −− = qb/b −− , so
2
2
+ qz−

)b/b −−
I = 21 (pz+

E10.8(b)

E10.9(b)

(a)

I (MgCl2 ) = 21 (1 × 22 + 2 × 1)b/b −− = 3b/b −−

(b)

I (Al2 (SO4 )3 ) = 21 (2 × 32 + 3 × 22 )b/b −− = 15b/b −−

(c)

I (Fe2 (SO4 )3 ) = 21 (2 × 32 + 3 × 22 )b/b −− = 15b/b −−

b(K3 [Fe(CN)6 ]) b(KCl) b(NaBr)
+
+
I = I (K3 [Fe(CN)6 ]) + I (KCl) + I (NaBr) = 21 (3 + 32 )
b −−
b −−
b −−
= (6) × (0.040) + (0.030) + (0.050) = 0.320
Question. Can you establish that the statement in the comment following the solution to Exercise
10.8a (in the Student’s Solutions Manual) holds for the solution of this exercise?
b

I = I (KNO3 ) = −− (KNO3 ) = 0.110
b
Therefore, the ionic strengths of the added salts must be 0.890.
(a)

b
I (KNO3 ) = −− , so b(KNO3 ) = 0.890 mol kg−1
b
and (0.890 mol kg−1 ) × (0.500 kg) = 0.445 mol KNO3
So (0.445 mol) × (101.11 g mol−1 ) = 45.0 g KNO3 must be added.


EQUILIBRIUM ELECTROCHEMISTRY

(b)

149

b
b
I (Ba(NO3 )2 ) = 21 (22 + 2 × 12 ) −− = 3 −− = 0.890
b
b
b=

0.890 −−
b = 0.2967 mol kg−1
3

and (0.2967 mol kg−1 ) × (0.500 kg) = 0.1484 mol Ba(NO3 )2

So (0.1484 mol) × (261.32 g mol−1 ) = 38.8 g Ba(NO3 )2
E10.10(b)

I (Al2 (SO4 )3 ) = 21 ((2 × 33 ) + (3 × 22 ))b/b −− = 15b/b −−
I (Ca(NO3 )2 ) = 21 (22 + 2)b/b −− = 3b/b −−
3(0.500 mol kg−1 ) = 15(b(Al2 (SO4 )3 ))
3
(0.500 mol kg−1 ) = 0.100 mol kg−1
b(Al2 (SO4 )3 ) = 15

E10.11(b)

p q

γ± = (γ+ γ− )1/s

s =p+q

For Al2 (SO4 )3 p = 2, q = 3, s = 5
γ± = (γ+2 γ−3 )1/5
E10.12(b) Since the solutions are dilute, use the Debye–H¨uckel limiting law
log γ± = −|z+ z− |AI 1/2
I = 21

i

zi2 (bi /b −− ) = 21 {1 × (0.020) + 1 × (0.020) + 4 × (0.035) + 2 × (0.035)}
= 0.125

log γ± = −1 × 1 × 0.509 × (0.125)1/2 = −0.17996

(For NaCl) γ± = 10−0.17996
= 0.661
E10.13(b)

I (CaCl2 ) = 21 (4 + 2)b/b −− = 3b/b −−
log γ± = −2 × 1 × 0.509 × (0.300)1/2 = −0.5576
γ± = 10−0.5576 = 0.2770 = 0.277

0.524 − 0.277
× 100 per cent = 47.1 per cent
0.524
A|z+ z− |I 1/2
E10.14(b) The extended Debye–H¨uckel law is log γ± = −
1 + BI 1/2
Solving for B
Error =

B=−

1
A|z+ z− |
+
log γ±
I 1/2

=−

1
0.509
+

log γ±
(b/b −− )1/2


INSTRUCTOR’S MANUAL

150

Draw up the following table
b/(mol kg−1 )
γ±
B

5.0 × 10−3
0.927
1.32

10.0 × 10−3
0.902
1.36

50.0 × 10−3
0.816
1.29

B = 1.3
E10.15(b)

KS = 1.4 × 10−8

PbI2 (aq)

PbI2 (s)
rG

−−

= −RT ln KS = −(8.314 J K−1 mol−1 ) × (298.15 K) × ln(1.4 × 10−8 )
= 44.83 kJ mol−1

rG

−−

=

fG

−−

fG

−−

(PbI2 , aq) −

(PbI2 , aq) =

rG

−−

+

fG

−−

(PbI2 , s)

fG
−1

−−

(PbI2 , s)

= 44.83 kJ mol

− 173.64 kJ mol−1

= −128.8 kJ mol−1
E10.16(b) The Nernst equation may be applied to half-cell potentials as well as to overall cell potentials.
E(H+ /H2 ) =

RT
a(H+ )
ln
F
(fH2 /p −− )1/2

E = E2 − E1 =

RT
RT
a2 (H+ )
γ± b2
[fH2 is constant] =
ln
ln
+
F
a1 (H )
F
γ± b 1

= (25.7 mV) × ln

(0.830) × (5.0 × 10−2 )
(0.929) × (5.0 × 10−3 )

= +56.3 mV

E10.17(b) Identify electrodes using species with the desired oxidation states.
L:

Cd(s) + 2OH− (aq) → Cd(OH)2 (s) + 2e−

R:

Ni(OH)3 (s) + e− → Ni(OH)2 (s) + OH− (aq)

Cd(s)|Cd(OH)2 (s)|OH− (aq)|Ni(OH)2 (s)|Ni(OH)3 (s)|Pt
E10.18(b) The cell notation specifies the right and left electrodes. Note that for proper cancellation we must
equalize the number of electrons in half-reactions being combined.
(a)

R: Ag2 CrO4 (s) + 2e− → 2Ag(s) + CrO2−
4 (aq)
L: Cl2 (g) + 2e− → 2Cl− (aq)
Overall (R − L): Ag2 CrO4 (s) + 2Cl− (aq) → 2Ag(s) + CrO2−
4 (aq) + Cl2 (g)

+0.45 V
+1.36 V
−0.91 V

(b)

R: Sn4+ (aq) + 2e− → Sn2+ (aq)
L: 2Fe3+ (aq) + 2e− → 2Fe2+ (aq)
Overall (R − L): Sn4+ (aq) + 2Fe2+ (aq) → Sn2+ (aq) + 2Fe3+ (aq)

+0.15 V
+0.77 V
−0.62 V

(c)

R: MnO2 (s) + 4H+ (aq) + 2e− → Mn2+ (aq) + 2H2 O(l)

L: Cu2+ (aq) + 2e− → Cu(s)
Overall (R − L): Cu(s) + MnO2 (s) + 4H+ (aq) → Cu2+ (aq) + Mn2+ (aq)
+ 2H2 O(l)

+1.23 V
+0.34 V
+0.89 V


EQUILIBRIUM ELECTROCHEMISTRY

151

Comment. Those cells for which E −− > 0 may operate as spontaneous galvanic cells under standard
conditions. Those for which E −− < 0 may operate as nonspontaneous electrolytic cells. Recall that
E −− informs us of the spontaneity of a cell under standard conditions only. For other conditions we
require E.
E10.19(b) The conditions (concentrations, etc.) under which these reactions occur are not given. For the purposes
of this exercise we assume standard conditions. The specification of the right and left electrodes is
determined by the direction of the reaction as written. As always, in combining half-reactions to form
an overall cell reaction we must write half-reactions with equal number of electrons to ensure proper
cancellation. We first identify the half-reactions, and then set up the corresponding cell.
(a)

R: 2H2 O(l) + 2e− → 2OH− (aq) + H2 (g) −0.83 V
L: 2Na+ (aq) + 2e− → 2Na(s)
−2.71 V
and the cell is
Na(s)|Na+ (aq), OH− (aq)|H2 (g)|Pt +1.88 V
or more simply

Na(s)|NaOH(aq)|H2 (g)|Pt

(b)

R: I2 (s) + 2e− → 2I− (aq)
+0.54 V
L: 2H+ (aq) + 2e− → H2 (g)
0
and the cell is
Pt|H2 (g)|H+ (aq), I− (aq)|I2 (s)|Pt +0.54 V
or more simply
Pt|H2 (g)|HI(aq)|I2 (s)|Pt

(c)

R: 2H+ (aq) + 2e− → H2 (g)
0.00 V
L: 2H2 O(l) + 2e− → H2 (g) + 2OH− (aq) −0.083 V
and the cell is
Pt|H2 (g)|H+ (aq), OH− (aq)|H2 (g)|Pt 0.083 V
or more simply
Pt|H2 (g)|H2 O(l)|H2 (g)|Pt

Comment. All of these cells have E −− > 0, corresponding to a spontaneous cell reaction under
standard conditions. If E −− had turned out to be negative, the spontaneous reaction would have been
the reverse of the one given, with the right and left electrodes of the cell also reversed.
E10.20(b) See the solutions for Exercise 10.18(b), where we have used E −− = ER−− − EL−− , with standard
electrode potentials from Table 10.7.
E10.21(b) See the solutions for Exercise 10.19(b), where we have used E −− = ER−− − EL−− , with standard
electrode potentials from Table 10.7.

E10.22(b) In each case find E −− = ER−− − EL−− from the data in Table 10.7, then use
rG

(a)

R:
L:

−−

= −νF E −− [10.32]
2−
−
S2 O2−
8 (aq) + 2e → 2SO4 (aq) +2.05 V
+ 1.51 V
−
−
I2 (s) + 2e → 2I (aq)
+0.54 V

rG

−−

= (−2) × (96.485 kC mol−1 ) × (1.51 V) = −291 kJ mol−1


INSTRUCTOR’S MANUAL

152

(b)

Zn2+ (aq) + 2e− → Zn(s) −0.76 V

Pb2+ (aq) + 2e− → Pb(s)
rG

−−

−0.13 V

E −− = −0.63 V

= (−2) × (96.485 kC mol−1 ) × (−0.63 V) = +122 kJ mol−1

E10.23(b) (a) A new half-cell may be obtained by the process (3) = (1) − (2), that is
(3) 2H2 O(l) + Ag(s) + e− → H2 (g) + 2OH− (aq) + Ag+ (aq)
But, E3−− = E1−− − E2−− , for the reason that the reduction potentials are intensive, as opposed to
extensive, quantities. Only extensive quantities are additive. However, the r G−− values of the
half-reactions are extensive properties, and thus
−−
r G3

−−
r G1

=

−

−−
r G2

−ν3 F E3−− = −ν1 F E1−− − (−ν2 F E2−− )
Solving for E3−− we obtain
E3−− =

ν1 E1−− − ν2 E2−−
(2) × (−0.828 V) − (1) × (0.799 V)
= −2.455 V
=
ν3
1

(b) The complete cell reactions is obtained in the usual manner. We take (2) × (2) − (1) to obtain
2Ag+ (aq) + H2 (g) + 2OH− (aq) → 2Ag(s) + 2H2 O(l)
E −− (cell) = ER−− − EL−− = E2−− − E1−− = (0.799 V) − (−0.828 V) = +1.627 V
Comment. The general relation for E −− of a new half-cell obtained from two others is
E3−− =
E10.24(b) (a)

ν1 E1−− ± ν2 E2−−
ν3

E = E −− −
Q=
J

RT
ln Q ν = 2
νF

2
2
aJνJ = aH
+a −
Cl

[all other activities = 1]

2 2
a− = (γ+ b+ )2 × (γ− b− )2
= a+

= (γ+ γ− )2 × (b+ b− )2 = γ±4 b4
Hence, E = E −− −
(b)
(c)

rG

b
b ≡ −− here and below
b
[16, b+ = b, b− = b]

2RT
RT

ln(γ± b)
ln(γ±4 b4 ) = E −− −
F
2F

= −νF E[10.32] = −(2)×(9.6485×104 C mol−1 )×(0.4658 V) = −89.89 kJ mol−1

log γ± = −|z+ z− |AI 1/2 [19] = −(0.509) × (0.010)1/2 [I = b for HCl(aq)] = −0.0509
γ± = 0.889
E −− = E +

2RT
ln(γ± b) = (0.4658 V) + (2) × (25.693 × 10−3 V) × ln(0.889 × 0.010)
F
= +0.223 V

The value compares favourably to that given in Table 10.7.


EQUILIBRIUM ELECTROCHEMISTRY

E10.25(b)

153

R:

Fe2+ (aq) + 2e− → Fe(s)

L:

2Ag+ (aq) + 2e− → 2Ag(s)
2Ag(s) + Fe2+ (aq) → 2Ag+ (aq) + Fe(s)

R − L:

E −− = ER−− − EL−− = (−0.44 V) − (0.80 V) = −1.24 V
rG

−−

= −νF E −− = −2 × (9.65 × 104 C mol−1 ) × (−1.24 V)
= +239 kJ mol−1

rH

−−

= 2 f H −− (Ag+ , aq) −

fH

−−

(Fe2+ , aq) = [(2) × (105.58) − (−89.1)] kJ mol−1

= +300.3 kJ mol−1
∂ r G−−
= − r S −− =
∂T

p
=
Therefore,

rG

−−

−
T

rH

−−

[ r G−− =

rH

− T r S]

(239 − 300.3) kJ mol−1
= −0.206 kJ mol−1 K −1
298.15 K

νF E −−
[10.36]
RT

Sn(s) + CuSO4 (aq)

Cu(s) + SnSO4 (aq)
2+
−
R: Cu (aq) + 2e → Cu(s) +0.34 V
+ 0.48 V
L: Sn2+ (aq) + 2e− → Sn(s) −0.14 V
ln K =

(b)

−−

(308 K) ≈ (239) + (10 K) × (−0.206 K −1 ) kJ mol−1 ≈ +237 kJ mol−1

E10.26(b) In each case ln K =
(a)

rG

(2) × (0.48 V)
= +37.4,
25.693 mV

K = 1.7 × 1016

Cu2+ (aq) + Cu(s)
2Cu+ (aq)
2+
−
R: Cu (aq) + e → Cu+ (aq) +0.16 V

− 0.36 V
+0.52 V
L: Cu+ (aq) + e− → Cu(s)
ln K =

−0.36 V
= −14.0,
25.693 mV

K = 8.2 × 10−7

E10.27(b) We need to obtain E −− for the couple
(3) Co3+ (aq) + 3e− → Co(s)
from the values of E −− for the couples
(1) Co3+ (aq) + e− → Co2+ (aq)

E1−− = 1.81 V

(2) Co2+ (aq) + 2e− → Co(s)

E2−− = −0.28 V

We see that (3) = (1) + (2); therefore (see the solution to Exercise 10.23(b))
E3 =

ν1 E1−− + ν2 E2−−
(1) × (1.81 V) + (2) × (−0.28 V)
=
= 0.42 V
ν3

3


INSTRUCTOR’S MANUAL

154

Then,
R:

Co3+ (aq) + 3e− → Co(s)

L: 3AgCl(s) + 3e− → 3Ag(s) + 3Cl− (aq)
R − L: Co3+ (aq) + 3Cl− (aq) + 3Ag(s) → 3AgCl(s) + Co(s)
E −− = ER−− − EL−− = (0.42 V) − (0.22 V) = +0.20 V

ER−− = 0.42 V
EL−− = 0.22 V

E10.28(b) First assume all activity coefficients are 1 and calculate KS◦ , the ideal solubility product constant.
AgI(s)
Ag+ (aq) + I− (aq)
S(AgI) = b(Ag+ ) = b(I− ) because all stoichiometric coefficients are 1.
b(Ag+ )b(I− )
S2
=
= (1.2 × 10−8 )2 = 1.44 × 10−16
Thus KS◦ =
b −−2
b −−2

2Bi3+ (aq) + 3S2− (aq)
(2)
Bi2 S3 (s)
(1)

b(Bi3+ ) = 2S(Bi2 S3 )
b(S2− ) = 3S(Bi2 S3 )
KS◦ =

(b(Bi3+ ))2 × (b(S2− ))3
(2S)2 × (3S)3
S 5
=
=
108
b −−
b −−5
b −−5

= 1.13 × 10−97
For AgI, KS = γ±2 KS◦
log γ± = −|z+ z− |AI 1/2
−−

I = Sb ,

A = 0.509

|z+ z− | = 1

so

log γ± = −(0.509) × (1.2 × 10−8 )1/2 = −5.58 × 10−5
γ± = 0.9999
KS = (0.9999)2 KS◦ = 0.9997KS◦
For Bi2 S3 , I = 15b/b −− = 15Sb −− , |z+ z− | = 6

so log γ± = −(0.509) × (6) × [15(1.6 × 10−20 )]1/2 = −1.496 × 10−9
γ± = 1.0
KS = γ±5 KS◦ = KS◦

Neglect of activity coefficients is not significant for AgI and Bi2 S3 .
E10.29(b) The Nernst equation applies to half-reactions as well as whole reactions; thus for
−
2+
8H+ + MnO−
4 (aq) + 5e → Mn (aq) + 4H2 O

E = E −− −
E10.30(b)

R:
L:

a(Mn2+ )
RT
ln
+ 8
5F
a(MnO−

4 )a(H )

2AgI(s) + 2e− → 2Ag(s) + 2I− (aq) −0.15 V
2H+ (aq) + 2e− → H2 (g)
0V

Overall(R − L):

2AgI(s) + H2 (g) → 2Ag(s) + 2H+ (aq) + 2I− (aq)

Q = a(H+ )2 a(I− )2
+

−

ν=2

Assume a(H ) = a(I ), Q = a(H+ )4


EQUILIBRIUM ELECTROCHEMISTRY

E = E −− −
pH =

155

RT
2RT
ln a(H+ )4 = E −− −

ln a(H+ ) = E −− + 2 × (2.303) ×
2F
F

F
2 × (2.303RT )

× (E − E −− ) =

RT
F

× pH

1.15 V
E + 0.15 V
=
= 9.72
0.1183 V
0.1183 V

E10.31(b) The electrode reactions are
L:

Ag+ (aq) + e− → Ag(s)

R:

AgI(s) + e− → Ag(s) + I− (aq)

Overall(R − L):

AgI(s) → Ag+ (aq) + I− (aq)

Since the cell reaction is a solubility equilibrium, for a saturated solution there is no further tendency
to dissolve and so E = 0
E10.32(b)

R:

2Bi3+ (aq) + 6e− → 2Bi(s)

L:

Bi2 S3 (s) + 6e− → 2Bi(s) + 3S2− (aq)

Overall(R − L):

2Bi3+ (aq) + 3S2− (aq) → Bi2 S3 (s)

ν=6

νF E −−
RT
6(0.96 V)
=
(25.693 × 10−3 V)

ln K =

= 224
K = e224
It is convenient to give the solution for (b) first.
(b) KS = K −1 = e−224 ≈ 10−98 , since the cell reaction is the reverse of the solubility equilibrium.
(a)

KS ≈ 10−98 =
S=

10−98
108

2
3
b
b
3+
2−
(Bi
)
×
(S
)
= (2S)2 × (3S)3 = 108S 5
b −−
b −−

1/5

≈ 10−20 mol L−1

Solutions to problems
Solutions to numerical problems
P10.1

We require two half-cell reactions, which upon subtracting one (left) from the other (right), yields the
given overall reaction (Section 10.4). The half-reaction at the right electrode corresponds to reduction,
that at the left electrode to oxidation, though all half-reactions are listed in Table 10.7 as reduction
reactions.
E −−
R: Hg2 SO4 (s) + 2e− → 2Hg(l) + SO2−
+0.62 V
4 (aq)
L: PbSO4 (s) + 2e− → Pb(s) + SO2−
(aq)
−0.36 V
4
R − L: Pb(s) + Hg2 SO4 (s) → PbSO4 (s) + 2Hg(l) +0.98 V


INSTRUCTOR’S MANUAL

156

Hence, a suitable cell would be
Pb(s)|PbSO4 (s)|H2 SO4 (aq)|Hg2 SO4 (s)|Hg(l)
or, alternatively,
Pb(s)|PbSO4 (s)|H2 SO4 (aq) H2 SO4 (aq) Hg2 SO4 (s)|Hg(l)
For the cell in which the only sources of electrolyte are the slightly soluble salts, PbSO4 and Hg2 SO4 ,
the cell would be

Pb(s)|PbSO4 (s)|PbSO4 (aq) Hg2 SO4 (aq)|Hg2 SO4 (s)|Hg(l)
The potential of this cell is given by the Nernst equation [10.34].
RT
ln Q [10.34]; ν = 2
νF
aPb2+ aSO2−
KS (PbSO4 )
4
Q=
=
aHg2+ aSO2−
KS (Hg2 SO4 )

E = E −− −

2

4

E = (0.98 V) −
= (0.98 V) −

RT
KS (PbSO4 )
ln
KS (Hg2 SO4 )
2F
25.693 × 10−3 V
2

× ln

1.6 × 10−8
6.6 × 10−7

[Table 10.6, 4th Edition, or CRC Handbook]
= (0.98 V) + (0.05 V) = +1.03 V
P10.6

Pt|H2 (g)|NaOH(aq), NaCl(aq)|AgCl(s)|Ag(s)
H2 (s) + 2AgCl(s) → 2Ag(s) + 2Cl− (aq) + 2H+ (aq)

ν=2

RT
ln Q, Q = a(H+ )2 a(Cl− )2
[f/p −− = 1]
2F
RT
RT
RT
Kw a(Cl− )
Kw γ± b(Cl− )
= E −− −
= E −− −
ln a(H+ )a(Cl− ) = E −− −
ln
ln
−
F

F
F
a(OH )
γ± b(OH− )

E = E −− −

= E −− −

Kw b(Cl− )
b(Cl− )
RT
RT
RT
−−
=
E
ln
ln
K
ln
−
−
w
F
F
F
b(OH− )
b(OH− )

= E −− + (2.303)

RT
RT
b(Cl− )
× pKw −
ln
F
F
b(OH− )
−

b(Cl )
ln b(OH
−
E − E −−
)
Hence, pKw =
+
2.303RT /F
2.303

=

pKw = − log Kw =

− ln Kw
2.303

E − E −−

+ 0.05114
2.303RT /F

E −− = ER−− − EL−− = E −− (AgCl, Ag) − E −− (H+ /H2 ) = +0.22 V − 0 [Table 10.7]
We then draw up the following table with the more precise value for E −− = +0.2223 V [Problem 10.8]
θ/◦ C
E/V
2.303RT
F

V
pKw

20.0
1.04774

25.0
1.04864

30.0
1.04942

0.05819

0.05918

0.06018

14.23

14.01

13.79


EQUILIBRIUM ELECTROCHEMISTRY

157

−−
d ln Kw
rH
[9.26]
=
2
dT
RT
rH

Hence,
then with
rH

−−

= −(2.303)RT 2

d pKw
≈
dT

−−

d
(pKw )
dT

pKw
T

≈ −(2.303) × (8.314 J K −1 mol−1 ) × (298.15 K)2 ×

13.79 − 14.23
10 K

= +74.9 kJ mol−1
−−

= −RT ln Kw = 2.303RT × pKw = +80.0 kJ mol−1

−−

=

rG
rS

rH

−−

−
T

rG

−−

= −17.1 J K−1 mol−1

See the original reference for a careful analysis of the precise data.
P10.7

The cells described in the problem are back-to-back pairs of cells each of the type
Ag(s)|AgX(s)|MX(b1 )|Mx Hg(s)
Hg

(Reduction of M+ and formation of amalgam)

R:

M+ (b1 ) + e− −→ Mx Hg(s)

L:

AgX(s) + e− → Ag(s) + X− (b1 )

R − L:

Hg

Ag(s) + M+ (b1 ) + X− (b1 ) −→ Mx Hg(s) + AgX(s)

ν=1

a(Mx Hg)
a(M+ )a(X− )
RT
E = E −− −
ln Q
F

Q=

For a pair of such cells back to back,
Ag(s)|AgX(s)|MX(b1 )|Mx Hg(s)|MX(b2 )|AgX(s)|Ag(s)
RT
RT
ER = E −− −
EL = E −− −
ln QR
ln QL
F
F
QL
(a(M+ )a(X− ))L
−RT
RT
ln
ln

E=
=
F
QR
F
(a(M+ )a(X− ))R
(Note that the unknown quantity a(Mx Hg) drops out of the expression for E.)
a(M+ )a(X− ) =

γ+ b+
b −−

γ− b−
b −−

= γ±2

b 2
b −−

(b+ = b− )

With L = (1) and R = (2) we have
E=

γ± (1)
2RT
b1
2RT
ln

+
ln
b2
F
γ± (2)
F

b1
Take b2 = 0.09141 mol kg−1 (the reference value), and write b = −−
b
E=

2RT
F

ln

b
γ±
+ ln
0.09141
γ± (ref)


INSTRUCTOR’S MANUAL

158

For b = 0.09141, the extended Debye–H¨uckel law gives
log γ± (ref) =

(−1.461) × (0.09141)1/2
+ (0.20) × (0.09141) = −0.2735
(1) + (1.70) × (0.09141)1/2

γ± (ref) = 0.5328
then E = (0.05139 V) × ln
ln γ± =

b
γ±
+ ln
0.09141
0.5328

E
b
− ln
(0.09141) × (0.05328)
0.05139 V

We then draw up the following table
b/(mol/kg−1 )
E/V
γ

0.0555
−0.0220
0.572

0.09141
0.0000
0.533

0.1652
0.0263
0.492

0.2171
0.0379
0.469

1.040
0.1156
0.444

1.350
0.1336
0.486

A more precise procedure is described in the original references for the temperature dependence of
E −− (Ag, AgCl, Cl− ), see Problem 10.10.
P10.10

The method of the solution is first to determine
1
2 H2 (g) + AgCl(s)

rG

−−

,

rH

−−

, and

rS

−−

for the cell reaction

→ Ag(s) + HCl(aq)

and then, from the values of these quantities and the known values of f G−− , f H −− , and S −− for
all the species other than Cl− (aq), to calculate f G−− , f H −− , and S −− for Cl− (aq).
rG

−−

= −νF E −−

At 298.15 K(25.00◦ C)
E −− /V = (0.23659) − (4.8564 × 10−4 ) × (25.00) − (3.4205 × 10−6 ) × (25.00)2
+ (5.869 × 10−9 ) × (25.00)3 = +0.22240 V
G−− = −(96.485 kC mol−1 ) × (0.22240 V) = −21.46 kJ mol−1

Therefore,
rS

−−

=−

∂E −−
∂θ p

V

∂ r G−−
=
∂T
p

∂E −−
× νF = νF
∂T p

∂E −− ◦ C
∂θ p K

[dθ/◦ C = dT /K]

= (−4.8564 × 10−4 /◦ C) − (2) × (3.4205 × 10−6 θ/(◦ C)2 )
+ (3) × (5.869 × 10−9 θ 2 /(◦ C)3 )

∂E −−
∂θ p
V/◦ C

= (−4.8564 × 10−4 ) − (6.8410 × 10−6 (θ/◦ C)) + (1.7607 × 10−8 (θ/◦ C)2 )

Therefore, at 25.00◦ C,
∂E −−
= −6.4566 × 10−4 V/◦ C
∂θ p
and
∂E −−
= (−6.4566 × 10−4 V/◦ C) × (◦ C/K) = −6.4566 × 10−4 V K−1
∂T p

(a)


EQUILIBRIUM ELECTROCHEMISTRY

159

Hence, from equation (a)
rS

−−

= (−96.485 kC mol−1 ) × (6.4566 × 10−4 V K−1 ) = −62.30 J K−1 mol−1

rH

and

−−

=

rG

−−

+ T r S −−

= −(21.46 kJ mol−1 ) + (298.15 K) × (−62.30 J K −1 mol−1 ) = −40.03 kJ mol−1
For the cell reaction
1
2 H2 (g) + AgCl(s) → Ag(s) + HCl(aq)
−−
= f G−− (H+ ) + f G−− (Cl− ) −
rG
= f G−− (Cl− ) − f G−− (AgCl)
fG

Hence,

−−

(Cl− ) =

rG

−−

+

fG

−−

fG

−−

(AgCl)
−−

[ f G (H+ ) = 0]

(AgCl) = [(−21.46) − (109.79)] kJ mol−1
= −131.25 kJ mol−1

fH

Similarly,

−−

(Cl− ) =

rH

−−

+

fH

−−

(AgCl) = (−40.03) − (127.07 kJ mol−1 )
= −167.10 kJ mol−1

For the entropy of Cl− in solution we use
−−
rS
−−

with S

= S −− (Ag) + S −− (H+ ) + S −− (Cl− ) − 21 S −− (H2 ) − S −− (AgCl)

(H+ ) = 0. Then,

S −− (Cl− ) =

rS

−−

− S −− (Ag) + 21 S −− (H2 ) + S −− (AgCl)

= (−62.30) − (42.55) + 21 × (130.68) + (96.2) = +56.7 J K−1 mol−1
P10.12

∂G
= V [5.10]
∂p T
∂ rG
we obtain
= rV
∂p T
Substituting r G = −νF E [10.32] yields

(a) From

∂E
rV
=−
∂p T ,n
νF
(b) The plot (Fig. 10.1) of E against p appears to fit a straight line very closely. A linear regression
analysis yields
Slope = 2.840 × 10−3 mV atm−1 ,
Intercept = 8.5583 mV,
R = 0.999 997 01 (an extremely good fit)
From r V

standard deviation = 3 × 10−6 mV atm−1
standard deviation = 2.8 × 10−3 mV

∂E
(−2.666 × 10−6 m3 mol−1 )
=−
∂p T ,n
1 × 9.6485 × 104 C mol−1
Since J = V C = Pa m3 , C =

m3
V
Pa m3
or
=
V
C
Pa


INSTRUCTOR’S MANUAL

160

13

12

11

10

9

8
0

500

1000

1500

Figure 10.1
Therefore
∂E
=
∂p T ,n

2.666 × 10−6
9.6485 × 104

V
1.01325 × 105 Pa
×
= 2.80 × 10−6 V atm−1
Pa
atm
= 2.80 × 10−3 mV atm−1

This compares closely to the result from the potential measurements.
(c) A fit to a second-order polynomial of the form
E = a + bp + cp 2

yields
a = 8.5592 mV,
b = 2.835 × 10−3 mV atm−1 ,
c = 3.02 × 10−9 mV atm−2 ,
R = 0.999 997 11

standard deviation = 0.0039 mV
standard deviation = 0.012 × 10−3 mV atm−1
standard deviation = 7.89 × 10−9 mV atm−1

This regression coefficient is only marginally better than that for the linear fit, but the uncertainty
in the quadratic term is > 200 per cent.
∂E
= b + 2cp
∂p T
The slope changes from

∂E
= b = 2.835 × 10−3 mV atm−1
∂p min

∂E
= b + 2c(1500 atm) = 2.836 × 10−3 mV atm−1
∂p max
∂E
are very good.
We conclude that the linear fit and constancy of
∂p
to

EQUILIBRIUM ELECTROCHEMISTRY

161

(d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c.
∂ rV
= 2c
∂p T
1 ∂ rV
2νcF
(κT )cell = −
=
V
∂p T
V
1
∂ 2E
=−
νF
∂p 2

(κT )cell =

cm3 atm
2(1) × (3.02 × 10−12 V atm−2 ) × (9.6485 × 104 C mol−1 ) × 82.058
8.3145 J
1 cm3
0.996 g

= 3.2 × 10−7 atm−1

g
× 18.016
1 mol

standard deviation ≈ 200 per cent

where we have assumed the density of the cell to be approximately that of water at 30◦ C.
Comment. It is evident from these calculations that the effect of pressure on the potentials of
cells involving only liquids and solids is not important; for this reaction the change is only
∼ 3 × 10−6 V atm−1 . The effective isothermal compressibility of the cell is of the order of magnitude
typical of solids rather than liquids; other than that, little significance can be attached to the calculated
numerical value.
P10.15

The equilibrium is
K=

a(H2 O)4 a(V4 O12 −4 )
γ (V4 O12 −4 )b(V4 O12 −4 )
≈
a(H2 VO4 − )4
γ (H2 VO4 − )4 b(H2 VO4 − )4

Let x be b(H2 VO4 − ); then b(V4 O12 −4 ) = (0.010 − x)/4. Then the equilibrium equation can be
expressed as
x4

Kγ (H2 VO4 − )4

= (0.010 − x)/4

γ (V4 O12 −4 )

which can be solved numerically once the constants are determined. The activity coefficients are
log γ (H2 VO4 − ) = −

0.5373
= −0.269
2

and log γ (V4 O12 −4 ) = −

so

0.5373(42 )
= −1.075
2

γ (H2 VO4 − ) = 0.538
so

γ (V4 O12 −4 ) = 0.0842

The equation is
x 4 (2.5 × 106 ) = (0.010 − x)/4
Its solution is
x = 0.0048 mol kg−1 = b(H2 VO4 − )
and b(V4 O12 −4 ) = 0.010 − (0.010 − 0.0048)/4 = 0.0013 mol kg−1

P10.18

The reduction reaction is
Sb2 O3 (s) + 3H2 O(l) + 6e− → 2Sb(s) + 6OH− (aq)

Q = a(OH− )6

ν=6

Therefore
(a)

RT
RT
2.303RT
ln a(OH− )6 = E −− −
ln a(OH− ) = E −− +
pOH
6F
F
F
[ln a(OH− ) = 2.303 log a(OH− ) = −2.303pOH]
E = E −− −


INSTRUCTOR’S MANUAL

162

(b) Since pOH + pH = pKw

E = E −− +

2.303RT
(pKw − pH)
F

(c) The change in potential is
E=

2.303RT
(pOHf − pOHi ) = (59.17 mV) × (pOHf − pOHi )
F

pOHf = − log(0.050γ± ) = − log 0.050 − log γ± = − log 0.050 + A (0.050) = 1.415
pOHi = − log(0.010γ± ) = − log 0.010 − log γ± = − log 0.010 + A (0.010) = 2.051
E = (59.17 mV) × (1.415 − 2.051) = −37.6 mV

Hence,
P10.19

We need to obtain

rH

−−

1
+
2 H2 (g) + Uup (aq)

for the reaction

→ Uup(s) + H+ (aq)

We draw up the thermodynamic cycle shown in Fig. 10.2.
Data are obtained from Table 13.4, 14.3, 2.6, and 2.6b. The conversion factor between eV and
kJ mol−1 is
1 eV = 96.485 kJ mol−1
The distance from A to B in the cycle is given by
rH

−−

= x = (3.22 eV) + 21 × (4.5 eV) + (13.6 eV) − (11.3 eV) − (5.52 eV) − (1.5 eV)
= 0.75 eV

rS

−−

=S

−−

(Uup, s) + S −− (H+ , aq) − 21 S −− (H2 , g) − S −− (Uup+ , aq)

= (0.69) + (0) − 21 × (1.354) − (1.34) meV K −1 = −1.33 meV K−1

i

Figure 10.2

rG

−−

=

rH

−−

− T r S −− = (0.75 eV) + (298.15 K) × (1.33 meV K −1 ) = +1.15 eV

which corresponds to +111 kJ mol−1
The electrode potential is therefore

− r G−−
, with ν = 1, or −1.15 V
νF


EQUILIBRIUM ELECTROCHEMISTRY

163

Solutions to theoretical problems
P10.21

MX(s)

M+ (aq) + X− (aq),

Ks ≈ b(M+ )b(X− )

b
b ≡ −−
b

b(X− ) = S + C

b(M+ ) = S,
Ks = S(S + C),

or

S 2 + CS − Ks = 0

4Ks 1/2 1
which solves to S = 21 (C 2 + 4Ks )1/2 − 21 C or S = 21 C 1 + 2
− 2C
C
C2,

If 4Ks

2Ks
S ≈ 21 C 1 + 2
C
P10.22

Ks
− 21 C (1 + x)1/2 ≈ 1 + 21 x + · · · ≈
C

Ks = a(M+ )a(X− ) = b(M+ )b(X− )γ±2 ;
log γ± = −AI 1/2 = −AC 1/2
γ± = e−2.303AC

Ks = S (S + C) × e−4.606AC
We solve S 2 + S C −

Ks
γ±2

1/2

1/2

=0

4Ks
1
to get S =
C2 + 2
2
γ±

1/2

Ks
1
[as in Problem 10.21]
− C≈
2
Cγ±2

Therefore, since γ±2 = e−4.606AC
P10.25

b(X− ) = S + C

ln γ± = −2.303AC 1/2

γ±2 = e−4.606AC

1/2

b(M+ ) = S ,

1/2

S ≈

Ks e−4.606AC
C

1/2

The half-reactions involved are:

−−
R: cyt ox + e− → cytred Ecyt
−−
−
L: Dox + e → Dred ED
The overall cell reaction is:

R − L = cyt ox + Dred

cytred + Dox

−−
−−
− ED
E −−− = Ecyt

(a) The Nernst equation for the cell reaction is
E=E−

RT [cyt red ][Dox ]
ln
F
[cyt ox ][Dred ]

at equilibrium, E = 0; therefore
ln
ln

[cytred ]eq [Dox ]eq
F

=
[cyt ox ]eq [Dred ]eq
RT
[Dox ]eq
[Dred ]eq

= ln

Therefore a plot of ln
intercept of

F
RT

−−
−−
− ED
Ecyt

[cyt]ox
[cyt]red
[Dox ]eq
[Dred ]eq

−−
−−
− ED
Ecyt

+

F
RT

−−
−−
− ED
Ecyt

against ln

[cyt]ox
[cyt]red

is linear with a slope of one and an


INSTRUCTOR’S MANUAL

164

(b) Draw up the following table:
[Dox ]eq
[Dred ]eq
[cytox ]eq
ln
[cyt red ]eq
ln

The plot of ln

−1.2124. Hence

−5.882

−4.776

−3.661

−3.002

−2.593

−1.436

−0.6274

−4.547

−3.772

−2.415

−1.625

−1.094

−0.2120

−0.3293

[Dox ]eq
[Dred ]eq

[cytox ]eq
[cyt red ]eq

against ln

is shown in Fig. 10.3. The intercept is

RT
× (−1.2124) + 0.237 V
F
= 0.0257 V × (−1.2124) + 0.237 V

−−
=
Ecyt

= +0.206 V

0

ln([Dox]eq / [Dred]eq)

–1

y = –1.2124 + 1.0116x

R = 0.99427

–2
–3
–4
–5
–6
–5

–4

–3

–2

–1

0

ln([cytox]eq / [cytred]eq)

1

Figure 10.3

Solutions to application
P10.27

(a)

molalityH2 SO4 = b(d) = a(d − d25 ) + c(d − d25 )2

where d is density in g cm−3 at 25 ◦ C, a = 14.523 mol kg−1 (g cm−3 )−1 ,
c = 25.031 mol kg−1 (g cm−3 )−2 , and d25 = 0.99707 g cm−3 .
For 1 kg solvent (mH2 O = 1 kg):
mass %H2 SO4 =

mH2 SO4
mH2 SO4 + mH2 O

100 =

b × 100
mH✘
2O
b + mH ✘
m✘
2 SO4✘H2 O

100 × b(d)

where mH2 SO4 = 0.09807 kg mol−1
b(d) + mH 1SO
2 4
an equation for the solution molarity is deduced with a unit analysis.
mass %H2 SO4 (d) =

molarityH2 SO4 (d) = b(d) × 1 −

3 cm 3
mass %H2 SO4 (d)

10✚
✚
d×
100
L

kg
3g
10✚
✚


EQUILIBRIUM ELECTROCHEMISTRY

165

Sulfuric Acid Solutions
Molality / (mol/kg)

10
8
6
4
2
0

1

1.1

1.2
Density/(g / mL)

1.3

1.4

Figure 10.4(a)

Sulfuric Acid Solutions

Mass Percentage
Sulfuric Acid

50
40
30
20
10
0

1

1.1

1.2
Density/(g / mL)

1.3

1.4

Figure 10.4(b)

Sulfuric Acid Solutions

Molarity/(mol / L)

7
6
5
4
3
2
1
0

(b)

1

1.1

1.2
Density/(g / mL)

1.3

1.4

Figure 10.4(c)

cell:
Pb(s) | PbSO4 (s) | H2 SO4 (aq) | PbO2 (s) | PbSO4 (s) | Pb(s)
−
cathode: PbO2 (s) + 3H+ (aq) + HSO−
4 (aq) + 2e → PbSO4 (s) + 2H2 O(l)
−−
Ecathode = 1.6913 V
anode: PbSO4 (s) + H+ (aq) + 2e− → Pb(s) + HSO−
4
−−
Eanode
= −0.3588 V
net:
PbO2 (s) + Pb(s) + 2H+ (aq) + 2HSO−
4 (aq) → 2PbSO4 (s) + 2H2 O(l)
−−
◦
E −− = Ecathode
− Eanode
= 2.0501V
(eqn 10.38)
rG

−−

= −νF E −− = −(2)(9.64853 × 104 C mol−1 )(2.0501 V)
= −3.956 × 105 C V mol−1 = −3.956 × 105 J mol−1 = −395.6 kJ mol−1

INSTRUCTOR’S MANUAL

166

−−
fH
−−
rH

values of Table 2.6 and the CRC Handbook of Chemistry and Physics are used in the
calculation.

rH

−−

= 2 f H −− (PbSO4 ) + 2 f H −− (H2 O(l)) −

fH

−−

(PbO2 ) −

fH

−−

(Pb)

−2 f H −− (H+ ) − 2 f H −− (HSO−
4)
= 2(−919.94 kJ mol−1 ) + 2(−285.83 kJ mol−1 ) − (−277.4 kJ mol−1 )
−2(−887.34 kJ mol−1 )
rH
rS

−−

−−

= −359.5 kJ mol−1
rH

=

−−

−
T

rG

−−

= 121 J K−1 mol−1
E −− (15◦ C) = E −− (25◦ C) +
= 2.0501 V +

=

−359.5 kJ mol−1 − (−395.6 kJ mol−1 )
298.15 K
(eqn 4.39)

E −− = E −− (25◦ C) +

✘
✟✘
✘−1
K−1
mol
(121 J ✟
)
(10  
K)
✘
−1
✘
✘
2(96485 C mol )

rS

−−

νF

T

(eqn 10.45)

= 2.0501V + 0.006V = 2.0507V
The temperature difference makes a negligibly small difference in the cell potential.
When Q = 6.0 × 10−5 ,
RT
ln Q (eqn 10.34)
νF
✘
✟✘
✘−1
K −1
mol
(8.31451 J ✟
)(298.15  
K)
= 2.0501 V −
ln(6.0 × 10−5 )
✘
✘−1
mol
2(96485 C ✘
)

E = E −− −

= 2.1750 V
(c) The general form of the reduction half-reaction is: ox + νe− + νH H+ + aA → red + xX using
eqn 10.34,

E = E −− −

x
ared aX
RT
RT
ln Q = E −− −
ln
νH a
νF
νF
aox aH
+ aA

1
RT
ln
νH
νF
aH+
(all species other than acids are at unit activity in a Pourboix diagram)

= E −− −

= E −− +

νH RT
νH RT ln(10)
ln aH+ = E −− +
log aH+

νF
νF

= E −− −

νH
ν

RT ln(10)
F

E = E −− − (0.05916V)

pH

νH
pH
ν

(eqn 9.29)


EQUILIBRIUM ELECTROCHEMISTRY

For the PbO2 | PbSO4 couple,
−
PbO2 (s) + 4H+ + SO2−
4 (aq) + 2e → PbSO4 (s) + 2H2 O(l)

E −− = 1.6913 V, νH = 4, ν = 2

E = 1.6913 V − (0.11832 V)pH
For pH = 5, E = 1.0997 V
For pH = 8, E = 0.7447 V
For the PbSO4 /Pb couple,
PbSO4 (s) + 2e− → Pb(s) + SO2−
4 (aq)
Since νH = O, E = E −− = −0.3588 V at all pH values in the Pourboix diagram.

167