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:

$$\ce{MnO4-(aq) + Br-(aq) -> MnO2(s) + BrO3-(aq)}$$

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.

  • $\begingroup$ You may want tp write down the expected answer, your answer and your reasoning for it. It may prevent closing your question as "not elaborated homowork class question". $\endgroup$ – Poutnik Aug 19 '19 at 10:20

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:

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

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.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.