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.

Answer 1

"The given reaction is as follows

$M{ n }^{ 3+ }(aq)\quad \rightarrow \quad M{ n }^{ 2+ }(s)\quad +\quad Mn{ O }_{ 2 }(s)\quad +\quad { H }^{ + }(aq)$

The oxidation half reaction is as follows

$\begin{matrix} { Mn }_{ (aq) }^{ 3+ } \\ +3 \end{matrix}\quad \rightarrow \quad \begin{matrix} { MnO }_{ 2(s) } \\ +4 \end{matrix}$

Let's balance the electrons

$M{ n }^{ 3+ }(aq)\quad \rightarrow \quad Mn{ O }_{ 2 }(s)\quad +\quad { e }^{ - }$

Balance the charges by adding $4{ H }^{ + } ions$

$M{ n }^{ 3+ }(aq)\quad \rightarrow \quad Mn{ O }_{ 2 }(s)\quad +\quad 4{ H }^{ + }\quad +\quad { e }^{ - }$

Balance the oxygen atoms by adding${ H }_{ 2 }O$ molecules

$M{ n }^{ 3+ }(aq)\quad +\quad 2{ H }_{ 2 }O\quad (l)\quad \quad \rightarrow \quad Mn{ O }_{ 2 }(s)\quad +\quad 4{ H }^{ + }\quad +\quad { e }^{ - }\quad$............(i)

The reduction half equation

${ Mn }^{ 3+ }(aq)\quad \quad \longrightarrow \quad M{ n }^{ 2+ }(aq)$

The oxidation number is balanced by adding one electron as

${ Mn }^{ 3+ }(aq)\quad +\quad { e }^{ - }\quad \quad \longrightarrow \quad M{ n }^{ 2+ }(aq)$.........(ii)

The balanced chemical equation can be obtained by adding equation (i) and (ii)

${ 2Mn }^{ 3+ }(aq)\quad +\quad 2{ H }_{ 2 }O\quad (l)\quad ^{ - }\quad \quad \longrightarrow \quad M{ n }{ O }_{ 2 }(s)\quad +\quad { 2Mn }^{ 2+ }(aq)\quad +\quad 4{ H }^{ + }$"