Zinc granules are added in excess to `500mL` OF `1.0m` nickel nitrate solution at `25^(@)C` until the equilibrium is reached. If the standard reduction potential of `Zn^(2+)|Zn` and `Ni^(2+)|Ni` are `-0.75V` and `-0.24V`, respectively, find out the concentration of `Ni^(2+)` in solution at equilibrium.

Explanation

The reaction to be considered is ,
`Zn(s)+Ni^(2+) (aq.) hArr Zn^(2+) (aq.) + Ni(s)`
The cell involving this reaction would be,
`Zn(s)|Zn^(2+) (aq.)||Ni^(2+) (aq.)|Ni(s)`
`E_(cell)^(@)=-0.24+0.75=0.51` volt
`log K_(eq)=(nFE^(@))/(2.303 RT)=(nE^(@))/0.0591=(2xx0.51)/(0.0591)=17.25`
So, `K_(eq)=1.78xx10^(17)`
Let x be the concentration of `Ni^(2+)` that have been reduced to nickel at equilibrium.
`Zn(s)+underset((1.0-x))(Ni^(2+) (aq.)) hArr underset(x)(Zn^(2+) (aq.)) +Ni(s)`
`K_(eq)=([Zn^(2+)])/([Ni^(2+)])=x/((1-x))=1.78xx10^(17)`
`x~~1.0 M`
So, `(1-x)=[Ni^(2+)]=1.0/(1.78xx10^(17))=5.6xx10^(-18) M`