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Given: E^(@) (Ag^(+)|Ag)=0.80 V and E^(...

Given: `E^(@) (Ag^(+)|Ag)=0.80 V and E^(@) (I^(-)|Agl Ag)-0.15 V`. The value of standard solubility product Agl would be about

A

`8.4 xx10^(-17)`

B

`8.9 xx10^(-15)`

C

`8.9 xx10^(-13)`

D

`8.9 xx10E^(-11)`

Text Solution

Verified by Experts

The correct Answer is:
B
We have `Agl+e^(-) =Ag+I^(-),E^(@)=-0.15 V`
`Ag^(+)+e^(-)=Ag,E^(@)=0.80 V`
`AgI=Ag^(+)+I^(-),E^(@)=(0.15 V)-0.80 V=-0.95 V`
`(-0.95) =(0.059 V)l,log k_(sp)^(@)log k_(sp)^(@)=0.95/0.059=-16.10 rarr k_(sp)^(@)=7.9 xx10^(17)`
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