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Question

Calculate the log10Ksp value for CdS at 25oC
Given:
Cd2+(aq) + 2eCd(s); Eo = 0.403 V
CdS(s) + 2eCd(s)+S2(aq); Eo = 1.21 V

(here Ksp is solubility product)

A
+13.65
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B
13.65
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C
+27.3
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D
27.3
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Solution

The correct option is D 27.3
It is a metal-metal insoluble salt electrode.

Here,
Cd2+(aq) + 2eCd(s)....(1); Eo = 0.403 V
CdS(s) + 2eCd(s)+S 2(aq).....(2); Eo = 1.21 V

Reversing equation (2), we get,
Cd(s) + S2(aq)CdS(s) + 2e......(3); Eo = 1.21 V
Add equation (1) and (3)
Cd2(aq) + S2(aq)CdS(s).....(4); E0 = 0.807

Reverse equation (4)
CdS(s)Cd2(aq) + S2(aq)....(5); Eo = 0.807 V

For equation (5), standard free energy will be (at equilibrium)
ΔG0=RT ln Ksp
Also,
ΔG0=nFE0

RT ln Ksp=nFE0

RT ln Ksp=nFE0

E0=2.303×RTnF log10 Ksp

E0 = 0.05922×log10Ksp

log10Ksp=0.807×20.059127.3

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