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Question

If Ksp(AgCl)=1012 Ksp(AgBr)=1015, then the emf of the cell Ag(s) | AgBr(s) | Br0.4M || Cl2M | AgCl(s) | Ag(s) is

A
0 V
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B
0.136 V
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C
0.04135 V
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D
Data insufficient
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Solution

The correct option is B 0.136 V
AgCl(s)Ksp=1012Ag+(aq)+Cl(aq)
when [Cl]=2M, Ag+=10122M
AgBr(s)Ksp=1015Ag+(aq)+Br(aq)
when [Br]=0.4M, Ag+=10150.4M
For cell reaction Ag(s) | AgBr(s) | Br(0.4M) || Cl(2M) | AgCl(s) | Ag(s)
Which is equivalent to Ag(s) | Ag+10150.4 || Ag+10122 | Ag(s)
So cell reaction Ag(s)+Ag+(10122) Ag+(10150.4)+Ag(s) which concentration cell
Ecell=Eocell0.0591nlogQc=00.05911log 10150.410122=0.0591log1022
Ecell=0.136 V

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