from data table:
Eo(Zn2+/Zn(s)) = -0.7618 V
Eo(Ni2+/Ni(s)) = -0.25 V
As per given reaction/cell notation,
cathode is (Ni2+/Ni(s))
anode is (Zn2+/Zn(s))
Eocell = Eocathode - Eoanode
= (-0.25) - (-0.7618)
= 0.5118 V
here, number of electrons being transferred, n = 2
Eo = (2.303*R*T)/(n*F) log Kc
At 25 oC or 298 K, R*T/F = 0.0592
So, Eo = (0.0592/n)*log Kc
0.5118 = (0.0592/2)*log Kc
log Kc = 17.2905
Kc = 1.952*10^17
Answer: 2.02*10^17
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