Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given EZn2+/Zn = +(-0.76 V]

Explanation

The electrode reaction written as
`Zn^(2+)+2e^(-)rarrZn`
Applying Nernst equation, we get
`E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.0591)/(2)log""(1)/([Zn^(2+)])`
As 0.1 M `ZnSO_(4)` solution is 95% disociated, this means that in the solution, conc. of `Zn^(2+)` is
`[Zn^(2+)]=(95)/(100)xx0.1M=0.095`
`E_(Zn^(2+)//Zn)=-0.76-(0.0591)/(2)log""(1)/(0.095)`
`=-0.76-0.02955(log1000-log95)=-076-0.02955(3-1.977)`
`=-0.76-0.03021=-0.79021` volt