1)\\\\(2MnO_4{^-}(aq)+Br-(aq) +H_2O(l) \\longrightarrow 2MnO_2(s)+BrO_3(aq) + 2OH^-(aq)\\\\)Here, instead of solving the overall reaction we solve as half reactions. The two half reactions occurring here are written as:\\\\(Br^- + 3H_2O \\longrightarrow BrO_3{^-} + 6H^+ + 6e^-\\\\) -----(Oxidation)\\\\(MnO_4{^-} + 4H^+ + 3e^- → MnO_2 +2H_2O\\\\)----- (Reduction)As in the reduction reaction no. of e- involved is half that of the one in the oxidation we will multiply the reaction by 2 so it becomes \\\\(2MnO_4{^-} + 8H^+ + 6e^- \\longrightarrow 2MnO_2 + 4H_2O\\\\)Thus, reaction can be balanced.