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Balance the redox reaction by oxidation number method. MnO_(4)^(-)(aq)+Br^(-)(aq)toMnO_(2)(s)+BrO_(2)^(-)(aq) (Basic medium) |
Answer» Solution : `MnO_(4)^(-) overset(1_(CANCEL(3))"UNITS")(to)MnO_(2)to` REDUCTION `Br^(-)overset(2_(cancel(6))"units")(to)Bro_(3)^(-)to`Oxidation `x^(1y)` The reduction EQN by 2 units and oxidation eqn by 1 unit `{:(MnO_(4)^(-)"" to ""2MnO_(2)),(Br^(-)""to""BrO_(3)^(-)),(______________________________),(2MnO_(4)^(-)+Br^(-)to2MnO_(2)+BrO_(3)^(-)):}` As the Reaction occurs in basic medium to equal ionic charges `20 H^(-)` are added on RHS. To balance hydrogen atom add `H_(2)O` on LHS. `2MnO_(4)^(-)+Br^(-)+H_(2)Oto2MnO_(2)+BrO_(3)^(-)+2OH^(-)` |
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