Q.A solution of $KMnO_4$ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

The key idea is that permanganate ($\text{MnO}_4^-$) is a powerful oxidising agent whose reduction product depends on the pH of the medium — acidic, neutral, or alkaline.

Step 1: In acidic medium (pH < 7), $\text{MnO}_4^-$ is reduced to colourless $\text{Mn}^{2+}$ ions.

The half-reaction is:

Step 2: In neutral or weakly alkaline medium (pH ≈ 7), reduction yields a brown precipitate of $\text{MnO}_2$.

The half-reaction is:

$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$ …

The key to understanding this lies in the variable oxidation states of manganese and how pH controls the stability of the intermediate species. Permanganate ion ($MnO_4^-$) is a powerful oxidising agent in all media, but the products differ because the reduction potential and the stability of manganese species change dramatically with pH.

Let me walk through each case systematically.

1. Acidic medium (pH < 1–2) In strong acid, the reduction goes all the way to $Mn^{2+}$, which is colourless in dilute solution. The half-reaction is:

The $E^\circ$ is +1.51 V, making it the most powerful oxidising condition.

How to carry it out: Add dilute $H_2SO_4$ to the $KMnO_4$ solution, then add a reducing agent like oxalic acid, $FeSO_4$, or $H_2O_2$. The purple colour fades to colourless as $Mn^{2+}$ forms.

2. Neutral or weakly alkaline medium (pH ~7–9) Here the reduction stops at $MnO_2$, a brown insoluble precipitate. The half-reaction is:

Notice that water provides the oxygen, and hydroxide ions are produced — so the solution becomes alkaline as the reaction proceeds.

How to carry it out: Simply add a reducing agent (like $Na_2SO_3$ or $KI$) to a neutral $KMnO_4$ solution. No acid or strong base is added. The purple colour turns brown as $MnO_2$ precipitates.

3. Strongly alkaline medium (pH > 12) In concentrated alkali, the reduction yields the green manganate ion $MnO_4^{2-}$ (oxidation state +6). The half-reaction is:

This is a one-electron reduction. The green colour is characteristic of $MnO_4^{2-}$.

How to carry it out: Add excess $KOH$ or $NaOH$ to $KMnO_4$ solution (making it strongly alkaline), then add a mild reducing agent like $KI$ or $Na_2SO_3$ in small amounts. Alternatively, you can heat solid $KMnO_4$ with $KOH$ — but that's a different method. …

Method: pH-Dependent Reduction of Permanganate

This problem is solved using the Redox Speciation Method — tracking how the oxidation state of manganese changes with pH.

Concept First (Why this happens)

$KMnO_4$ contains manganese in its +7 oxidation state. The reduction product depends on the $H^+$ concentration because:

• In acidic medium, $H^+$ ions are available to stabilize lower oxidation states as cations.
• In neutral/weakly basic medium, $MnO_2$ (insoluble) forms.
• In strongly basic medium, the manganate ion ($MnO_4^{2-}$) is stable.

Steps of the Method

Step 1: Identify the three reduction stages

Step 2: Write the half-reactions for each case

Acidic medium (colourless $Mn^{2+}$):

Neutral/weakly basic (brown $MnO_2$):

Strongly basic (green $MnO_4^{2-}$):

Step 3: How to carry out each reduction

| Desired product | Reducing agent | Conditions |

|----------------|----------------|------------| …

Here is a breakdown of the common mistakes students make on this classic inorganic synthesis question, along with how to avoid each.

The Core Concept (The "Why")

The key is that $KMnO_4$ (manganese in +7 oxidation state) is a powerful oxidising agent. The pH of the solution dictates the final reduction product of manganese because the reduction half-reaction involves $H^+$ ions.

• In Acidic Medium ($pH < 7$): $MnO_4^-$ is reduced to the colourless $Mn^{2+}$ ion.
• In Neutral/Faintly Alkaline Medium ($pH \approx 7-9$): $MnO_4^-$ is reduced to a brown precipitate of $MnO_2$.
• In Strongly Alkaline Medium ($pH > 10$): $MnO_4^-$ is reduced to a green solution of $MnO_4^{2-}$ (manganate ion).

Common Mistake #1: Confusing the Colour of the Products

The Mistake: Students often mix up which product is formed in which medium. For example, they might say "green solution in acidic medium" or "colourless solution in alkaline medium."

How to Avoid It:

• Memorise the "pH-Colour" Triad: Create a simple mental map.
  • Acid → Colourless ($Mn^{2+}$)
  • Neutral → Brown ($MnO_2$)
  • Alkaline → Green ($MnO_4^{2-}$)
• Use a Mnemonic: "Acid gives Clear, Neutral gives Brown, Alkaline gives Green." (A-C, N-B, A-G).
• Visualise the Ions: Remember that $Mn^{2+}$ is a very pale pink (appears colourless in dilute solution), $MnO_2$ is a solid brown precipitate, and $MnO_4^{2-}$ is a distinct green colour in solution.

Common Mistake #2: Writing the Wrong Half-Reactions

The Mistake: Students write the reduction half-reaction incorrectly, especially forgetting to balance $H^+$ and $H_2O$ or using the wrong number of electrons.

How to Avoid It:

• Always Balance by the "ION-ELECTRON" Method: For each medium, write the balanced half-reaction. This is non-negotiable for exam accuracy.

  1. Acidic Medium (to $Mn^{2+}$):

*Note: 5 electrons are gained. The solution becomes colourless.*

**2. Neutral/Faintly Alkaline Medium (to $MnO_2$):**

*Note: 3 electrons are gained. The brown precipitate is $MnO_2$.*

**3. Strongly Alkaline Medium (to $MnO_4^{2-}$):**

*Note: Only 1 electron is gained. The green colour is due to the manganate ion.*

• Check the Number of Electrons: The number of electrons gained decreases as the pH increases (5 → 3 → 1). This is a good sanity check.

Common Mistake #3: Forgetting the "How" (The Reagents)

The Mistake: Students can state the products but cannot describe how to carry out the reduction (e.g., what reagent to add).

How to Avoid It:

• Learn the Specific Reducing Agents: The question asks "how are they carried out?" You must know the common reagents.

  • For Colourless $Mn^{2+}$ (Acidic): Add a reducing agent like oxalic acid ($H_2C_2O_4$) or ferrous sulphate ($FeSO_4$) in the presence of dilute $H_2SO_4$.

    • Example: $2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$

  • For Brown $MnO_2$ (Neutral): Add a reducing agent like sodium sulphite ($Na_2SO_3$) or hydrogen peroxide ($H_2O_2$) in neutral or faintly alkaline conditions.

    • Example: $2KMnO_4 + 3Na_2SO_3 + H_2O \rightarrow 2MnO_2 + 3Na_2SO_4 + 2KOH$

  • For Green $MnO_4^{2-}$ (Strongly Alkaline): Add a reducing agent like potassium sulphite ($K_2SO_3$) or potassium iodide ($KI$) in a concentrated solution of KOH or NaOH. …