Question

Calculate Δ_rG⁰ and log K_c for the following cell:

\[\ce{Ni(s) + 2Ag^+(aq) -> Ni^{2+}(aq) + 2Ag(s)}\]

Given that \[\ce{E^0_{cell}}\] = 1.05 V, 1F = 96,500 C mol^–1.

Answer 1

According to the equation,

\[\ce{Ni + 2Ag^+ -> Ni^{2+} + 2Ag}\]

ΔG = –nFE⁰

Where ΔG = Gibb's free energy

ΔG = –2 × 96500 × 1.05

N = No. of electrons gained or lost = 2

ΔG = –202.650 kJ

F = Faraday's constant = 96500

E⁰ = Standard emf = 1.05V

The equation shows the relationship between Gibb's free energy and Equilibrium constant.

`E_(cell)^0 = 0.0591/n log K_c`

`log K_c = - (1.05 xx 2)/0.0591`

log K_c = 35.3

K_c =  3.41 × 10³⁵