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Question

If the solution of CuSO4 in which copper rod is immersed is diluted to 10 times , the reduction electode potential:
ECu+2/Cu = 0.34V

A
increased by 0.03 V
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B
decreased by 0.03 V
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C
increased by 0.059 V
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D
decreased by 0.059 V
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Solution

The correct option is B decreased by 0.03 V
Cu2++2eCu

Ecell=Eocell0.05912log1[Cu2+]

The new concentration of Cu2+ is 110 of [Cu2+]

E1cell=Eocell0.05912log10[Cu2+]

E1cell=Eocell0.05912log1[Cu2+]0.05912log10

ΔEcell=0.05912log10

E1cell=0.0592

Thus the electrode potential decrease by 0.059V2=0.03V

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