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Cu2++2e⟶Cu; log Cu2+ vs . Ered graph is of the type as shown in figure where OA = 0.34V, then electrode potential of the half-cell of Cu ;Cu2+ (0.1 M) will be:

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a

−0.34+0.0592V

b

0.34 +0.059V

c

0.34V

d

None of these

answer is A.

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Detailed Solution

ECu/Cu2+=ECu/Cu2+∘−0.0592log⁡Cu2+If log⁡Cu2+=0 i.e. Cu2+=1then ECu/Cu2+=E0Cu/Cu2+or OA=ECu/Cu+2o=−ECu+2/Cu∘=−0.34Now, ECu/Cu2⋅=−0.34−0.0592log 0.1=−0.34+0.0592

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Cu2++2e⟶Cu; log Cu2+ vs . Ered graph is of the type as shown in figure where OA = 0.34V, then electrode potential of the half-cell of Cu ;Cu2+ (0.1 M) will be: