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Question

For an electrochemical cell Sn(s)|Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s), the ratio [Sn2+][Pb2+] when this cell attains equilibrium is ____.
(Given : E0Sn2+|Sn=0.14 V, E0Pb2+|Pb=0.13 V, 2.303RTF=0.06)

A
2.15
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B
2.154
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Solution

The cell reaction is
Sn+Pb2+Sn2++Pb
In the given reaction, Pb2+ get reduced and Sn get oxidised.
Hence,
E0cell=E0cathode (red)E0anode (red)
E0cell=0.13(0.14)
E0cell=+0.01
At equilibrium,
Ecell=0

Sn+Pb2+Sn2++Pb
Ecell=E0cell0.062log Q

0=0.010.062log[Sn2+][Pb2+]

0.01=0.062log[Sn2+][Pb2+]

13=log[Sn2+]Pb2+

[Sn2+][Pb2+]=101/3=2.154

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