`Mn^(3+)` ions are unstable in solution and undergo disproportionation to give `Mn^(2+)`, `MnO_(2)` and `H^(+)` ions. What will be the balanced equation for the reaction ?

Explanation

The skeletal equation is
`Mn^(3+)(aq)toMn^(2+)(aq)+MnO_(2)(s)toH^(+)(aq)`
The equation can be balanced as follows.
(i) Writing the oxidation numbers of all atoms, we have
`overset(+3)(Mn^(3+))(aq)tooverset(+2)(Mn^(2+))(aq)+overset(+4-2)(MnO_(2))(s)+overset(+1)(H^(+))(aq)`
Obviously, `Mn^(3+)` is simultaneously reducing to `Mn^(2+)` and oxidising to `MnO_(2)`
(ii) Therefore, the half reactions corresponding to reduction and oxidation processes are as follows:
`Mn^(3+)(aq)toMn^(2+)(aq)` (reduction half reaction)
`Mn^(3+)(aq)toMnO_(2)(s)` (oxidatio half reaction )
(iii) Balancing of reduction half reaction :
(a) The number of Mn atoms is the same on the two sides.
(b) There is no O atom involved in the reaction.
(c) Balancing charge by adding electron, we have
`Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)` (balanced reduction half reaction)
(iv) Balancing of oxidation half reaction :
(a) The number of Mn atoms is the same on the two sides.
Since, the reaction proceeds in acidic medium, O atom can be balanced by the addition of two `H_(2)O` molecules on the left hand side.
`Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)`
Balancing H atoms, we have
`Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)`
(c) Balancing charge by adding electrons, we have
`Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)` (balanced oxidation half reaction)
(d) Adding balanced half reactions, we get
`{:(" "Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)),(Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)),(bar(2Mn^(3+)(aq)+2H_(2)O(l)toMn^(2+)(aq)+MnO_(2)(s)+4H^(+))):}`
This is the balanced ionic equation for the given reaction.