Electrochemistry

Question

1. What is the change in the cell voltage when the ion concentrations in the Cathode half-cell are

increased by a factor of 10?

2. 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

(a). First, determine the cell potential of the cell as shown.

(b). Then, use the Nernst equation to find cell potential when cathode half cell concentrations are 10x

(c). Then, use the Nernst equation to find cell potential when anode half cell concentrations are 10x

(a). cathode = Ag; Anode = Fe; both @ 1 M

Ag+ + e- ==> Ag(s) Eº = 0.80 V

Fe2+ + 2e- ==> Fe(s) Eº = -0.44 V

2Ag+ + Fe(s) ==> 2Ag(s) + Fe2+ ... cell reaction

Eºcell = 0.80 + 0.44 = 1.24 V

(b). cathode Ag = 10 M; anode Fe = 1 M

E_cell = Eº_cell - 0.0592 / n log [Fe²⁺] / [Ag⁺]²

Ecell = 1.24 - (0.0296)(-2) = 1.24 + 0.0592

Ecell = 1.30 V

Change in cell voltage = 1.30 - 1.24 = 0.06 V

(c). cathode Ag = 1 M; anode Fe = 10 M

E_cell = Eº_cell - 0.0592 / n log [Fe²⁺] / [Ag⁺]²

Ecell = 1.24 - (0.0296)(1) = 1.24 - 0.0296

Ecell = 1.21 V

Change in cell voltage = 1.21 - 1.24 = -0.03 V