Q.$KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with KI, iodide ion is oxidised to ____________.

Concept: Inorganic Synthesis — redox behaviour of $KMnO_4$ in alkaline medium.

Reasoning:

1. In alkaline medium, $KMnO_4$ is reduced to $MnO_2$ (manganese dioxide, +4 state) and the half-reaction is: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ …

The key to this question lies in understanding how the oxidising power of $KMnO_4$ changes with pH. In acidic medium, $MnO_4^-$ is reduced to $\ce{Mn^{2+}}$ (a 5-electron change) and is a very strong oxidant — it can oxidise $\ce{I-}$ to $I_2$ easily. But in alkaline medium, the reduction product is different, and so is the extent of oxidation it can achieve.

When the medium is alkaline, $MnO_4^-$ typically reduces to $MnO_2$ (manganese dioxide, oxidation state +4) or, in strongly alkaline conditions, to manganate ion $MnO_4^{2-}$ (oxidation state +6). The number of electrons gained per $MnO_4^-$ is smaller (3 electrons for $MnO_2$, 1 electron for $MnO_4^{2-}$), so the oxidising power per mole is less intense. However, the reaction is still vigorous enough to push iodide beyond elemental iodine.

Iodide ion $\ce{I-}$ (oxidation state -1) can be oxidised stepwise: first to $I_2$ (0), then to hypoiodite $IO^-$ (+1), then to iodite $IO_2^-$ (+3), then to iodate $IO_3^-$ (+5), and finally to periodate $IO_4^-$ (+7). In alkaline medium, $KMnO_4$ is strong enough to take it to the +5 state — iodate — but not to periodate (which requires even stronger oxidants or specific conditions like hot alkaline $KMnO_4$ with a catalyst).

Let’s walk through the actual reaction.

1. Identify the half-reactions. In alkaline medium, the reduction half-reaction for permanganate is:

(This is the most common version; in very concentrated alkali, $MnO_4^{2-}$ forms instead, but the principle is the same.)

2. Oxidation half-reaction for iodide. Iodide is oxidised to iodate:

3. Balance the electrons. The reduction consumes 3 electrons per $MnO_4^-$, the oxidation produces 6 electrons per $I^-$. To balance, we need 2 $MnO_4^-$ for every 1 $I^-$:

4. Add the two half-reactions. Cancel water and hydroxide where possible: …

Concept: Oxidation States & Redox Reactions in Alkaline Medium

The key idea is that the product formed depends on the medium (acidic, neutral, or alkaline). In alkaline medium, $KMnO_4$ is reduced to $MnO_2$ (not $Mn^{2+}$), and the iodide ion is oxidised to a higher state than just $I_2$.

Method: Half-Reaction Balancing in Alkaline Medium

Steps:

1. Identify the half-reactions

   • Reduction: $MnO_4^- \rightarrow MnO_2$ (in alkaline medium)
   • Oxidation: $I^- \rightarrow$ unknown product

2. Balance the reduction half-reaction (alkaline conditions)

   • $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$
   • This shows 3 electrons gained per $MnO_4^-$.

3. Determine the oxidation product of iodide

   • In alkaline medium, $I^-$ is oxidised to $IO_3^-$ (iodate), not $I_2$.
   • Balanced oxidation: $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$

4. Combine the half-reactions …

Common Mistakes & How to Avoid Them

Mistake 1: Assuming the product is always $I_2$

Why students make this mistake:

Students often remember that $KI$ reacts with oxidising agents to give iodine ($I_2$), especially from reactions in acidic medium. They apply this memory without checking the medium.

How to avoid:

Always check the medium (acidic, alkaline, neutral) before deciding the product. In alkaline medium, $KMnO_4$ is a stronger oxidising agent and can oxidise $I^-$ beyond $I_2$.

Mistake 2: Forgetting the oxidation state change of Mn in alkaline medium

Why students make this mistake:

Students memorise that $KMnO_4$ in acidic medium gives $Mn^{2+}$ (change from +7 to +2). They incorrectly assume the same change in alkaline medium.

How to avoid:

Remember the three different reductions of $KMnO_4$:

In alkaline medium, $KMnO_4$ is reduced to $K_2MnO_4$ (manganate, green), gaining only 1 electron per $MnO_4^-$ ion.

Mistake 3: Not balancing the half-reactions to find the correct product

Why students make this mistake:

Students guess the product without writing balanced equations. They may pick $IO_4^-$ thinking "more oxidation is better" without checking electron balance.

How to avoid:

Write and balance the half-reactions:

Reduction half (alkaline medium):

Oxidation half:

Iodide ($I^-$, oxidation state -1) is oxidised. Check possible products:

• $I_2$ (0): loses 1 electron per I atom → 2 electrons per $I_2$ molecule
• $IO^-$ (+1): loses 2 electrons per I atom
• $IO_3^-$ (+5): loses 6 electrons per I atom
• $IO_4^-$ (+7): loses 8 electrons per I atom

Balance electrons:

Each $MnO_4^-$ gains only 1 electron (→ $MnO_4^{2-}$), so the electrons lost in the oxidation half must be supplied by the right number of permanganate ions. In cold alkaline medium the iodide is oxidised all the way to iodate, $IO_3^-$ (oxidation state $+5$), which means each iodine loses 6 electrons:

$I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$ …