(a) Calculate the electrode potential of silver electrode dipped in 0.1 M solution of silver nitrate of 298 K assuming `AgNO_(3)` to be completely dissociated. The standard electrode potential of `Ag^(+)|Ag` is 0.80V at 298K.
(b) At what concentration of `Ag^(+)` ions will this electrode have a potential of 0.0 volt?

Answer

(a) `0.741`V
(b) `2.9xx10^(-14)M`

Explanation

(a) `E_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)^(@)-(0.0591)/(1)"log"(1)/([Ag^(+)])`. On solving we get `E_(Ag^(+)//Ag)=0.7409V`
(b) `E_(Ag^(+)//Ag)=0therefore0=0.80-(0.0591)/(1)"log"(1)/([Ag^(+)])` or `log[Ag^(+)]=-(0.80)/(0.0591)=-13.5364=overline(14).4636`
or `[Ag^(+)]=2.9xx10^(-14)M`