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

Concept: Redox Reaction Stoichiometry — balancing by the ion-electron method in basic medium.

Step 1: Write the half-reactions.

Reduction: $\text{MnO}_4^- \rightarrow \text{MnO}_2$

Oxidation: $\text{Br}^- \rightarrow \text{BrO}_3^-$

Step 2: Balance atoms other than H and O, then balance O with $\text{H}_2\text{O}$ and H with $\text{H}^+$ (in basic medium, add $\text{OH}^-$ to neutralise $\text{H}^+$).

Reduction: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

Oxidation: $\text{Br}^- + 6\text{OH}^- \rightarrow \text{BrO}_3^- + 3\text{H}_2\text{O} + 6e^-$

Step 3: Equalise electrons and add. Multiply reduction by 2:

$2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^-$

Add to oxidation half: …

Why This Approach Works

Balancing redox reactions in basic medium follows a systematic logic. The key idea: electrons lost in oxidation must exactly equal electrons gained in reduction. But in basic solution, we can't just add H⁺ to balance hydrogen — we use water and hydroxide ions instead. The half-reaction method breaks the messy overall reaction into two clean, manageable pieces: one for oxidation, one for reduction. Each half-reaction is balanced separately for atoms and charge, then combined.

Let's walk through it step by step.

Step-by-Step Solution

1. Identify the oxidation and reduction half-reactions

First, assign oxidation states to see what changes.

• In $\text{MnO}_4^-$: Mn is +7. In $\text{MnO}_2$: Mn is +4. So Mn goes from +7 to +4 — that's a reduction (gain of 3 electrons per Mn).
• In $\text{Br}^-$: Br is -1. In $\text{BrO}_3^-$: Br is +5. So Br goes from -1 to +5 — that's an oxidation (loss of 6 electrons per Br).

So:

• Reduction half-reaction: $\text{MnO}_4^- \rightarrow \text{MnO}_2$
• Oxidation half-reaction: $\text{Br}^- \rightarrow \text{BrO}_3^-$

2. Balance the reduction half-reaction (in basic medium)

Start with atoms other than H and O. Mn is already balanced (1 on each side).

Balance oxygen by adding water. Left has 4 O, right has 2 O. Add 2 $\text{H}_2\text{O}$ to the right:

Now balance hydrogen. Right has 4 H, left has none. In basic medium, add $\text{OH}^-$ to the side that needs H, and water to the other side. Here, add 4 $\text{OH}^-$ to the left:

Check charge balance. Left: -1 + 4(-1) = -5. Right: neutral. Add 3 electrons to the left to make charge -5 on both sides:

3. Balance the oxidation half-reaction (in basic medium)

Start: $\text{Br}^- \rightarrow \text{BrO}_3^-$

Balance oxygen. Left has 0 O, right has 3 O. Add 3 $\text{H}_2\text{O}$ to the left:

Balance hydrogen. Left has 6 H, right has 0. Add 6 $\text{OH}^-$ to the right:

Check charge. Left: -1. Right: -1 + 6(-1) = -7. Add 6 electrons to the right to balance:

4. Equalise electrons and add the half-reactions …