Question

In the neutral or faintly alkaline medium, KMnO₄ oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from

• A. (+7 to +4)
• B. (+6 to +4)
• C. (+7 to +3)
• D. (+6 to +5)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Potassium permanganate (\(KMnO_{4}\)) is a strong oxidizing agent. Its reduction product depends on the pH of the medium.
Detailed Explanation:
In neutral or faintly alkaline medium, the reaction is:
\(2MnO_{4}^{-} + H_{2}O + I^{-} \rightarrow 2MnO_{2} + 2OH^{-} + IO_{3}^{-}\)
1. In \(KMnO_{4}\) (\(MnO_{4}^{-}\)), the oxidation state of \(Mn\) is \(+7\).
2. The product formed in neutral/alkaline medium is Manganese dioxide (\(MnO_{2}\)).
3. In \(MnO_{2}\), let the oxidation state of \(Mn\) be \(x\): \(x + 2(-2) = 0 \Rightarrow x = +4\).
The change is from \(+7\) to \(+4\).
Step 2: Final Answer:
The oxidation state of Manganese changes from +7 to +4.