Consider the following four electrodes:A = Cu2+ (0.0001 M)/Cu (s)B = Cu2+ (0.1 M)/Cu (s)C = Cu2+ (0.01 M)/Cu (s)D = Cu2+ (0.001 M)/Cu (s)If the standard reduction potential of Cu is + 0.34V, the reduction potentials (in volts) of the above electrodes follow the order

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A > D > C > B

B > C > D > A

C > D > B > A

A > B > C > D

answer is B.

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Detailed Solution

For metal electrodeReduction potential α [M+n]With increase in concentration of metal ion Reduction potential increases.B > C > D > A.

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