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?

Both matels Mg and Fe can reduce copper metal from a solution having c

1.	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?
Answer» 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 `Fe` does

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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?

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