Question

The `Mn^(3+)` ion is unstable in solution and undergoes disproportionation reaction to give `Mn^(+2), MnO_(2)`, and `H^(o+)` ion. Write a balanced ionic equation for the reaction.

Answer 1

The give reaction can be represented as :
`Mn_((aq)^(3+) to Mn_((aq))^(2+) + MnO_(2 (s)) + H_((aq))^(+)`
The oxidation half equation is :
`overset(+3)(M)n_((aq))^(3+) to overset(+4)(M)n O_(2 (s))`
The oxidation number is balanced by adding one electron as :
`Mn_((aq))^(3+) to MnO_(2 (s)) + e^(-)`
The charge is balanced by adding `4H^(+)` ions as :
`Mn_((aq))^(3+) to MnO_(2 (s)) + 4 H_((aq))^(+) + e^(-)`
The O atoms and `H^(+)` ions are balanced by adding `2H_(2)O` molecules as :
`Mn_((aq))^(3+) + 2H_(2)O_((l)) to MnO_(2 (s)) + 4 H_((aq))^(+) + e^(-) .... (i)`
The reduction half equation is :
`Mn_((aq))^(3+) to Mn_((aq))^(2+)`
The oxidation number is balanced by adding one electron as :
`Mn_((aq))^(3+) e^(-) to Mn_((aq))^(2+) ....... (ii) `
The balanced chemical equation can be obtained by adding equation (i) and (ii) as :
`2Mn_((aq))^(3+) + 2H_(2)O_((l)) to Mn_((aq))^(2+) + 4 H_((aq))^(+)`