Question

\(MnO_4^- + Br^- + H_2O \rightarrow MnO_2 + BrO_3^- + OH^-\). In the balanced reaction, the coefficients of \(MnO_4^-\), \(BrO_3^-\) and \(OH^-\) are respectively:

• A. \(1,\,1,\,2\)
• B. \(2,\,1,\,4\)
• C. \(2,\,1,\,2\)
• D. \(1,\,2,\,2\)

Hint:

For redox reactions in \textbf{basic medium}:
Balance atoms other than O and H first
Balance O using \(H_2O\)
Balance H using \(OH^-\)
Balance charge using electrons

Answer 1

The Correct Option is C

Step 1: Identify oxidation states. \[ Mn: +7 \rightarrow +4 \quad (\text{reduction}) \] \[ Br: -1 \rightarrow +5 \quad (\text{oxidation}) \]
Step 2: Write half-reactions in basic medium. Reduction half-reaction: \[ MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^- \] Oxidation half-reaction: \[ Br^- + 6OH^- \rightarrow BrO_3^- + 3H_2O + 6e^- \]
Step 3: Equalize electrons by multiplying the reduction half-reaction by 2. \[ 2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^- \]
Step 4: Add the two half-reactions and cancel common terms. \[ 2MnO_4^- + Br^- + H_2O \rightarrow 2MnO_2 + BrO_3^- + 2OH^- \]
Step 5: Hence, the coefficients of \(MnO_4^-\), \(BrO_3^-\), and \(OH^-\) are: \[ \boxed{2,\;1,\;2} \]