Q.When $KMnO_4$ solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because

The key idea is autocatalysis — a reaction where one of the products itself acts as a catalyst, speeding up the reaction as it accumulates.

Reasoning:

1. $KMnO_4$ (purple) oxidises oxalic acid to $CO_2$ in acidic medium. The initial reaction is slow because the activation energy is high. …

The key here is autocatalysis. In an autocatalytic reaction, one of the products itself catalyses the reaction. So as the reaction proceeds, the catalyst concentration builds up, and the rate increases dramatically — hence the "slow in the beginning, instantaneous later" behaviour.

Let’s see why this happens specifically with permanganate and oxalic acid.

1. The overall reaction In acidic medium, permanganate ($\text{MnO}_4^-$, purple) oxidises oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) to $\text{CO}_2$, and is itself reduced to $\text{Mn}^{2+}$ (colourless in dilute solution).

The purple colour fades as $\text{MnO}_4^-$ is consumed.

2. Why the slow start?

   The direct reaction between $\text{MnO}_4^-$ and $\text{H}_2\text{C}_2\text{O}_4$ has a high activation energy. Both ions are negatively charged, so they repel each other. Collisions that lead to reaction are rare initially. That’s why the first few drops of permanganate take time to decolourise.

3. What changes?

   As soon as a tiny amount of $\text{Mn}^{2+}$ is formed, something interesting happens. $\text{Mn}^{2+}$ can act as a catalyst for the reduction of $\text{MnO}_4^-$. It does so by providing an alternative, lower-energy pathway — often via intermediate oxidation states like $\text{Mn}^{3+}$ or $\text{Mn}^{4+}$.

4. Autocatalysis in action

   The $\text{Mn}^{2+}$ produced in the reaction speeds up the very reaction that makes more $\text{Mn}^{2+}$. This is the textbook definition of autocatalysis: a product catalyses its own formation.

As $\text{Mn}^{2+}$ concentration rises, the rate shoots up — hence the decolourisation becomes nearly instantaneous.

5. Why not the other options?
   • (i) $\text{CO}_2$ is formed — True, but $\text{CO}_2$ does not catalyse this reaction. It’s just a product. …

Method: Autocatalysis Analysis

This question tests the concept of autocatalysis — where one of the reaction products itself acts as a catalyst for the same reaction.

Step-by-step reasoning:

1. Identify the reaction The reaction between $KMnO_4$ (purple) and oxalic acid ($H_2C_2O_4$) in acidic medium is:

2. Observe the colour change

   • $MnO_4^-$ is purple (coloured).
   • $Mn^{2+}$ is colourless (pale pink in concentrated solutions, but effectively colourless in dilute conditions).
   • Decolourisation means $MnO_4^-$ is being consumed.

3. Why is the decolourisation slow at first?

   • Initially, no $Mn^{2+}$ is present.
   • The direct reaction between $MnO_4^-$ and $H_2C_2O_4$ is slow.

4. Why does it become instantaneous later?

   • As the reaction proceeds, $Mn^{2+}$ is produced.
   • $Mn^{2+}$ catalyses the reaction — it speeds up the reduction of $MnO_4^-$ by oxalic acid.
   • This is autocatalysis: the product ($Mn^{2+}$) catalyses its own formation. …

Here’s a breakdown of the common mistakes students make on this classic autocatalysis question, and how to avoid each one.

Mistake 1: Choosing (iii) — “$MnO_4^-$ catalyses the reaction”

Why students pick it:

They see the purple $MnO_4^-$ ion is involved and assume it must be the catalyst because it “disappears” and the reaction speeds up.

Why it’s wrong:

A catalyst is a substance that speeds up a reaction without being consumed in the overall process. Here, $MnO_4^-$ is consumed (it gets reduced to $Mn^{2+}$). A consumed species cannot be a catalyst.

How to avoid:

Always check: Is the proposed catalyst chemically changed during the reaction? If yes, it’s not a true catalyst — it might be a reactant or an autocatalyst precursor.

Mistake 2: Choosing (i) — “$CO_2$ is formed as the product”

Why students pick it:

They know $CO_2$ is produced, and they think product formation somehow “drives” the reaction faster.

Why it’s wrong:

$CO_2$ is just a product — it does not catalyse the reaction. The decolourisation speed-up is not because $CO_2$ appears, but because a different species ($Mn^{2+}$) builds up.

How to avoid:

Distinguish between products and catalysts. A product can be a catalyst only if it actively participates in speeding up the reaction (autocatalysis). $CO_2$ is inert here.

Mistake 3: Choosing (ii) — “Reaction is exothermic”

Why students pick it:

They think heat from an exothermic reaction “speeds things up” (which is true in a general sense), so they assume that’s the reason for the sudden acceleration.

Why it’s wrong:

While the reaction is exothermic, the initial slow phase happens even though heat is being released from the start. The sudden jump in rate is not due to gradual heating — it’s due to the appearance of $Mn^{2+}$, which then catalyses the reaction. Temperature rise is a secondary effect, not the primary cause.

How to avoid:

Ask: Does the explanation account for the delay? Exothermicity would affect the rate from the beginning, not suddenly after a lag. Autocatalysis explains the lag + sudden speed-up perfectly.

Mistake 4: Not recognising the autocatalysis mechanism

Why students miss (iv):

They don’t know the term “autocatalyst” or the specific reaction steps.

The correct mechanism (why (iv) is right):

1. Slow step (initial): …