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Question

Given, ECr3+/Cr=0.72 V,EFe2+/Fe=0.42 V.

The potential for the cell Cr|Cr3+(0.1 M)||Fe2+(0.01 M)|Fe is:

A
0.26 V
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B
0.399 V
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C
0.399 V
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D
0.26 V
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Solution

The correct option is D 0.26 V
Cr|Cr3+(0.1 M)||Fe2+(0.01 M)|Fe
Oxidation half-cell
CrCr3++3e]×2
Reduction half-cell
Fe2++2eFe]×3
Net cell reaction
2Cr+3Fe2+2Cr3++3Fe,n=6
Ecell=Eoxi+Ered
=0.720.42=0.30 V
Ecell=Ecell0.0591nlog[Cr3+]2[Fe2+]3
=0.300.05916log(0.1)2(0.01)3
=0.300.05916log102106
=0.300.05916log104
Ecell=0.2606 V.

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