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Question

Given, E0Cr3+/Cr=−0.72V,E0Fe2+/Fe=−0.42V

The potential for the cell

Cr|Cr3+(0.1M)||Fe2+(10.0M)|Fe is :

A
0.339V
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B
0.26V
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C
0.349V
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D
0.339V
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Solution

The correct option is C 0.349V
The standard emf of cell is
E0cell=E0Fe2+|FeE0Cr3+|Cr=0.42V(0.72V)=0.3V
The emf of cell is
Ecell=E0cell0.0592nlog[Cr3+]2[Fe2+]3=0.3V0.05926log[0.1]2[10]3=0.349V.

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