What is the potential of a half-cell consisting of zinc electrode in `"0.01 M ZnSO_(4)` solution `25^(@)C. E^(@) = 0.763 V.`

Explanation

The half-cell reaction is
`Znrarr Zn^(2+)+2e^(-)`
The Nernst equation for the oxidation half-cell reaction is
`E=E^(@)-(0.0591)/(n)log[Zn^(2+)]`
The number of electrons transferred n = `2 and E^(@) = 0.763 V`
Substituting these values in the Nernst equation we have
`E=0.763-(0.0591)/(2)(-2)`
`=0.763+0.0591=0.8221V`
Calculation of Cell potential
The Nernst equation is applicable to cell potentials as well. Thus,
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK`
K is the equilibrium constant of the redox cell reaction.