If excess of` Zn` is added to `1.0M` solution of `CuSO_(4)` , find the concentration of `Cu^(2+)` ions at equilibrium.
Given`:E^(c-)._((Zn^(2+)|Zn))=-0.76V`
`E^(c-)._(cell)=(E^(c-)._((Cu^(2+)|Cu))=0.34V`

Explanation

At equilibrium, `E_(cell=0.0V` and reaction coefficietn `Q=K_(eq)`.
`E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)`
`=0.34-(-0.76)=1.10V`
At equilibrium,
`E^(c-)._(cell)=(0.059)/(n_(cell))logK_(eq)`
`:. log K_(eq)=(E^(c-)._(cell)xxn_(cell))/(0.059)=(1.10xx2)/(0.059)`
`=37.30`
`:. K _(eq)=Antilod(37.30)=2.02xx10^(37)`
`(` Antilog of `0.30~~2)`
Let `a M` is `[Cu^(2+)]` at equuilibirum
`Zn(s)+Cu^(2+)(aq) hArr Zn^(2+)+Cu`
`K_(eq)=([Zn^(2+)])/([Cu^(2+)])=(alpha)/(1-alpha)~~(1)/(1-alpha)` `[:'a~~1]`
Since the value of equilibrium constant `(2.0xx10^(37))` is very high and activity of `Zn(s)` and activity of `Cu(s)=1`
`implies (1)/(1-alpha)=K_(eq)`
`implies 1-alpha=(1)/(K_(eq))=(1)/(2.0xx10^(37))=5xx10^(-38)M`
`implies[Cu^(2+)]_(eq)=(1-alpha)=5xx10^(-38)M`