InterviewSolution
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InterviewSolution
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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. |
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Answer» 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))^(+)` |
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