A silver electrode is immersed in a 0.1 (M) `AgNO_(3)` solution at `25^(@)C`. If `AgNO_(3)` dissociates almost completely in the solution, then determine the potential of the silver electrode.
Given : `E_(Ag^(+)|Ag)^(@)=0.80V`.

Explanation

The reduction electrode reaction is :
`Ag^(+) (aq) + e^(-) hArr Ag (s)`
`E(Ag^(+) |Ag) = E^(Theta) (Ag^(+) |Ag) - ((0.059))/(1) "log" ([Ag])/([Ag^(+)])`
`[Ag^(+)] = 0.1 M , [Ag] = 1 , E^(Theta) (Ag^(+) | Ag) = 0.80 V`
`E_(Ag^(+) |Ag)^(Theta) = 0.80 - ((0.059))/(1) "log" (1)/((0.1))`
`= 0.80 - 0.059 = 0.741 V`.