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Question

The emf (in V) of a Daniel cell containing 0.1M ZnSO4 and 0.01M CuSO4 solutions at their respective electrodes are:
(EoCu2+/Cu=0.34V;EoZn2/Zn=0.76V)

A
1.10
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B
1.16
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C
1.13
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D
1.07
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Solution

The correct option is D 1.07
Ecell=Eocell0.0592log[Zn2+][Cu2+]

Ecell=1.10.0592log[Zn2+][Cu2+]

Now,

Eocell=EoCu2+/CuEoZn2+/Zn =1.1 V

Again

Ecell=1.10.0592log[0.1][0.01]

Ecell=1.10.029 log [10]

Ecell=1.10.029 (as log [10]=1)

Ecell=1.10.029=1.07 V

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