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

$\text{Step 1: Skeletal ionic equation}$

The skeletal ionic equation is:

$Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)$

$\text{Step 2: Assigning oxidation numbers for Mn and Br}$

$+7$ $-1$ $+4$ $+5$
$Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)$

This indicates that permanganate ion is the oxidant and bromide ion is the reductant.

$\text{Step 3: Calculate the increase and decrease of oxidation number}$

Calculate the increase and decrease of oxidation number, and make them equal. From step 2 we can notice that there is change in oxidation state of Mn, It changes from $+7$ to $+4.$ There is decrease of $+3$ in oxidation state of $Mn$ on right hand side of the equation. Oxidation state of $Br$ changes from $-1$ to $+5$.

$+7$ $-1$ $+4$ $+5$
$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to 2Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)$

$\text{Step 4: Make ionic charges equal on both sides of the reaction}$

As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add $2O{H}^{–}$ ions on the right to make ionic charges equal

$+7$ $-1$ $+4$ $+5$
$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)\to 2Mn{O}_2\left(s\right)+Br{O}_4^{-}\left(aq\right)+2O{H}^{-}\left(aq\right)$

$\text{Step 5: Count the hydrogen atoms and add appropriate number of water molecules}$

Finally, count the hydrogen atoms and add appropriate number of water molecules $(i.e.one{H}_{2}Omolecule)$ on the left side to achieve balanced redox change

$2Mn{O}_4^{-}\left(aq\right)+B{r}^{-}\left(aq\right)+H_{2}O\left(l\right)\to 2Mn{O}_2\left(s\right)+Br{O}_3^{-}\left(aq\right)+2O{H}^{-}\left(aq\right)$