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:

\[\ce{Mn^{(3+)}_{(aq)} -> Mn^{(2+)}_{(aq)} + MnO_{2(s)} + H+_{(aq)}}\]

The oxidation half-equation is:

\[\ce{^{+3}Mn^{3+}_{(aq)} -> ^{+4}MnO_{2(s)}}\]

The oxidation number is balanced by adding one electron as:

\[\ce{Mn^{3+}_{(aq)} -> MnO_{2(s)} + e-}\]

The charge is balanced by adding 4H⁺ ions as:

\[\ce{Mn^{3+}_{(aq)} -> MnO_{2(s)} + 4H_{(aq)}^ + e-}\]

The O atoms and H⁺ ions are balanced by adding 2H₂O molecules as:

\[\ce{Mn^{3+}_{(aq)} + 2H2O_{(l)} -> MnO_{2(s)} + 4H+_{(aq)} + e-}\]    .....(i)

The reduction half equation is:

\[\ce{Mn^{3+}_{(aq)} -> Mn^{2+}_{(aq)}}\]

The oxidation number is balanced by adding one electron as:

\[\ce{Mn^{3+}_{(aq)} + e- -> Mn^{2+}_{(aq)}}\]  ....(ii)

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

\[\ce{2Mn^{3+}_{(aq)} + 2H_2O_{(l)} -> MnO_{2(s)} + 2Mn^{2+}_{(aq)} + 4H+_{(aq)}}\]