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Question

Calculate the reduction potential for the given half cells at 25oC.
Pt(s)|Fe2+(aq, 0.1 M),Fe3+(aq, 0.01 M);E0Fe3+/Fe2+=+0.77 V

A
EFe3+/Fe2+=0.77 V
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B
EFe3+/Fe2+=0.83 V
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C
EFe3+/Fe2+=0.83 V
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D
EFe3+/Fe2+=0.711 V
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Solution

The correct option is D EFe3+/Fe2+=0.711 V
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)+eFe2+(aq) (reduction)

Nernst equation for
aA+bBcC+dD

E=E02.303RTnFlog[C]c[D]d[A]a[B]b.....(Eqn.1)

Substituting,

R=8.314 J K1 mol1

F=96500 C/mol

T=25+273=298 K, we get,

E=E00.059nlog[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+]

EFe3+/Fe2+=0.770.0591log0.10.01

EFe3+/Fe2+=0.770.0591log 10

EFe3+/Fe2+=0.711 V

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