Zinc granules are added in excess to 500 mL of 1M `Ni(NO_(3))_(2)` solution of `25^(@)C` untill the equilibrium is reached. If `E_(Zn^(2+)//Zn)^(@)` and `E_(Ni^(2+)//Ni)^(@)` are `-0.75V` and `-0.24V` respectively, find out the `[Ni^(2+)]` at equilibrium.

Explanation

The cell reaction for the cell
`Zn,Zn^(2+)||Ni^(2+), Ni` is
`Zn+Ni^(2+) leftrightarrow Zn^(2+)+Ni` for which
`E_(cell)=E_(cell)^@-0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])`
At equilibrium `E_(cell)=0`
`therefore E_(cell)^@=0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])`
`E_(Ni^(2+))^@,Ni-E_(Zn^(2+))^@, Zn=0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])`
`=-0.24-(-0.75)= 0.0591/2 log""([Zn^(2+)])/([Ni^(2+)])`
`log""([Zn^(2+)])/([Ni^(2+)])=17.259`
Taking antilog
`([Zn^(2+)])/([Ni^(2+)])=1.816 times 10^17`
This concentration ratio shows that almost whole of the `Ni^(2+)` ions are reduced to Ni and therefore the concentration of `Zn^(2+)` produced from Zn would be nearly 1M `[because Ni(NO_3)_2=1M]` Thus,
`1/([Ni^(2+)])=1.816 times 10^17`
or `[Ni^(2+)]=5.5 times 10^-18 M`