What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Exercises · 4.22

Q. What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

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Concept understanding — Disproportionation Reaction

Disproportionation Reactions: The Self-Oxidation-Reduction

The Intuition

Imagine you have a group of friends who are all equally wealthy — each has exactly ₹100. Now suppose one friend decides to give ₹50 to another. After this transaction, one friend has ₹50 (lost money), another has ₹150 (gained money), and the rest are unchanged. Notice something: the same action — transferring money — made one person poorer and another richer.

A disproportionation reaction works on a similar principle, but with electrons instead of money. One atom of an element simultaneously gets oxidised (loses electrons) and reduced (gains electrons). The same element ends up in two different oxidation states — one higher, one lower — starting from a single intermediate oxidation state.

Note

The word "disproportionation" literally means "breaking apart into unequal parts." The original state splits into two different states.

The Precise Definition

A disproportionation reaction is a redox reaction in which a single substance (element or compound) in an intermediate oxidation state is simultaneously oxidised and reduced, producing two different products — one with a higher oxidation state and one with a lower oxidation state.

The general form looks like this:

\text{Element in intermediate state} \longrightarrow \text{Higher oxidation state} + \text{Lower oxidation state}

The Key Condition

For disproportionation to occur, the element must be in an intermediate oxidation state — meaning it can both increase and decrease its oxidation number. If the element is already in its highest possible oxidation state, it can only be reduced. If it's in its lowest, it can only be oxidised. No disproportionation possible.

Important

Disproportionation requires the element to have at least three accessible oxidation states: one lower, one intermediate (the starting point), and one higher.

Classic Example: Hydrogen Peroxide

Hydrogen peroxide ( $H_2O_2$ ) is the textbook example. Oxygen in $H_2O_2$ has an oxidation state of -1. This is intermediate — oxygen can go to 0 (in $O_2$ ) or to -2 (in $H_2O$ ).

When $H_2O_2$ decomposes:

2H_2O_2 \longrightarrow 2H_2O + O_2

Let's track the oxygen:

In $H_2O_2$ : oxidation state = -1
In $H_2O$ : oxidation state = -2 (reduction — gained an electron)
In $O_2$ : oxidation state = 0 (oxidation — lost an electron)

The same oxygen atoms (from the same molecule) undergo both oxidation and reduction. That's disproportionation.

Another Common Example: Copper(I) in Solution

Copper(I) ion ( $Cu^+$ ) is unstable in aqueous solution and disproportionates:

2Cu^+ \longrightarrow Cu + Cu^{2+}

$Cu^+$ (oxidation state +1) is the intermediate
$Cu$ (oxidation state 0) is the reduced product
$Cu^{2+}$ (oxidation state +2) is the oxidised product

Watch out

A common mistake is to think that a single atom does both oxidation and reduction. In reality, two atoms of the same element are involved — one gets oxidised, the other gets reduced. The reaction requires at least two formula units of the starting substance.

How to Identify a Disproportionation Reaction

Look for a single reactant that contains an element in an intermediate oxidation state.
Check the products — the same element must appear in two different oxidation states (one higher, one lower than the starting state). …

Why this formula?

Disproportionation Reaction — Understanding the Why

A disproportionation reaction is a redox reaction where the same element in one oxidation state simultaneously undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number).

The key formula that governs whether such a reaction is spontaneous is based on the standard electrode potentials ( $E^\circ$ ) of the two half-reactions.

The Core Idea: Why Does Disproportionation Happen?

For an element in an intermediate oxidation state, it can be both oxidised and reduced.

Whether this happens spontaneously depends on the relative ease of these two processes.

Consider an element $X$ in oxidation state $+n$ :

Oxidation half-reaction:

$X^{+n} \rightarrow X^{+(n+1)} + e^-$

(loss of electron, oxidation number increases)
Reduction half-reaction:

$X^{+n} + e^- \rightarrow X^{+(n-1)}$

(gain of electron, oxidation number decreases)

The overall disproportionation reaction is:

2X^{+n} \rightarrow X^{+(n+1)} + X^{+(n-1)}

The Key Formula: Spontaneity Condition

For a disproportionation reaction to be spontaneous (under standard conditions), the overall cell potential $E^\circ_{\text{cell}}$ must be positive.

Derivation:

Identify the two half-reactions and their standard reduction potentials ( $E^\circ$ ):
- Reduction half-reaction (the one that gains electrons):
  
  $X^{+n} + e^- \rightarrow X^{+(n-1)}$
  
  Let its standard reduction potential be $E^\circ_{\text{red}}$ .
- Oxidation half-reaction (the one that loses electrons):
  
  $X^{+n} \rightarrow X^{+(n+1)} + e^-$
  
  This is the reverse of a reduction. So its standard oxidation potential is $-E^\circ_{\text{ox}}$ , where $E^\circ_{\text{ox}}$ is the standard reduction potential for:
  
  $X^{+(n+1)} + e^- \rightarrow X^{+n}$
Overall cell potential is:

E^\circ_{\text{cell}} = E^\circ_{\text{reduction half-cell}} - E^\circ_{\text{oxidation half-cell}}

But careful: The oxidation half-cell is the reverse of a reduction. So we write:

E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}}

where:

$E^\circ_{\text{red}}$ = standard reduction potential for $X^{+n} \rightarrow X^{+(n-1)}$
$E^\circ_{\text{ox}}$ = standard reduction potential for $X^{+(n+1)} \rightarrow X^{+n}$

Spontaneity condition:

\boxed{E^\circ_{\text{cell}} > 0 \quad \Rightarrow \quad E^\circ_{\text{red}} > E^\circ_{\text{ox}}}

In words: Disproportionation is spontaneous if the reduction potential for the lower oxidation state is greater than that for the higher oxidation state.

Why This Makes Sense — A Conceptual Explanation

$E^\circ_{\text{red}}$ tells you how easily $X^{+n}$ gets reduced to $X^{+(n-1)}$ . …

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Disproportionation is a redox reaction where the same element in one oxidation state simultaneously oxidises and reduces itself. Two aqueous examples are the decomposition of hydrogen peroxide ( $\ce{H2O2}$ ) and the reaction of copper(I) ion ( $\ce{Cu+}$ ).

Understanding Disproportionation

A disproportionation reaction is a special type of redox reaction where a single substance acts as both the oxidising agent and the reducing agent. The key idea: one atom of the element gains electrons (gets reduced) while another atom of the same element loses electrons (gets oxidised). This is possible only when the element has at least three accessible oxidation states — one intermediate state that can go both up and down.

Think of it like a tug-of-war within the same molecule or ion: one part pulls electrons in, the other pushes them out. The net result is that the starting species transforms into two different products — one with a higher oxidation number, one with a lower oxidation number.

Watch out

A common mistake is to think that any reaction producing two products from one reactant is disproportionation. It must be a redox change where the same element undergoes both oxidation and reduction. For example, $\ce{2H2O2 -> 2H2O + O2}$ is disproportionation, but $\ce{CaCO3 -> CaO + CO2}$ is not — no change in oxidation numbers.

Step-by-Step Reasoning

Identify the essential condition

For a reaction to be disproportionation, the reacting species must contain an element in an intermediate oxidation state. That element must be able to increase its oxidation number (oxidation) and decrease its oxidation number (reduction) simultaneously.
Example 1: Hydrogen peroxide ( $\ce{H2O2}$ )
- In $\ce{H2O2}$ , oxygen has an oxidation number of $-1$ (intermediate between $0$ in $\ce{O2}$ and $-2$ in $\ce{H2O}$ ).
- In aqueous solution, $\ce{H2O2}$ decomposes slowly:

\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}

Check the oxidation numbers:
- Oxygen in $\ce{H2O2}$ : $-1$
- Oxygen in $\ce{H2O}$ : $-2$ (reduction, gain of electrons)
- Oxygen in $\ce{O2}$ : $0$ (oxidation, loss of electrons)
So one oxygen atom is reduced from $-1$ to $-2$ , while another is oxidised from $-1$ to $0$ . This is a classic disproportionation.

Example 2: Copper(I) ion ( $\ce{Cu+}$ )
- Copper(I) has an oxidation state of $+1$ , which is intermediate between $0$ (metallic copper) and $+2$ (copper(II) ion). …

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Disproportionation Reactions — Common Mistakes & How to Avoid Them

What is Disproportionation?

A disproportionation reaction is a redox reaction in which the same element in one oxidation state is simultaneously oxidised (oxidation number increases) and reduced (oxidation number decreases).

Key idea: One substance acts as both oxidising and reducing agent.

✗ Common Mistake #1: Confusing disproportionation with comproportionation

The error: Students think any reaction where one element changes oxidation state is disproportionation.

Example of confusion:

$2H_2O_2 \rightarrow 2H_2O + O_2$ — this is disproportionation (oxygen in $H_2O_2$ goes from -1 to -2 and 0).

But $Cu^{2+} + Cu \rightarrow 2Cu^+$ is comproportionation (two different oxidation states of Cu combine to one intermediate state).

How to avoid:

Check the starting species: Disproportionation starts with one oxidation state of an element.
Check the products: The element must appear in two different oxidation states — one higher, one lower.

✗ Common Mistake #2: Forgetting to check the medium (aqueous vs non-aqueous)

The error: Giving examples that do not occur in aqueous solution, or ignoring that the reaction medium affects feasibility.

Example of wrong answer:

$2KClO_3 \rightarrow 2KCl + 3O_2$ — this is a thermal decomposition, not an aqueous disproportionation.

How to avoid:

The question explicitly asks for aqueous solution examples.
Stick to well-known aqueous disproportionations:
- Chlorine in water: $Cl_2 + H_2O \rightarrow HCl + HOCl$ (Cl: 0 → -1 and +1)
- Hydrogen peroxide: $2H_2O_2 \rightarrow 2H_2O + O_2$ (O: -1 → -2 and 0)

✗ Common Mistake #3: Incorrectly assigning oxidation numbers

The error: Miscalculating oxidation states leads to wrong identification of disproportionation.

Example of error:

In $P_4 + 3OH^- + 3H_2O \rightarrow 3H_2PO_2^- + PH_3$ , some students think P goes from 0 to +1 and -3 — correct. But they might mistakenly say P goes from 0 to +5 and -3 (wrong).

How to avoid:

Memorise rules:
- Oxidation number of O is usually -2 (except peroxides).
- Oxidation number of H is usually +1 (except hydrides).
- Sum of oxidation numbers in a neutral molecule = 0; in an ion = charge.
Practice with common elements: Cl, O, S, P, Mn — these frequently show disproportionation.

✗ Common Mistake #4: Giving only one example or non-standard examples

The error: Providing examples that are not clearly disproportionation, or giving only one when two are asked.

How to avoid:

Memorise at least 3 standard examples from your NCERT/board textbook:
1. $Cl_2 + H_2O \rightarrow HCl + HOCl$
2. $2H_2O_2 \rightarrow 2H_2O + O_2$ …

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Q.What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Concept understanding — Disproportionation Reaction

Disproportionation Reactions: The Self-Oxidation-Reduction

The Intuition

The Precise Definition

The Key Condition

Classic Example: Hydrogen Peroxide

Another Common Example: Copper(I) in Solution

How to Identify a Disproportionation Reaction

Why this formula?

Disproportionation Reaction — Understanding the Why

The Core Idea: Why Does Disproportionation Happen?

The Key Formula: Spontaneity Condition

Derivation:

Why This Makes Sense — A Conceptual Explanation

Concept: Disproportionation Reaction

Understanding Disproportionation

Step-by-Step Reasoning

Disproportionation Reaction — Concept & Method

What is Disproportionation?

Method to Identify a Disproportionation Reaction

Two Examples in Aqueous Solution

Example 1: Chlorine with cold dilute NaOH

Disproportionation Reactions — Common Mistakes & How to Avoid Them

What is Disproportionation?

✗ Common Mistake #1: Confusing disproportionation with comproportionation

✗ Common Mistake #2: Forgetting to check the medium (aqueous vs non-aqueous)

✗ Common Mistake #3: Incorrectly assigning oxidation numbers

✗ Common Mistake #4: Giving only one example or non-standard examples

Q.What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Understanding Disproportionation

Step-by-Step Reasoning

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