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Question

Emf of the cell Ni/Ni+2(0/1M)||Au+3(1.0M)/Au at 298K will be:
[EoNi/Ni+2=0.25V,EoAu/Au+3=1.5V]

A
1.75V
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B
1.7795V
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C
+1.779V
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D
+0.7795V
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Solution

The correct option is B 1.75V
Ni/N2+ ||Au3+/Au
Eo Ni/Ni2+=0.25 V
Eo Au/Au3+=1.5 V
Eo Au3+/Au=1.5 V
Eocell=Eo Ni/Ni2++Eo Au3+/Au
=0.25+1.5=1.75 V

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