Question

Use tabulated electrode potentials to calculate the equilibrium constant (K) at 25∘C for the following reaction. Answer in units of x10¹⁷

Zn (s) + Ni²⁺ (aq) Zn²⁺ (aq) + Ni (s)

Answer 1

Half reaction:

At anode (oxidation) :

Zn Zn⁺² + 2e^-   E_A° = +0.76 V

At cathode:

Ni⁺² + 2e^- Ni E_c° = - 0.26 V

So, E_cell° = 0.76 -0.26 = 0.50 V

∆G = -nFE_cell°

where, n = 2 as 2 electon are involved.

1 F = 96500 C

So, ∆G = -(2 mol) (96500 C) (0.50 V)

= -96500 J/mol

∆G = -RT ln K

R = 8.314 J/K.mol

T = 25°C = 298 K

So, 2.303 log K = - ∆G/ RT

log K = -(-96500 J/mol) /(8.314 J/K.mol) (298 K) (2.303)

log K = 16.91

So, K = 0.817 x 10¹⁷

_​​​