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

`Ag^(+)(aq)+erarrAg(s).` The to Nernst equation for this reaction is,
`E_(Ag^(+)|Ag)=E_(Ag^(+)|Ag)^(@)-(0.059)/(n)log.([Ag])/([Ag^(+)])`
Since 1 mol of electrons is involved in the electrode reaction, n = 1. For pure `Ag, [Ag]=1`.
`therefore" "E_(Ag^(+)|Ag)=E_(Ag^(+)|Ag)^(@)+(0.059)/(1)log[Ag^(+)]`
Substiting the values of `E_(Ag^(+)|Ag)^(@) and [Ag^(+)]` gives
`E_(Ag^(+)|Ag)=0.80+0.059 log(0.15)=0.7513V`