Ncert Exercises & Solutions

8.21 : 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.

NEETprep Answer:

The given reaction can be represented as:

{Mn}^{3 +}_{(aq)} \to {Mn}^{2 +}_{(aq)} + {MnO}_{2}_{(g)} + {H^{+}}_{(aq)}

The oxidation half equation is:

\overset{+ 3}{M} {n^{3 +}}_{(a q)} \to \overset{+ 4}{M} {nO}_{2 (s)}

The oxidation number is balanced by adding one electron as:

\overset{}{M} {n^{3 +}}_{(a q)} \to \overset{}{M} {nO}_{2 (s)} + e^{-}

The charge is balanced by adding 4

H^{+}

ions as:

\overset{}{M} {n^{3 +}}_{(a q)} \to \overset{}{M} {nO}_{2 (s)} + 4 {H^{+}}_{(aq)} + e^{-}

The O atoms and

H^{+}

ions are balanced by adding

2 H_{2} O

molecules as:

\overset{}{M} {n^{3 +}}_{(a q)} + 2 H_{2} O_{(l)} \to \overset{}{M} {nO}_{2 (s)} + 4 {H^{+}}_{(aq)} + e^{-} . . . . . . . . (i)

The reduction half equation is:

\overset{}{M} {n^{3 +}}_{(a q)} \to \overset{}{M} {n^{2 +}}_{(a q)}

The oxidation number is balanced by adding one electron as:

\overset{}{M} {n^{3 +}}_{(a q)} + e^{-} \to \overset{}{M} {n^{2 +}}_{(aq)} . . . . . . . (ii)

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

2 {Mn}^{3 +}_{(aq)} + 2 H_{2} O_{(l)} \to {MnO}_{2 (s)} + 2 {Mn}^{2 +} (aq) + 4 {H^{+}}_{(aq)}

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