Lets find Eo 1st
from data table:
Eo(Ni2+/Ni(s)) = -0.25 V
Eo(Cu2+/Cu(s)) = 0.337 V
As per given reaction/cell notation,
cathode is (Cu2+/Cu(s))
anode is (Ni2+/Ni(s))
Eocell = Eocathode - Eoanode
= (0.337) - (-0.25)
= 0.587 V
Number of electron being transferred in balanced reaction is 2
So, n = 2
use:
E = Eo - (2.303*RT/nF) log {[Ni2+]^1/[Cu2+]^1}
Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591
So, above expression becomes:
E = Eo - (0.0591/n) log {[Ni2+]^1/[Cu2+]^1}
E = 0.587 - (0.0591/2) log (0.001^1/0.01^1)
E = 0.587-(-2.956*10^-2)
E = 0.62 V
Answer: 0.62 V
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