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 redox change is "Zn+ni^(2+)hArrZn^(2+)+Ni),("mM before equilibrium 500 0"),("mM at equilibrium a (500-a)"):}`
`E_(cell)^(0)=E_(OP_(ZN//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))`
`E_(cell)^(0)=E_(Nn//Zn^(2+))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)+(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])`
At equilibrium `E_(cell)=0`
`thereforeE_(OP_(Zn//Zn^(2+)))^(0)+E_(RP_(Ni^(2+)//Ni))^(0)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])`
`or 0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)])`
`([Ni^(2+)])/([Zn^(2+)])="anti log"((-0.51xx2)/(0.059))=5.15xx10^(-18)`
`therefore(a)/(500-a)=5.15xx10^(-18)`
`thereforea=500xx5.15xx10^(-18)`
`therefore[Ni^(2+)]=(mV)/(V)=(500xx5.15xx10^(-18))/(500)`
`=5.15xx10^(-18)M`