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Given Ag^(+)+e^(-)rarrAg,E^(@)=0.799 ...

Given
`Ag^(+)+e^(-)rarrAg,E^(@)=0.799 V`
Dissociation constant for `[Ag(NH_(3)_(2)]` into `Ag^(+)` and `NH_(3) "is" 6xx10^(-14)` Then for the folllowing half cel reaction
`[Ag(NH_(3)^(2))]^(+)e^(-)rarrAg+2NH_(3) "Calculate" E^(@)`

Text Solution

Verified by Experts

The correct Answer is:
`0.02 V`
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