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Question

Given:
(i) Cu2++2eCu,E=0.337 V
(ii) Cu2++eCu+,E=0.153 V
Electrode potential, E for the reaction, Cu++eCu, will be

A
0.38 V
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B
0.52 V
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C
0.90 V
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D
0.30 V
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Solution

The correct option is D 0.52 V
Cu2++eCu can be written from

At anode: Cu+Cu2++e;Eoox=0.153V (Oxidation)
At cathode: Cu2++2eCu;EoRedn=0.337V (Reduction)
______________________
Cu++eCu (half cell reduction)

For half cell reduction in general

n1×FEoCu+|Cu=n2FEoCu+|Cu2++n3FECu2+|Cu

EoCu+|Cu=EoCu+|Cu2++2×EoCu2+|Cu

EoCu+|Cu=0.153+2×0.337=0.521V

n no. of e change

Option B is correct.

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