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Question

During the electrolysis of 0.1 M CuSO4 solution using copper electrodes, a depletion of [Cu2+] occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the concentrations of [Cu2+] near the anode & cathode are respectively 0.12 M & 0.08 M, calculate the back e.m.f. developed. Temperature = 298 K.

A
22 mV
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B
5.2 mV
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C
29 mV
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D
59 mV
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Solution

The correct option is A 5.2 mV
Back emf = emf due to different concentrations of ions near cathode and anode.
using the relation
Eelectrode=E00.059nlog([product][reactant])
Reaction at cathode:
Cu2+((aq)+2eCu(s)
Ecathode=E00.0592log([Cu][Cu2+])
Ecathode=0.0592log(10.08)
Ecathode=0.032 V
Anode reaction:
$Cu(s)\rightarrow Cu^{ 2+ }(aq)+2e^-$
Eanode=E00.0592log([Cu2+][Cu])
Eanode=0.0592log(0.121)
Eanode=0.027 V
Since potential at cathode is not equal to that at anode the residual potential is the Back emf i.e.:
Eb=|Ecathode||Eanode|
Eb=0.0320.027
Eb=5×103 V
option B is correct

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