Balancing the chemical equation using Oxidation Number Method Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.


Please balance the equation using the oxidation number method. The skeletal ionic equation is:

$$\ce{MnO4-(aq) + Br-(aq) -&gt; MnO2(s) + BrO3-(aq)}$$

This question is asked in Problem 8.9, NCERT Textbook Class XI, Pg. no. 267. My answer to this question is different from the answer given in the book. I'd love to know the correct answer to this question.
It's hard to pinpoint what the problem here is since I don't have the mentioned book to compare an answer with.
The simplest way is write down half-reactions for reduction (red) and oxidation (ox) processes once you've assigned oxidation numbers (denoted above the symbols of the elements which are participating in a redox reaction), then balance the number of the transferred electrons and the algebraic sum of both half-reactions will yield in a complete balanced redox reaction:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &amp;→ \overset{+4}{Mn}O2 + 4 OH-} &amp;\quad |\cdot 2 \tag{red}\\
\ce{\overset{-1}{Br}^- + 6 OH- &amp;→ \overset{+5}{Br}O3- + 3 H2O + 6 e-} &amp;\quad \tag{ox}\\
\hline
\ce{2 MnO4- + Br- + H2O &amp;→ 2 MnO2 + BrO3- + 2 OH-} \tag{redox}
\end{align}
$$

As a sanity check, you can see that there are no protons $\ce{H+}$ participating in either of the reactions, and the resulting medium is indeed basic (slight excess of $\ce{OH-}$), which is in a good agreement with the conditions mentioned in the problem.