Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

(N/A) Step $1$: The skeletal ionic equation is: $MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow MnO_{2(s)} + BrO_3^{-}{(aq)}$
Step $2$: Assign oxidation numbers for $Mn$ and $Br$: $\mathop{Mn}\limits^{+7}O_4^{-}{(aq)} + \mathop{Br^{-}}\limits^{-1}{(aq)}$ $\rightarrow \mathop{Mn}\limits^{+4}O_{2(s)} + \mathop{Br}\limits^{+5}O_3^{-}{(aq)}$. This indicates that permanganate ion is the oxidant and bromide ion is the reductant.
Step $3$: Calculate the increase and decrease of oxidation number,and make the increase equal to the decrease: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)}$.
Step $4$: As the reaction occurs in the basic medium,and the ionic charges are not equal on both sides,add $2OH^{-}$ ions on the right to make ionic charges equal: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)} + 2OH^{-}{(aq)}$.
Step $5$: Finally,count the hydrogen atoms and add appropriate number of water molecules (i.e.,one $H_2O$ molecule) on the left side to achieve balanced redox change: $2MnO_4^{-}{(aq)} + Br^{-}{(aq)} + H_2O_{(l)} \rightarrow 2MnO_{2(s)} + BrO_3^{-}{(aq)} + 2OH^{-}{(aq)}$.