Determine the ΔE of an electrochemical cell.

Question

Determine ΔE (in V) for an electrochemical cell at 16.39 °C composed of a Zn electrode in a solution of 0.0331 M Zn²⁺ coupled to a Zn electrode in a solution of 0.666 M Zn²⁺.

Zn²⁺(aq) + 2e− ⇌ Zn(s)    E° = -0.762 V

Answer 1

The cell described is a concentration cell where both electrodes and solutions are the same.

Use the Nernst equation:

E_cell + Eº_cell -RT/nF lnQ

E_cell = ?

Eº_cell = 0 since both electrodes are the same

R = 8.314 J/Kmole

T = 16.39 + 273.15 = 289.54K

n = 2 mols electrons

F = 96485 C / mol electrons

Q = 0.0331/0.666

E_cell = 0 - (8.314)(289.54)/(2)(96485) ln (0.0331/0.666)

= - 0.01247 x - 3.00

Ecell = 0.0374 V

(be sure to check all of the math)