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Question
The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.
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Solution
The given reaction can be represented as:
\[\ce{Mn^{(3+)}_{(aq)} -> Mn^{(2+)}_{(aq)} + MnO_{2(s)} + H+_{(aq)}}\]
The oxidation half-equation is:
\[\ce{^{+3}Mn^{3+}_{(aq)} -> ^{+4}MnO_{2(s)}}\]
The oxidation number is balanced by adding one electron as:
\[\ce{Mn^{3+}_{(aq)} -> MnO_{2(s)} + e-}\]
The charge is balanced by adding 4H+ ions as:
\[\ce{Mn^{3+}_{(aq)} -> MnO_{2(s)} + 4H_{(aq)}^ + e-}\]
The O atoms and H+ ions are balanced by adding 2H2O molecules as:
\[\ce{Mn^{3+}_{(aq)} + 2H2O_{(l)} -> MnO_{2(s)} + 4H+_{(aq)} + e-}\] .....(i)
The reduction half equation is:
\[\ce{Mn^{3+}_{(aq)} -> Mn^{2+}_{(aq)}}\]
The oxidation number is balanced by adding one electron as:
\[\ce{Mn^{3+}_{(aq)} + e- -> Mn^{2+}_{(aq)}}\] ....(ii)
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
\[\ce{2Mn^{3+}_{(aq)} + 2H_2O_{(l)} -> MnO_{2(s)} + 2Mn^{2+}_{(aq)} + 4H+_{(aq)}}\]
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