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A silver wire dipped in 0.1M HCI solutio...

A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will be

A

`2.95 xx 10^(-11)`

B

`5.1 xx 10^(-11)`

C

`3.95 xx 10^(-11)`

D

`1.95 xx 10^(-11)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Ag +CI^(-) rarr AgCI +e^(-)`
`E = E^(@) + 0.0591 log [CI^(-)]`
`-0.25 = E^(@) +0.0591 log 0.1 E^(@) =- 0.1909`
Now for reaction `AgCI +CI^(-)rarrAgCI +e^(-)`
`{:(Ag^(+)+e^(-)rarrAg),(bar(Ag^(+)+I^(-)rarrAgCI)):}`
`E = E_(Ag//AgCI//CI^(-))^(@) +E_(Ag^(+)//Ag)^(@) +0.0591 log K_(sp)`
`O =- 0.1909 +0.799 +0.0591 log K_(sp)`
`K_(sp) = 5.13 xx 10^(-11)`
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