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Question

Given ECr3+Cr=0.72VEFe2+Fe=0.42V
The potential for the cell
CrCr3+(0.1M)||Fe2+(0.01M)|Fe is

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

The correct option is C 0.26 V
E0cell=0.720.72
=0.3
Ecell=0.30.05916 log 0.120.013
=0.30.1×4
= 0.3 – 0.04
= 0.26

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