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.

Answer

`5.15xx10^(-18)M`

Explanation

The redox change is


`E_(cell)=E_(o x i d (Zn|Zn))+E_(red(Ni^(2+)|Ni))`
`E_(cell)=E^(c-)._(o x i d (Zn|Zn^(2+)))+E^(c-)._(red(Ni^(2+)|Ni))+(0.059)/(2)log.([Ni^(2+)])/([Zn^(2+)])`
At equilibrium, `E_(cell)=0`
`:. E^(c-)._(o x i d(Zn|Zn^(2+)))+E^(c-)._(red(Ni^(2+)|Ni))=-(0.059)/(2)log.([Ni^(2+)])/([Zn^(2+)])`
or `or 0.75+(-0.24)=-(0.059)/(2)log.([Ni^(2+)])/([Zn^(2+)])`
`or([Ni^(2+)])/([Zn^(2+)])=antilog(-(0.51xx2)/(0.059))`
`=5.15xx10^(-18)`
`:. (a)/((500-a))=5.15xx10^(-18) (:'500-a~~500)`
`:. a=500xx5.15xx10^(-18)`
`:. [Ni^(2+)]=(a)/(V)=(500xx5.15xx10^(-18))/(500)`
`=5.15xx10^(-18)M`