CHEMISTRY TEXTBOOK

if [A] = [B] = [C] = [D] = 1M,
Ecell = E^0 cell
If a gaseous substance is present in the
cell reaction its concentration term is replaced
by the partial pressure of the gas.

The Nernst equation can be used to
calculate cell potential and electrode potential.

i. Calculation of cell potential : Consider the
cell
Cd(s)Cd^2 ⊕(aq) Cu^2 ⊕^ (aq) Cu.
Let us first write the cell reaction

Cd (s) Cd^2 ⊕ (aq) + 2 e

(oxidation at anode)

Cu^2 ⊕ (aq) + 2 e Cu (s)

(reduction at cathode)

Cd^ (s) + Cu^2 ⊕^ (aq) Cd^2 ⊕ (aq) + Cu (s)

(overall cell reaction)

Here n = 2

The potential of cell is given by Nernst
equation,

Ecell = E^0 cell - 0.0592

2

log 10

[Cd^2 ⊕]^
[Cu^2 ⊕]^
at 25^0 C.
(Concentration of solids and pure liquids
are taken to be unity.)

ii. Calculation of electrode potential

Consider Zn^2 ⊕(aq)Zn(s)

The reduction reaction for the electrode is

Zn^2 ⊕ (aq) + 2 e Zn (s)

Applying Nernst equation, electrode potential
is given by

EZn = E^0 Zn - 0.0592

2

log 10

1

[Zn^2 ⊕]^

= E^0 Zn +

0.0592

2

log 10 [Zn^2 ⊕] at 25^0 C

Problem 5.10 : Calculate the voltage

of the cell, Sn (s)Sn^2 ⊕ (0.02M)Ag⊕

(0.01M)Ag (s) at 25^0 C.

E^0 Sn = - 0.136 V, E^0 Ag = 0.800 V.

Solution :

First we write the cell reaction.

Sn (s) Sn^2 ⊕ (0.02M) + 2 e

(oxidation at anode)

[Ag⊕ (0.01M) + e Ag (s)] × 2

(reduction at cathode)

Sn^ (s) + 2 Ag⊕ (0.01M)

Sn^2 ⊕ (0.02 M) + 2 Ag (s)

(overall cell reaction)

The cell potential is given by

Ecell = E^0 cell -

0.0592V

2

log 10

[Sn^2 ⊕]

[Ag⊕]^2

E^0 cell = E^0 Ag - E^0 Sn = 0.8 V + 0.136 V

= 0.936 V

Hence,

Ecell = 0.936V - 0.0592V

2

log 10

0.02

(0.01)^2

= 0.936V -

0.0592V

2

log 10 200

= 0.936V -

0.0592V

2

× 2.301

= 0.936 V - 0.0681V = 0.8679 V

Problem 5.11 : The standard potential

of the electrode, Zn^2 ⊕ (0.02 M) Zn (s)

is - 0.76 V. Calculate its potential.

Solution :

Electrode reaction :

Zn^2 ⊕ (0.02M) + 2 e Zn (s)

EZn = E^0 Zn - 0.0592V

n

log 10

1

[Zn^2 ⊕]

= - 0.76 V + 0.0592V

2

log 10 (0.02)

= - 0.76 V +

0.0592V

2

× (-1.6990)

= - 0.76 V - 0.0503V = - 0.81 V