Question

The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺ , MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.

Answer 1

The given reaction can be represented as:
\(Mn^{3+}(aq)\rightarrow Mn^{2+}(aq)+MnO_2(s)+H^+(aq)\)

The oxidation half equation is:
\(\overset{+3}Mn^{3+}_{(aq)}\rightarrow \overset {+4}MnO_2(s)\)

The oxidation number is balanced by adding one electron as:
\(Mn^{3+}(aq )\rightarrow+MnO_2(s)+e^−\)

The charge is balanced by adding \(4H ^+\) ions as:
\(Mn^{3+}(aq)\rightarrow+MnO_2(s)+4H^+(aq)+e^-\)

The \(O\) atoms and \(H ^+\) ions are balanced by adding \(2H_2O\) molecules as:
\(Mn^{3+}(aq)+2H_2O\rightarrow +MnO_2(s)+4H^+e^−\)…..(i)

The reduction half equation is:
\(Mn^{3+}(aq)\rightarrow Mn^{2+}(aq)\)

The oxidation number is balanced by adding one electron as:
\(Mn^{3+}(aq)+e^−\rightarrow Mn^{2+}(aq)\)

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
\(Mn^{3+}(aq)+2H_2O(l)\rightarrow MnO_2(s)+2Mn^{2+}(aq)+4H^+(aq)\)