(D) Iron $(Fe)$ is used to obtain metallic $Cu$ from $CuSO_{4(aq)}$ because $Fe$ is more reactive than $Cu$ (it has a more negative standard reduction

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(D) Iron $(Fe)$ is used to obtain metallic $Cu$ from $CuSO_{4(aq)}$ because $Fe$ is more reactive than $Cu$ (it has a more negative standard reduction potential), thus it displaces copper from $CuSO_{4}$ solution to give metallic copper $(Cu)$.$CuSO_{4(aq)} + Fe_{(s)} \longrightarrow FeSO_{4(aq)} + Cu_{(s)}$