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

Explanation

Step 1 : The skeletal ionic equation is :
`MnO_(4)^(-) (aq) + Br^(-)(aq) rarr MnO_(2)(s) + BrO_(3)^(-)(aq)`
Step 2 : Assign oxidation numbers for Mn and Br
`overset(+7)(Mn)O_(4)^(-)(aq) + overset(-1)(Br^(-))(aq) rarr overset(+4)(Mn)O_(2) (S) + overset(+5)(BrO_(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.
`2 overset(+7)(Mn)O_(4)^(-) (aq) + overset(-1)(Br^(-))(aq) rarr 2 overset(+4)(Mn)O_(2)(s) + overset(+5)(Br)O_(3)^(-) (aq)`
Step 4 : As the reaction occurs in the basic medium, and the ionic charges are not equal on both sies, add 2 `OH^(-)` ions on the right to make ionic charges equal.
`2 MnO_(4)^(-) (aq) + Br^(-) (aq) rarr 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_(2)O` molecule) on the left side to achieve balanced redox change.
`2 MnO_(4)^(-) (aq) + Br^(-) (aq) + H_(2)O (I) rarr 2MnO_(2) (s) + BrO_(3) (aq) + 2OH^(-) (aq)`