The $M n^{3 +}$ ion is unstable in solution and undergoes disproportionation to give $M n^{2 +}, M n O_{2}$ and $H^{+}$ ion. Write a balanced ionic equation for the reaction.

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Solution

The unbalanced chemical reaction is:
$M n^{3 +} (a q) \to M n^{2 +} (a q) + M n O_{2} (s) + H^{+} (a q)$
The oxidation half reaction is,
$M n^{3 +} (a q) \to M n O_{2} (s)$ .
To balance oxidation number, one electron is added on R.H.S.
$M n^{3 +} (a q) \to M n O_{2} (s) + e^{-}$
4 protons are added to balance the charge.
$M n^{3 +} (a q) \to M n O_{2} (s) + 4 H^{+} (a q) + e^{-}$
2 water molecules are added to balance O atoms.
The reduction half reaction is $M n^{3 +} (a q) \to M n^{2 +} (a q)$ .
An electron is added to balance oxidation number.
$M n^{3 +} (a q) + e^{-} \to M n^{2 +} (a q)$
Two half cell reactions are added to obtain balanced chemical equation.
$2 M n^{3 +} (a q) + 2 H_{2} O (l) \to M n^{2 +} (a q) + M n O_{2} (s) + 4 H^{+} (a q)$

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