Zinc rod is dipped in 0.1 M solution of `ZnSO_4`. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[Given, `E_(Zn^(2+)//Zn)^(@) = -0.76 V`]

Explanation

Applying Nernst Equation
`E_(Zn^(2+)//Zn) = E_(Zn^(2+)//Zn)^(@) -0.0591/2 log 1/([Zn^(2+)])`
`[Zn^(2+)] =0.1 xx 95/100 = 0.095 M`
`Zn^(2+) + 2e^(-) to Zn` or n=2
`E_(Zn^(2+)//Zn) = -0.76 V - 0.0591/2 [log1000- log 95]=-0.76 V - (0.0591)/2 [ 3.000 - 1.9777]`
`=-0.76 V - (0.0591)/2 xx 1.0223 = -0.76 V - 0.00604/2`
`=-0.76 V - 0.0302 = -0.7902` V
Electrode potential `=-0.7902` V