Question

a) What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10?

b) What is the change in the cell voltage when the ion concentrations in the anode half-cell are increased by a factor of 10?

Answer 1

We know that the cell reactions is:

2 Ag⁺ (aq) + Fe (s) ------> Fe²⁺ (aq) + 2 Ag (s)

a)

When the ion concentration at cathode half cell is increased by a factor of 10,

[Ag⁺] = 10* 1 M = 10 M

E_cell = E_cell⁰ - 0.0592/n (log Q)

= 1.25 V - 0.0592/2 (log` [Fe²⁺]/[Ag⁺]²)

= 1.25 V - 0.0296 {log [(1.00)/(10)²]}

= 1.25 V - (-0.0592)

= 1.3092 V

E_cell - E_cell⁰ = 1.3092 V - 1.25 V = 0.0592 V

b)

When the ion concentration at anode half cell is increased by a factor of 10,

[Fe²⁺] = 10* 1 M = 10 M

E_cell = E_cell⁰ - 0.0592/n (log Q)

= 1.25 V - 0.0592/2 (log` [Fe²⁺]/[Ag⁺]²)

= 1.25 V - 0.0296 {log [(10)/(1.00)²]}

= 1.25 V - 0.0296

= 1.2204 V

E_cell - E_cell⁰ = 1.2204 V - 1.25 V = - 0.0296 ​ V