Calculate the reduction potential for the given half cells at 25oC. Pt(s)|Fe2+(aq,0.1M),Fe3+(aq,0.01M);E0Fe3+/Fe2+=+0.77V
A
EFe3+/Fe2+=0.711V
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B
EFe3+/Fe2+=0.83V
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C
EFe3+/Fe2+=−0.83V
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D
EFe3+/Fe2+=0.77V
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Solution
The correct option is AEFe3+/Fe2+=0.711V The given half cell electrode is a redox electrode.
Here, electrode potential results due to the presence of ions of the same substance in different oxidation states.
The given redox electrode reaction is, Fe3+(aq)+e−⇌Fe2+(aq) (reduction)
Nernst equation for aA+bB⇋cC+dD
E=E0−2.303RTnFlog[C]c[D]d[A]a[B]b.....(Eqn.1)
Substituting,
R=8.314JK−1mol−1
F=96500C/mol
T=25+273=298K, we get,
E=E0−0.059nlog[C]c[D]d[A]a[B]b....(Eqn.2)
where,
n is the number of electrons inovolved in the reaction.
For given half cell reaction, EFe3+/Fe2+=E0Fe3+/Fe2+−0.059nlog[Fe2+][Fe3+]