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Question

The emf of the cell is given as:

Ni|Ni2(1.0)||Au+3(0.1M)|Au
[Eo for Ni2/Ni=0.25V,Eo for Au+3/Au=1.50V]

A
+1.25V
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B
1.75V
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C
+1.75V
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D
+2.50V
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Solution

The correct option is B +1.75V
Ni|Ni2(1.0)||Au3+(0.1M)|Au
EMF=EoRHSEoLHS
=1.5(0.25)
=1.5+0.25=1.75V

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