Animated Solution for Chemistry - d and f-Block Elements: One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

Permanganate's Power in Neutral Waters

Welcome to the fascinating world of redox reactions, where electrons are the currency and transition metals are the master bankers. Today, we are diving deep into a classic reaction that frequently appears in competitive exams: the oxidation of iodide ions by the mighty permanganate ion. But there is a twist—the medium is neutral.

Potassium Permanganate

Potassium permanganate ($KMn{O}_4$) is a staple in any chemistry lab, famous for its intense purple color. It is a powerful oxidizing agent, meaning it loves to snatch electrons from other species. However, its behavior is not one-size-fits-all; it is highly dependent on the pH of the solution it finds itself in.

In an acidic medium, permanganate is an absolute beast. It gets reduced all the way from a $+7$ oxidation state to a $+2$ oxidation state ($M{n}^{2+}$), gaining 5 electrons in the process.

But what happens when we remove the acid? In a neutral or faintly alkaline medium, permanganate becomes a bit milder. It only manages to grab 3 electrons, reducing from $+7$ to $+4$, and precipitating out as the dark brown solid, manganese dioxide ($Mn{O}_2$).

Now, let's look at our victim—the iodide ion ($I^{-}$). Iodide is a good reducing agent; it is quite willing to give up its extra electron.

If we were in an acidic medium, the permanganate would oxidize the iodide ion to elemental iodine ($I_2$). The reaction would stop there.

However, the neutral medium changes the thermodynamic landscape entirely. Because the permanganate is not being reduced as drastically (only to $Mn{O}_2$), the reaction pathway shifts. The permanganate continues to oxidize the iodine species further and further, stripping away a total of 6 electrons per iodine atom.

This extensive oxidation pushes the iodine from a $-1$ oxidation state all the way up to a $+5$ oxidation state, resulting in the formation of the iodate ion ($I{O}_3^{-}$).

Let's put it all together by looking at the half-reactions.

Oxidation Half:

Reduction Half:

To balance the electrons, we must multiply the reduction half-reaction by 2. Adding them together yields the final, beautifully balanced redox equation:

This reaction is a perfect example of why you must always read the question carefully. The word "neutral" completely changes the outcome. While acidic permanganate gives you $I_2$, neutral permanganate gives you $I{O}_3^{-}$.

By understanding the underlying electron transfer and the influence of the medium, you don't just memorize a fact; you master the chemical logic. Keep this principle in mind, and you will never fall for this classic trap again!