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)
Half reaction:
At anode (oxidation) :
Zn
Zn+2 + 2e- EA° =
+0.76 V
At cathode:
Ni+2 + 2e-
Ni Ec° = - 0.26 V
So, Ecell° = 0.76 -0.26 = 0.50 V
∆G = -nFEcell°
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
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