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Question

Construct galvanic cells from the following pairs of half cells and calculate their emf at 25oC.

Fe3+(aq, 0.1 M),Fe2+(aq, 1 M)|Pt(s); E0Fe3+/Fe2+=0.77 V

Ag(s)|AgCl(s)∣Cl−(aq, 0.001M); E0AgCl/Ag=0.22 V

A
Ecell=0.567 V
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B
Ecell=0.313 V
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C
Ecell=0.023 V
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D
Ecell=1.13 V
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Solution

The correct option is B Ecell=0.313 V
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)+eFe2+(aq)
By Nernst equation,

EFe3+/Fe2+=EFe3+/Fe2+0.05911log[Fe2+][Fe3+]

=0.770.05911log10.1=0.71 V

And, for the half - cell reaction,

AgCl(s)+eAg(s)+Cl(aq)

ECl/AgCl/Ag=E0Cl/AgCl/Ag0.05911log [Cl

ECl/AgCl/Ag=0.220.05911log 0.001

ECl/AgCl/Ag=0.220.05911log 103

ECl/AgCl/Ag=0.22+3×0.05911

ECl/AgCl/Ag=0.397 V

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

Ecell=0.710.397Ecell=0.313 V


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