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E^(@) of Ag electrode is 0.8 V and solu...

`E^(@)` of Ag electrode is 0.8 V and solubility product of AgI is `1 xx 10^(-16)`. Calculate the potential of Ag electrode at `25^(@)C` in a saturated AgI solution in water.

Text Solution

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`E_(Ag)^(@) = 0.8 V, K_(sp) = 10^(-16)`
At 298 K, `E_("cell") = E_("cell")^(@) - ( 0.0591)/( n) log K_(sp)`
`= 0.8 - ( 0.0591)/( 1) log ( 10^(-16))`
`= 0.8 + 16 xx 0.0591 = 0.8 + 0.95 = 1.75V`
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