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The pK(sp) of AgI is 16.07 if the E^(@) ...

The `pK_(sp)` of `AgI` is `16.07` if the `E^(@)` value for `Ag^(+)//Ag` is `0.7991V`, find the `E^(@)` for the half cell reaction `AgI(s) +e^(-) rarr Ag+I^(-)`

Text Solution

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The correct Answer is:
`E^(@) =- 0.149V`

`E_(I^(-)//AgI//Ag)^(@) = E_(Ag^(+)//Ag)^(@) +(0.0591)/(1)log K_(sp) (AgI)`
`= 0.7991 +0.991 xx 16.07`
`rArr 0.149 V`
`(1)/(2) H_(2) rarr H^(+) +e^(-)`
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