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Question

Consider the cell;
V2++VO2++2H+2V3++H2O; Eo=0.616V

V3++Ag++H2OVO2++2H++Ag; Eo=0.439V
Potential of EoV3+/V2+ is :
(Given that EoAg+/Ag=0.799V)

A
0.256 V
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B
0.256 V
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C
1.055 V
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D
0.177 V
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Solution

The correct option is B 0.256 V
The e.m.f of the cell for the second reaction is 0.439 V.

Hence,

E0cell=E0cathodeE0anode=E0Ag+|AgE0VO2+|V3+

0.439V=0.799VEVO2+|V3+

EVO2+|V3+=0.36V
The e.m.f of the cell for the first reaction is 0.616 V.

Hence,

E0cell=E0cathodeE0anode=E0VO2+|V3+E0V3+|V2+

0.616V=0.36VE0V3+|V2+

E0V3+|V2+=0.256V.
Hence, option A is the right answer.

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