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 given reaction can be represented as:
$M n_{(a q)}^{3 +} ⟶ M n_{(a q)}^{2 +} + M n O_{2 (s)} + H_{(a q)}^{+}$
The oxidation half equation is:
$+ 3 M n_{(a q)}^{3 +} ⟶ + 4 M n O_{2 (s)}$
The oxidation number is balanced by adding on electron as:
$M n_{(a q)}^{3 +} ⟶ M n O_{2 (s)} + e^{-}$
The charge is balanced by adding $4 H^{+}$ ions as:
$M n_{(a q)}^{3 +} + 2 H_{2} O_{(l)} ⟶ M n O_{2 (s)} + 4 H_{(a q)}^{+} + e^{-} \dots (i)$
The reduction half equation is:
$M n_{(a q)}^{3 +} ⟶ M n_{(a q)}^{2 +}$
The oxidation number is balanced by adding one electron as:
$M n_{(a q)}^{3 +} + e^{-} ⟶ M n_{(a q)}^{2 +} \dots (i i)$
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
$2 M n_{(a q)}^{3 +} + 2 H_{2} O_{(l)} ⟶ M n O_{2 (s)} + M n_{(a q)}^{2 +} + 4 H_{(a q)}^{+}$

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