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Question

Cu2eCu2+,E0=0.347V

Sn2eSn2+,E0=+0.143V

The EMF of the cell constructed with these electrodes is :


A
+0.066V
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B
0.066V
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C
+0.490V
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D
0.82V
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Solution

The correct option is B +0.490V
The given potentials are standard oxidation potentials. so, EMF of the cell =
E(cathode)- E(anode)
The electrode with higher value of potential is chosen as cathode so the the sign of EMF is positive,only then the cell reaction can be spontaneous. so, in the above cell,
EMF= 0.143-(-0.347)=0.49V

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