Zinc electrode is constituted at 298kby placing Zn - rod in 0.1 (M) aqueous solution of `ZnSO_(4)` which is `95%`. dissociated at this concentration. What will be the electrode potential `(E_(Zn^(2+)|Zn))` of the electrode?
`("Given that, "E_(Zn^(2+)|Zn)^(@)=-0.76V)`.

Explanation

Electrode reaction: `Zn^(2+)(aq)+2erarrZn`
For this half- reaction, the reduction potential of the electrode is
`E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zn)^(0)-(0.059)/(2)log.([Zn])/([Zn^(2+)])`
As `[Zn]=1, E_(Zn^(2+)|Zn)=E_(Zn^(2+)|Zm)^(@)-(0.059)/(2)log.(1)/([Zn^(2+)])`
`=E_(Zn^(2+)|Zn)+(0.059)/(2)log[Zn^(2+)]`
If `ZnSO_(4)` in its 0.1 M solution dissociates to an extent of `95%`, then `[Zn^(2+)]=0.095`. Therefore,
`E_(Zn^(2+)|Zn)=[-0.76+(0.059)/(2)log0.095]V=-0.79V`