Write reactions of KMnO₄ as oxidizing agent in acidic, neutral, and basic medium

medium CBSE JEE-MAIN NCERT Class 12 4 min read

Question

Write the reactions of KMnO₄ as an oxidising agent in (i) acidic medium, (ii) neutral medium, and (iii) basic medium.


Solution — Step by Step

The key is what product manganese forms — that depends entirely on the medium. The oxidation state of Mn changes differently in each case, and the number of electrons transferred changes too.

MediumMn oxidation stateProductColour change
Acidic+7 → +2Mn²⁺ (MnSO₄)Purple → colourless
Neutral+7 → +4MnO₂Purple → brown ppt
Basic+7 → +6MnO₄²⁻Purple → green

This is the memory anchor for the entire question.

In the presence of dilute H2SO4\text{H}_2\text{SO}_4, KMnO4\text{KMnO}_4 is the strongest oxidising agent of the three cases, gaining 5 electrons per Mn atom.

Half-reaction:

MnO4+8H++5eMn2++4H2O\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Reaction with FeSO₄ (NCERT example):

2KMnO4+10FeSO4+8H2SO42MnSO4+5Fe2(SO4)3+K2SO4+8H2O2\text{KMnO}_4 + 10\text{FeSO}_4 + 8\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + 8\text{H}_2\text{O}

Reaction with oxalic acid (PYQ favourite):

2KMnO4+5H2C2O4+3H2SO42MnSO4+K2SO4+10CO2+8H2O2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + \text{K}_2\text{SO}_4 + 10\text{CO}_2 + 8\text{H}_2\text{O}

The solution turns from purple to colourless — the visual confirmation that Mn²⁺ has formed.

With no acid or alkali, KMnO4\text{KMnO}_4 only gains 3 electrons. The product is MnO2\text{MnO}_2, which precipitates as a brown solid.

Half-reaction:

MnO4+2H2O+3eMnO2+4OH\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-

Reaction with Na₂SO₃:

2KMnO4+3Na2SO3+H2O2MnO2+3Na2SO4+2KOH2\text{KMnO}_4 + 3\text{Na}_2\text{SO}_3 + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2\downarrow + 3\text{Na}_2\text{SO}_4 + 2\text{KOH}

You can also write the simpler NCERT form showing nascent oxygen release:

2KMnO4+H2O2MnO2+2KOH+3[O]2\text{KMnO}_4 + \text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 2\text{KOH} + 3[\text{O}]

In the presence of alkali, KMnO4\text{KMnO}_4 is the weakest oxidising agent — it only gains 1 electron, converting to green manganate (MnO42\text{MnO}_4^{2-}).

Half-reaction:

MnO4+eMnO42\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}

Reaction with Na₂SO₃:

2KMnO4+Na2SO3+2KOH2K2MnO4+Na2SO4+H2O2\text{KMnO}_4 + \text{Na}_2\text{SO}_3 + 2\text{KOH} \rightarrow 2\text{K}_2\text{MnO}_4 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

The solution turns from purple to green — the signature of K2MnO4\text{K}_2\text{MnO}_4.


Why This Works

The medium controls the availability of H+\text{H}^+ ions, which directly affects how many electrons MnO4\text{MnO}_4^- can accept. More H+\text{H}^+ available → more electrons transferred → stronger oxidising power.

In acidic medium, the 8 H+\text{H}^+ ions stabilise the Mn2+\text{Mn}^{2+} product by pulling the equilibrium hard to the right. In neutral medium, there’s only enough proton availability to reach MnO2\text{MnO}_2 (+4 state). In basic medium, the OH⁻ actually competes with the reduction, so only a 1-electron change to MnO42\text{MnO}_4^{2-} (+6 state) occurs.

This is why the titration with oxalic acid in Mohr’s method requires acidification — without H2SO4\text{H}_2\text{SO}_4, the reaction stops at MnO2\text{MnO}_2 and the endpoint is obscured by the brown precipitate.


Alternative Method — Electron Counting Shortcut

Instead of balancing by inspection, use the electron transfer method to verify your equation quickly.

For acidic medium with FeSO₄:

  • Mn gains 5 electrons (per formula unit): +7+2+7 \rightarrow +2
  • Fe loses 1 electron: +2+3+2 \rightarrow +3
  • To balance: 2 KMnO₄ (10e⁻ gained) needs 10 FeSO₄ (10e⁻ lost) ✓

For board exams, always write the colour change alongside each reaction. Examiners specifically award marks for: acidic = colourless, neutral = brown precipitate, basic = green. A correct equation with no colour mention often loses half a mark.


Common Mistake

Students write MnO42\text{MnO}_4^{2-} (manganate) as the product in neutral medium instead of MnO2\text{MnO}_2. Remember: manganate is the basic medium product. In neutral medium, Mn goes to the +4 state (MnO2\text{MnO}_2), not the +6 state. The oxidation states in order are: acidic → +2, neutral → +4, basic → +6. They increase as the medium becomes less acidic. That’s your anchor.

This question carries 3 marks in CBSE Class 12 boards and appears almost every alternate year. Write all three reactions with the correct balancing and colour changes — don’t skip the neutral medium one, which students often forget under exam pressure.

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