Oxidation Product of Iodide Ion with Permanganate in Neutral Medium

To determine the product formed from the oxidation of iodide ion ($\text{I}^-$) by permanganate ion ($\text{MnO}_4^-$) in a neutral (or faintly alkaline) aqueous medium, we analyze the redox behavior of potassium permanganate in different media:

1. In Acidic Medium: In acidic solution, $\text{MnO}_4^-$ acts as a strong oxidizing agent and is reduced to $\text{Mn}^{2+}$, while $\text{I}^-$ is oxidized to molecular iodine ($\text{I}_2$): $2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{I}_2 + 8\text{H}_2\text{O}$

2. In Neutral or Faintly Alkaline Medium: In a neutral or weakly alkaline medium, $\text{MnO}_4^-$ is reduced to manganese dioxide ($\text{MnO}_2$), while the iodide ion ($\text{I}^-$) undergoes further oxidation to form the iodate ion ($\text{IO}_3^-$).

The half-reactions involved are:

• Reduction Half-Reaction: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \quad (\times 2)$

• Oxidation Half-Reaction: $\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$

Combining these two half-reactions gives the overall balanced redox equation: $2\text{MnO}_4^- + \text{I}^- + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + \text{IO}_3^- + 2\text{OH}^-$

Hence, the oxidation product of the iodide ion ($\text{I}^-$) in a neutral aqueous medium is the iodate ion ($\text{IO}_3^-$).

Correct Option: (B) $\text{IO}_3^-$