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Question

The emf of the cell, Ni|Ni2+(1.0 M)||Ag+(1.0 M)|Ag
(E for Ni2+/Ni=0.25 volt,E for Ag+/Ag=0.80 volt) is given by__________.

A
0.25+0.80=0.55 volt
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B
0.25(+0.80)=1.05 volt
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C
0+0.80(0.25)=+1.05 volt
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D
0.80(0.25)=0.55 volt
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Solution

The correct option is C 0+0.80(0.25)=+1.05 volt
In the given cell we have,
Ni+2Ag+Ni2++Ag
so,
Ecell=E0cell0.05912logNi2+(Ag+)2
Ecell=0.80(0.25)- 0.05912log(1)

Ecell=0.80(0.25)=1.05volt

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