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 redox change is
`{:(,Zn+,Ni^(2+),hArr,Zn^(2+)+,Ni),("mM before equilibrium",,500,,0,),("mM at equilibrium",,a,,(500-a),):}`
`:. E_(cell) = E_(OP_(Zn//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))`
`E_(cell) = E_(Zn//Zn^(2+))^(@)+E_(RP_(Ni^(2+)//Ni))^(@)+(0.059)/(2)"log"_(10)([Ni^(2+)])/([Zn^(2+)])`
At equilibrium `E_(cell) = 0`
`:. E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Ni^(2+)//Ni))^(@) =-(0.059)/(1)log_(10).([Ni^(2+)])/([Zn^(2+)])`
or `0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)]) `
`([Ni^(2+)])/([Zn^(2+)]) = "antilog" (-(0.51 xx 2)/(0.059))=5.15 xx 10^(-18)`
`:. (a)/(500-a) = 5.15 xx 10^(-18)`
`:. a = 500 xx 5.15 xx 10^(-18)`
`:. [Ni^(2+)] = (mM)/(V) = (500 xx 5.15 xx 10^(-18))/(500)`
`= 5.15 xx 10^(-18)M`