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Question

The standard reduction potential of Pb and Zn electrodes are -0.126 and -0.763 volts respectively. The e.m.f. of the cell
Zn|Zn2+(0.1M)Pb2+(1M)|Pb is:

A
0.637V
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B
<0.637V
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C
>0.637V
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D
0.889V
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Solution

The correct option is D 0.889V
We know that Ecell is given by the equation,
Ecell=E0cell0.059nlog(K)
Here for the given reaction,
Zn+Pb2+Zn2++Pb
n=2 for this reaction.
[Zn2+]=0.1M and [Pb2+]=1M
E0cell=EcathodeEanode=0.126(0.763)=0.889V
Hence,
Ecell=0.8890.0592log0.11=0.889V
Hence,option D is correct answer.

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