A zinc rod is dipped in 0.1 M `ZnSO_(4)` solution. The salt is 95% dissociated of this dilution at 298 K. Calculate electrode potential.
`(E_(Zn^(2+)//Zn)=-0.76 V)`.

Explanation

The electrode reaction is
`Zn^(2+) + 2e^(-) hArr Zn (s)`
According to Nernst equation , at 298 K
`E(Zn^(2+)|Zn) = E^(Theta) (Zn^(2+)|Zn) - (0.059)/(n) "log" ([Zn])/([Zn^(2+)(aq)])`
`E^(Theta) (Zn^(2+) |Zn) = -0.76 V , [Zn] = 1`,
`[Zn^(2+) (aq)] = 0.1 xx 95//100 = 0.095 M`
`therefore E (Zn^(2+)|Zn) = -0.76 - (0.059)/(n) "log"(1)/(0.095)`
`=-0.76 - 0.03 =- 0.79 V`.