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 skeleton ionic equation is :
`MnO_(4)^(-)(aq) + Br^(-) (aq) to 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) to overset(+4)(Mn)O_(2)(s) + overset(+5)(Br)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.
`2 overset(+7)(Mn)O_(4)^(-) (aq) + overset(-1)(Br^(-)) (aq) to 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 sides, add 2 `OH^(-)` ions on the right to make ionic charges equal.
`2MnO_(4)^(-) (aq) + Br^(-)(aq) to 2MnO_(2) (s) + BrO_(3)^(-) (aq) + 2 OH^(-) (aq)`
Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e. one `H2_(O)` molecule) on the left side to achieve balanced redox change.
`2 MnO_(4)^(-) (aq) + Br^(-) (aq) + H_(2)O(1) to 2 MnO_(2) (s) + BrO_(3)^(-) (aq) + 2 OH^(-) (aq)`