Both matels Mg and Fe can reduce copper metal from a solution having copper ions `(Cu^(2+))`. According to the equilibria:
`Mg(s)+Cu^(2+)hArrMg^(2+)+Cu(s), K_(1)=6xx10^(90)`
`Fe(s)+Cu^(2+)hArrFe^(2+)+Cu(s), K_(2)=3xx10^(26)`
Which metal will remove cupric ion from the solution to a greater extent?

Explanation

Since `K_(1) gt K_(2)`, the product in the first reaction is much more favoured than in the second one. Mg thus removes more `Cu^(2+)` from solution than does Fe.
`K_(1) and K_(2)` include the constant concentration of the solid species).