Question

what is the balanced net ionic equation for the electrolysis of molten manganese(ii) iodide?\\(\ce{mni2(l) -> mn(s) + i2(s)}\\)\\(\ce{mn^{2+}(aq) + 2i^{-}(aq) -> mn(s) + i2(s)}\\)\\(\ce{mni2(l) -> mn(s) + 2i(s)}\\)\\(\ce{mn^{2+}(aq) + 2i^{-}(aq) -> mn(s) + 2i(s)}\\)

Explanation:

1. First, recall the state of the substance in electrolysis of molten compound: Manganese(II) iodide (\(MnI_2\)) in molten state is \(MnI_2(l)\), not in aqueous (\(aq\)) state. So options with \(aq\) (aqueous) can be eliminated.
2. Then, recall the products of electrolysis of molten \(MnI_2\): The cation \(Mn^{2+}\) will be reduced to \(Mn(s)\), and the anion \(I^-\) will be oxidized. The oxidation of \(I^-\) (iodide ion) produces \(I_2\) (molecular iodine, solid) rather than \(I(s)\) (atomic iodine, which is not stable in this context). So the correct formula for the product of iodide oxidation is \(I_2(s)\), and the reactant is molten \(MnI_2(l)\).

Answer:

A. \(MnI_2(l)
ightarrow Mn(s)+I_2(s)\)