Construct galvanic cells from the following pairs of half cells and calculate their emf at 25oC.
Fe3+(aq,0.1M),Fe2+(aq,1M)|Pt(s);E0Fe3+/Fe2+=0.77V
Ag(s)|AgCl(s)∣Cl−(aq,0.001M);E0AgCl/Ag=0.22V
A
Ecell=0.567V
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B
Ecell=0.313V
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C
Ecell=0.023V
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D
Ecell=1.13V
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Solution
The correct option is BEcell=0.313V For the half - cell reaction,
From the standard reduction potential values we can see Ag(s) will act as anode in the galvanic cell.
Applying nearst eq. for individual half cells: Fe3+(aq)+e−⇌Fe2+(aq)
By Nernst equation,
EFe3+/Fe2+=EFe3+/Fe2+−0.05911log[Fe2+][Fe3+]
=0.77−0.05911log10.1=0.71V
And, for the half - cell reaction,
AgCl(s)+e−⇌Ag(s)+Cl−(aq)
ECl−/AgCl/Ag=E0Cl−/AgCl/Ag−0.05911log[Cl−
ECl−/AgCl/Ag=0.22−0.05911log0.001
ECl−/AgCl/Ag=0.22−0.05911log10−3
ECl−/AgCl/Ag=0.22+3×0.05911
ECl−/AgCl/Ag=0.397V
∴ Cell may be represented as
Ag(s)|AgCl(s)∣Cl−(aq)∣∣Fe3+(aq),Fe2+(aq)|Pt(s)
Ecell= reduction potential of RHS electrode −reduction potential of LHS electrode