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Question

Given EoCr3+| Cr =0.72V, EoFe2+|Fe=0.42 V.


The potential for the cell

Cr |Cr3+(0.1M)Fe2+(0.01M)| Fe is:

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

The correct option is C 0.26V
Cell reaction;
2Cr(s)+3Fe2+(aq)2Cr3+(aq)+3Fe(s)

Now,
Ecell=EoFe2+|FeEoCr3+|Cr0.0591n log[Cr3+]2[Fe2+]3

Ecell=0.42(0.72)0.05916 log(0.1)2(0.01)3=0.26V

Option C is correct.

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