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Question

Consider the following cell reaction:
2Fe(s)+O2(g)+4H+(aq)2Fe2+(aq)+2H2O(l),Eo=1.67V
At [Fe2+]=103M,P(O2)=0.1 atm and pH=3, the cell potential at 25oC is:

A
1.47V
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B
1.77V
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C
1.87V
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D
1.57V
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Solution

The correct option is C 1.57V
In the given reaction Fe is oxidised and O2 is reduced.

2Fe2Fe2++4e

4e+O2+4H+2H2O

Ecell=EoOPFe0.0594log[Fe2+]2+EoRPO2+0.0594logPO2×[H+]4

=Eocell+0.0594logPO2×[H+]4[Fe2+]2

=1.67+0.0594log0.1×(103)4(103)2

=1.67+0.0594log107

=1.67+0.059×(7)4

=1.670.103=1.57V

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