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:
$Mn^{3+}(aq)+2H_2O(l)\rightarrow MnO_2(s)+2Mn^{2+}(aq)+4H^+(aq)$