1.

(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»


Solution :(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`


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