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Question

Given that E0Cl2/Cl=1.35 V and Ksp(AgCl)=1010 at 25oC.

E0 corresponding to the electrode reaction is
12Cl2(g)+Ag+(aq)+eAgCl(s) is

A
0.75 V
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B
2.65 V
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C
1.05 V
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D
1.94 V
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Solution

The correct option is D 1.94 V
For sparingly soluble AgCl
AgCl(s)Ag+(aq)+Cl(aq)...(eq 1)
Ksp(AgCl)=1010
ΔG1=RTlnKsp
For :
12Cl2(g)+eCl(aq)......(eq 2); E02=1.35 V
ΔG2=nFE02
Reversing eqn 1 and adding with equation 2, we get the desired equation
12Cl2(g)+Ag+(aq)+eAgCl(s)...(eq3)
ΔG3=nFE03=ΔG2ΔG1
nFE03=nFE02(RTlnKsp)
E03=E02RTFln1010E03=1.35+(0.059×10)=1.94 V


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