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Calculate ΔrG0 and log Kc for the following cell: Ni(s)+2AgA+(aq)⟶NiA2+(aq)+2Ag(s) Given that EAcell0 = 1.05 V, 1F = 96,500 C mol–1. - Chemistry

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Question

Calculate ΔrG0 and log Kc 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.

Numerical
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Solution

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 Kc = 35.3

Kc =  3.41 × 1035

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