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Question

The standard reduction potentials for two reactions are given below:
AgCl(s) + e - Ag(s)+Cl - (aq) E=0.22V
Ag+(aq)+eAg(s) E=0.80V
The solubility product of AgCl under standard conditions of temperature (298 K) is given by:

A
1.6×105
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B
1.5×108
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C
3.2×1010
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D
1.5×1010
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Solution

The correct option is C 1.5×1010
Solution:- (D) 1.5×1010
AgCl(s)+eAg(s)+Cl(aq.)E=0.22V.....(1)
Ag+(aq.)+eAg(s)E=0.80V
Ag(s)Ag+(aq.)+e.....E=0.80V.....(2)
Adding equation (1)&(2), we get
AgCl(s)Ag+(aq.)+Cl(aq.)E=0.58V
From Nernst equation,
Ecell=EcellRTnFlnK
0=0.588.314×298×2.3031×96500logK
logK=0.580.0591=9.81
K=1.5×1010
Hence the solubility product of AgCl is 1.5×1010.

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