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Mn^(3+) ions are unstable in solution and undergo disproportionation to give Mn^(2+), MnO_(2) and H^(+) ions. What will be the balanced equation for the reaction ? |
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Answer» Solution :The skeletal equation is `Mn^(3+)(aq)toMn^(2+)(aq)+MnO_(2)(s)toH^(+)(aq)` The equation can be BALANCED as follows. (i) Writing the oxidation numbers of all atoms, we have `overset(+3)(Mn^(3+))(aq)tooverset(+2)(Mn^(2+))(aq)+overset(+4-2)(MnO_(2))(s)+overset(+1)(H^(+))(aq)` Obviously, `Mn^(3+)` is simultaneously reducing to `Mn^(2+)` and OXIDISING to `MnO_(2)` (ii) THEREFORE, the half reactions corresponding to reduction and oxidation processes are as follows: `Mn^(3+)(aq)toMn^(2+)(aq)` (reduction half reaction) `Mn^(3+)(aq)toMnO_(2)(s)` (oxidatio half reaction ) (iii) Balancing of reduction half reaction : (a) The number of Mn atoms is the same on the two sides. (B) There is no O ATOM involved in the reaction. (c) Balancing charge by adding electron, we have `Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)` (balanced reduction half reaction) (iv) Balancing of oxidation half reaction : (a) The number of Mn atoms is the same on the two sides. Since, the reaction proceeds in acidic medium, O atom can be balanced by the addition of two `H_(2)O` molecules on the left hand side. `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)` Balancing H atoms, we have `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)` (c) Balancing charge by adding electrons, we have `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)` (balanced oxidation half reaction) (d) Adding balanced half reactions, we get `{:(""Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)),(Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)),(bar(2Mn^(3+)(aq)+2H_(2)O(l)toMn^(2+)(aq)+MnO_(2)(s)+4H^(+))):}` This is the balanced ionic equation for the given reaction. |
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