Both $ $ Mg $ $ and $ $ Fe $ $ metal can reduce copper from a solution having $ $ C{u^{2 + }} $ $ ion, according to equilibria.
 $ $ M{g_{(s)}} + \,C{u^{2 + }} \leftrightharpoons M{g^{2 + }} + C{u_{(s)}}\,;{K_1}\, = 6 \times {10^{90}} $ $
 $ $ F{e_{(s)}} + C{u^{2 + }} \leftrightharpoons F{e^{2 + }} + C{u_{(s)}}\,;\,{K_2} = 3 \times {10^{26}} $ $
Choose the correct option regarding above equilibrium.
(A) $ Mg $ removes more $ C{u^{2 + }} $ from solution
(B) $ Fe $ removes more $ C{u^{2 + }} $ from solution
(C) Both will equally remove $ C{u^{2 + }} $ from solution
(D) Both metals cannot remove $ C{u^{2 + }} $ from solution

Hint: To check the reduction of ion by the metal in the solution we have to compare the value of the equilibrium constant of both solutions. The solution that has a higher value of the equilibrium constant will displace more ions from the solution.

Complete Step-by-step Answer:
In the electrochemical series, those elements that have more negative value can replace those elements that have a less negative value of reduction potential. The value of reduction potential of $ Mg $ and $ Fe $ are $ - 2.37 $ and $ - 0.44 $ respectively. The reduction potential of copper is $ $ + 0.34 $ $ . Now as we can see that both $ Mg $ and $ Fe $ are able to reduce copper. Now we are given the value of equilibrium constants for both these solutions and we will compare them to find the correct option.
The equilibrium constant is dependent on the backward rate reaction and the forward rate reaction i.e. $ {k_f} $ and $ {k_r} $ . It is given as:
 $ {K_c} = \dfrac{{{k_f}}}{{{k_r}}} $
From the relation, we can deduce that the equilibrium constant is directly proportional to the forward rate reaction i.e. the product formation. A higher value of $ {K_c} $ means the forward rate reaction is higher and more product formation is favored i.e. ions of that solution are getting displaced at a higher rate. In the given question the value of $ {K_1} $ is $ 6 \times {10^{90}} $ and value of $ {K_2} $ is $ 3 \times {10^{26}} $ . Hence we can say that $ {K_1} > {K_2} $ . It means ions are displaced more by the $ Mg $ ions. Hence, $ Mg $ removes more $ C{u^{2 + }} $ from the solution.
Therefore, option (A) is correct.

Note:
To replace a metal ion from the solution with another metal we have to look for their position in the electrochemical series. This series is made up of the arrangement of various equilibrium of redox reactions based on their reduction potential value.