Chemistry-
General
Easy
Question
Cu2+ + 2e– ¾® Cu; log[Cu2+] vs Ered graph is of the type shown in figure where OA = 0.34V, then electrode potential of the half cell of Cu/Cu2+ (0.1 M) will be
- – 0.34 +
V
- 0.34 + 0.059V
- 0.34 V
- None
V
The correct answer is: – 0.34 + V
if log[Cu2+] = 0, i.e. [Cu2+] = 1 M then 
or OA = 
= – 0.34V
Now 
= – 0.34 – 
= – 0.34 + V
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