The $\mathrm{Mn}^{3+}$ Ion is unstable in solution and undergoes disproportionation to give $\mathrm

Question

The $\mathrm{Mn}^{3+}$ Ion is unstable in solution and undergoes disproportionation to give $\mathrm{Mn}^{2+}, \mathrm{MnO}_2$ and $\mathrm{H}^{+}$lon. Write a balanced ionic equation for the reaction and stoichiometric coefficients for $\mathrm{Mn}^{3+}, \mathrm{MnO}_2$ and $\mathrm{H}^{+}$May be

Professor PC Thomas Classes · Answer

The skeletal equation is,
$$
\mathrm{Mn}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{MnO}_2(\mathrm{~s})+\mathrm{H}^{+}(\mathrm{aq})
$$

Reduction half reaction
$$
\mathrm{Mn}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}
$$

Oxidation half reaction
$$
\mathrm{Mn}^{3+}(\mathrm{aq}) \longrightarrow \stackrel{+4}{\mathrm{MnO}_2}+\mathrm{e}^{-}
$$

Balance charge by adding $4 \mathrm{H}^{+}$to RHS and then balance O atoms by adding $2 \mathrm{H}_2 \mathrm{O}$ to LHS
$$
\mathrm{Mn}^{3+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{MnO}_2(\mathrm{~s})+\mathrm{e}^{-}+4 \mathrm{H}^{+}(\mathrm{aq})
$$

By adding (i) and (iii) we get
$$
2 \mathrm{Mn}^{3+}(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Mn}^{2+}+\mathrm{MnO}_2(\mathrm{~s})+4 \mathrm{H}^{+}(\mathrm{aq})
$$
This represents the final balanced redox reaction (disproportionation reaction).