Solution:
For the given reaction,
Sn2+ (aq) + Cd (s) = Sn (s) + Cd2+(aq)
E° = E°(Sn2+/Sn) - E°(Cd2+/Cd)
= - 0.19 V - (-0.40 V)
= 0.21 V
Standard free energy change (ΔG°) is related with equilibrium constant (K) and potential (E°) as,
ΔG° = - nFE°
And
ΔG° = - 2.303 RT log K
Thus,
nFE° = 2.303 R T log K
Given,
E° = 0.21 V
F = Faraday number = 96500 C
R = 8.314 J K-1 mol-1
T = 298 K ( standard state temperature)
n = 2 ( number of electrons flow)
Thus,
log K = n F E° / 2.303 R T
= 2 x 96500 C x 0.21 V / 2.303 x 8.314 J K-1 mol-1 x 298 K
log K = 7.10324
K = antilog 7.10324
K = 12683525.9
K = 1.2684 x 10^7
The value of K is positive, hence reaction is feasible, ΔG is less than zero.
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