Q.
Given the data at 250C , Ag+I−→AgI+e−;E0=+0.152 VAg→Ag++e−;E0=−0.8 VWhat is the value of log Ksp for AgI ?(2.303RTF=0.059V)
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a
−16.13
b
−8.12
c
8.612
d
−37.83
answer is A.
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Detailed Solution
Given Ag→Ag++e−;E0=−0.8⇒(1) Ag+I−→AgI+e−;E0=+0.152⇒(2)_ (1)−(2)⇒AgI →Ag++I−;E0=−0.952 V E0=2.303RTnF.logKsp −0.952=0.0591.logKsp log.Ksp=−0.9520.59≈−16.108
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