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Given the data at 25^(@)C. Ag+I^(-) to...

Given the data at `25^(@)C`.
`Ag+I^(-) to Agl+e^(-) , E^(@)=-0.152" V "`
`Ag to Ag^(+)+e^(-) , E^(@)=-0.800" V "`.
The value of log `K_(sp)` for AgI is :

A

-8.12

B

8.612

C

-37.83

D

-16.13

Text Solution

Verified by Experts

The correct Answer is:
D
(d) `Agl+e^(-) to Ag+I^(-) , E^(@)=-0.152" V "`
`Ag to Ag^(+)+e^(-) , E^(@)=-0.800" V "`
`Agl to Ag^(+)+I^(-) , E^(@)=-0.952" V "`
`E_(cell)^(@)=(0.059)/(n)logK=(0.059)/(n)logK_(sp)`
`logK_(sp)=(E_(cell)^(@)xxn)/(0.059)=((-0.952)xx1)/((0.059))=-16.135`
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