Zn - granules are added in excess to 500 mL of 1.0 M nicke nitrate solution at `25^(@)C` until the equilibrium is reached. If 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

Since `E_(Zn^(2+)|Zn)^(@)le E_(Ni^(2+)|Ni)^(@)`, Zn will undergo oxidation and `Ni^(2+)` reduciton.
The reaction that occurs due to addition of Zn - granules to the solution of nickel nitrate is :
`Zn(s)+Ni^(2+)(aq)+Zn^(2+)(aq)+Ni(s)" ...[1]"`
Suppose, the concentration of `Zn^(2+)` ions at equilibrium = x M. Hence, at equilibrium concentration of `Ni^(2+)=(1-x)M" "[because " initial concentration of "Ni^(2+)="1 M and 1 mol of Zn reacts with 1 mol of "Ni^(2+)]`
`therefore" "[Zn^(2+)]=xM and [Ni^(2+)]=(1-x)M`
Nernst equation for the eaction (1) is :
`E_("cell")=E_("cell")^(@)-(0.059)/(2)log.([Zn^(2+)])/([Ni^(2+)])`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=[-0.24-(-0.75)]V=0.51V`
At equilibrium, `E_("cell")=0" "therefore" "0=0.51-(0.059)/(2)log.(x)/(1-x)`
`"or, "log.(x)/(1-x)=17.28" or, "(x)/(1-x)=1.9xx10^(17)`
`"or, "(1)/(1-x)-1=1.9xx10^(17)" "therefore" "(1-x)=5.26xx10^(-18)`
`therefore" Concentration of "Ni^(2+)" at equilibrium "=5.26xx10^(-18)M`