1.

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.

Answer»

Solution :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.


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