Question
The $Mn^{3+}$ ion is unstable in solution and undergoes disproportionation to give $Mn^{2+}, MnO_2,$ and $H^+$ ion. Write a balanced ionic equation for the reaction.
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Answer
The given reaction can be represented as: $\text{Mn}_{(\text{aq})}^{3+}\rightarrow\text{Mn}_{(\text{aq})}^{2+}+\text{MnO}_{2(\text{s})}+\text{H}_{(\text{aq})}^+$ The oxidation half equation is: $\stackrel{{+3}}{\ \ \ \ \ \ \ \hbox{Mn}^{3+}}_{(\text{aq})}\rightarrow\stackrel{{+4}}{\ \ \ \ \ \ \ \hbox{MnO}}_{2(\text{s})}$ The oxidation number is balanced by adding one electron as: $\text{Mn}_{(\text{aq})}^{3+}\rightarrow\text{MnO}_{2(\text{s})}+\text{e}^{-}$ The charge is balanced by adding $4H^+$^ ions as: $\text{Mn}_{(\text{aq})}^{3+}\rightarrow\text{MnO}_{2(\text{s})}+4\text{H}_{(\text{aq})}^++\text{e}^{-}$ The O atoms and H^+ ions are balanced by adding $2H_2O$ molecules as: $\text{Mn}_{(\text{aq})}^{3+}+2\text{H}_2\text{O}_{(\text{l})}\rightarrow\text{MnO}_{2(\text{s})}+4\text{H}_{(\text{aq})}^++\text{e}^{-}\ .....(\text{i})$ The reduction half equation is: $\text{Mn}^{3+}_{(\text{aq})}\rightarrow\text{Mn}^{2+}_{(\text{aq})}$ The oxidation number is balanced by adding one electron as: $\text{Mn}^{3+}_{(\text{aq})}+\text{e}^-\rightarrow\text{Mn}^{2+}_{(\text{aq})}\ .....(\text{ii})$The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
$2\text{Mn}_{(\text{aq})}^{3+}+2\text{H}_2\text{O}_{(\text{l})}\rightarrow\text{MnO}_{2(\text{s})}+2\text{Mn}^{2+}_{(\text{aq})}+4\text{H}^{+}_{(\text{aq})}$
$2\text{Mn}_{(\text{aq})}^{3+}+2\text{H}_2\text{O}_{(\text{l})}\rightarrow\text{MnO}_{2(\text{s})}+2\text{Mn}^{2+}_{(\text{aq})}+4\text{H}^{+}_{(\text{aq})}$
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