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

To determine which metal, magnesium (Mg) or iron (Fe), will remove cupric ions (Cu²⁺) from a solution to a greater extent, we can analyze the given equilibrium constants for the reactions involving these metals.

### Step-by-Step Solution:

1. **Understand the Reactions**:
- The reactions for the reduction of cupric ions by magnesium and iron are:
- For magnesium:
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