Disproportionation: When One Substance Is Oxidised and Reduced

Bubble chlorine into cold, dilute sodium hydroxide. Two products form: chloride ions, \(\mathrm{Cl}^{-}\), and hypochlorite ions, \(\mathrm{ClO}^{-}\).

Here is the interesting part. Chlorine starts with an oxidation state of 0. In chloride ions, its oxidation state becomes −1. In hypochlorite ions, it becomes +1.

So, can the same substance be oxidised and reduced in one reaction?

Yes. This type of reaction is called disproportionation. Different atoms of the same element in the starting substance undergo oxidation and reduction simultaneously.

Let’s work through why this happens, how to recognise it, and how to prove that the electrons, atoms, and charges all balance.

01How can one substance undergo oxidation and reduction?

Start with the reaction between chlorine and cold, dilute sodium hydroxide.

The balanced net ionic equation is:

\[
\ce{Cl2(aq) + 2OH^-(aq) -> Cl^-(aq) + ClO^-(aq) + H2O(l)}
\]

The conditions matter: this reaction produces hypochlorite in cold, dilute alkaline solution. Sodium ions are spectators, so they have been omitted from the net ionic equation.

Before looking at the oxidation states, make a prediction.

Both chlorine atoms in \(\ce{Cl2}\) start in the same situation. Do they both become chloride ions, or does something else happen?

The answer is that they follow different paths. One chlorine atom becomes chloride, while another becomes part of a hypochlorite ion.

Step 1: Work out the starting oxidation state

Chlorine exists as an elemental molecule, \(\ce{Cl2}\).

An element in its uncombined form always has an oxidation state of zero.

Therefore:

\[
\ce{Cl2}: \quad \mathrm{Cl}=0
\]

Both chlorine atoms start at zero.

Step 2: Work out the oxidation states in the products

Consider the chloride ion first.

\[
\mathrm{Cl}^{-}: \quad \mathrm{Cl}=-1
\]

Chlorine has changed from 0 to −1.

Now consider hypochlorite, \(\mathrm{ClO}^{-}\).

Oxygen usually has an oxidation state of −2 in compounds like this. The total oxidation states must equal the overall ionic charge.

Let \(x\) represent the oxidation state of chlorine.

\[
x+(-2)=-1
\]

\[
x=+1
\]

Therefore, chlorine in hypochlorite has an oxidation state of +1.

Step 3: Identify oxidation and reduction

Compare the starting and final oxidation states.

Remember the two rules:

• Oxidation: oxidation state increases; electrons are lost.
• Reduction: oxidation state decreases; electrons are gained.

In this reaction:

• One chlorine atom loses one electron and is oxidised.
• Another chlorine atom gains one electron and is reduced.

The electrons lost and gained balance.

This is disproportionation: the same starting substance supplies both the oxidised and reduced products.

Importantly, one chlorine atom does not simultaneously undergo both processes. Different chlorine atoms follow different reaction pathways.

02How do the electrons balance?

It is worth checking this properly. Identifying oxidation states is only part of the job. A valid redox equation must also conserve atoms and charge.

We can demonstrate the electron transfer using half-equations.

Reduction half-equation

Chlorine gains electrons to form chloride ions:

\[
\ce{Cl2 + 2e^- -> 2Cl^-}
\]

Oxidation half-equation

Chlorine loses electrons to form hypochlorite ions in alkaline solution:

\[
\ce{Cl2 + 4OH^- ->}
\]

\[
\ce{2ClO^- + 2H2O + 2e^-}
\]

Notice that both half-equations involve two electrons.

When we add them, the electrons cancel:

\[
\ce{2Cl2 + 4OH^- ->}
\]

\[
\ce{2Cl^- + 2ClO^- + 2H2O}
\]

Divide every coefficient by two to obtain the simplest whole-number equation:

\[
\boxed{\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}}
\]

Check atom and charge conservation

A balanced redox equation must satisfy two conditions.

1. Every type of atom must be conserved.

2. Total electrical charge must be conserved.

On the reactant side:

\[
2(-1)=-2
\]

On the product side:

\[
(-1)+(-1)=-2
\]

Both sides have a total charge of −2.

Everything balances.

The half-equations are an accounting method for tracking electrons. They do not mean that free electrons need to travel through an external wire during this reaction.

03Does disproportionation always produce the same products?

No. Reaction conditions can change which oxidation product forms.

Chlorine provides a particularly useful comparison.

Cold, dilute sodium hydroxide

Under cold, dilute alkaline conditions, chlorine forms chloride and hypochlorite.

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

Chlorine changes from 0 to:

• −1 in chloride, through reduction.
• +1 in hypochlorite, through oxidation.

Hot, concentrated sodium hydroxide

Under hot, concentrated alkaline conditions, the overall reaction instead produces chloride and chlorate.

\[
\ce{3Cl2 + 6OH^- ->}
\]

\[
\ce{5Cl^- + ClO3^- + 3H2O}
\]

The chlorine oxidation state in chlorate is +5.

We can calculate it:

Let \(x\) be the oxidation state of chlorine.

\[
x+3(-2)=-1
\]

\[
x=+5
\]

Now compare the two products.

Both reactions are disproportionation reactions.

The electron balance is also different.

In the cold, dilute reaction, one chlorine atom gains one electron while another loses one electron.

In the hot, concentrated reaction, five chlorine atoms each gain one electron, while one chlorine atom loses five electrons.

The electron transfer still balances.

The important lesson is that you cannot identify the products from the starting substance alone. You must also consider the reaction conditions.

If you need to revise the names and formulae of ions such as hypochlorite and chlorate, see IUPAC Naming of Inorganic Substances for HSC Chemistry.

04A contrasting case: ordinary redox is not necessarily disproportionation

Here is a common source of confusion.

Consider zinc reacting with copper(II) ions:

\[
\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}
\]

This is a redox reaction because oxidation and reduction both occur.

However, it is not disproportionation.

Look at the oxidation states:

Zinc loses two electrons.

Copper(II) gains two electrons.

The two processes involve different elements from different reactants. Neither starting element undergoes both oxidation and reduction.

Compare this with chlorine disproportionation:

• In the zinc reaction, one element is oxidised and another is reduced.
• In the chlorine reaction, the same element from the same starting substance is both oxidised and reduced.

That is the distinction you need to recognise.

05A reliable method for identifying disproportionation

When you encounter an unfamiliar redox equation, use this sequence.

Step 1: Identify the element you suspect might be undergoing disproportionation.

Step 2: Calculate its oxidation state in the reactant.

Step 3: Calculate its oxidation state in every product containing that element.

Step 4: Compare the changes.

• If some atoms of that element increase their oxidation state, oxidation has occurred.
• If other atoms of the same element decrease their oxidation state, reduction has occurred.

Step 5: Check that the overall equation conserves atoms and charge.

If both oxidation and reduction occur to the same element from the same starting species, you have identified disproportionation.

Do not simply look for one reaction producing two products. That is not enough.

You must prove that the same element moves in opposite oxidation-state directions.

06Worked examples

07Questions and solutions

The following questions move from straightforward oxidation-state identification to more demanding problems involving stoichiometry, reaction conditions, and multi-stage redox reactions.

Question 1

Chlorine reacts with cold, dilute sodium hydroxide according to the following equation:

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

Identify:

a) The oxidation state of chlorine in each reactant and product.

b) Which chlorine species undergoes oxidation and which undergoes reduction.

c) Whether the reaction is disproportionation.

Solution 1

Yes, this is a disproportionation reaction. Chlorine from the same starting substance undergoes both oxidation and reduction.

a) Oxidation states

• \(\ce{Cl2}\): 0, because chlorine is in its elemental form.
• \(\mathrm{Cl}^{-}\): −1, because it is a chloride ion.
• \(\mathrm{ClO}^{-}\): +1, because oxygen is −2 and the overall charge is −1.

b) Oxidation and reduction

Chlorine forming chloride changes from 0 to −1.

Its oxidation state decreases, so it gains an electron and undergoes reduction.

Chlorine forming hypochlorite changes from 0 to +1.

Its oxidation state increases, so it loses an electron and undergoes oxidation.

c) Classification

The same starting substance, \(\ce{Cl2}\), produces chlorine atoms that undergo oxidation and reduction.

Therefore, the reaction is disproportionation.

Question 2

A 100.0 g sample of aqueous hydrogen peroxide solution contains 6.80 g of \(\ce{H2O2}\).

The hydrogen peroxide decomposes completely according to:

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

A manganese dioxide catalyst is present.

Given:

• Molar mass of \(\ce{H2O2}\) = 34.0 g mol\(^{-1}\)
• Molar mass of \(\ce{O2}\) = 32.0 g mol\(^{-1}\)

Calculate:

a) The amount of hydrogen peroxide initially present.

b) The amount and mass of oxygen gas produced.

c) The number of moles of electrons gained and lost during the reaction.

Explain why this reaction is disproportionation.

Solution 2

The reaction produces 0.100 mol of oxygen gas, equivalent to 3.20 g, and hydrogen peroxide undergoes disproportionation.

a) Initial amount of hydrogen peroxide

Use:

\[
n=\frac{m}{M}
\]

Substitute:

\[
n(\ce{H2O2})=
\frac{6.80\ \mathrm{g}}
{34.0\ \mathrm{g\,mol^{-1}}}
\]

\[
\boxed{n(\ce{H2O2})=0.200\ \mathrm{mol}}
\]

b) Amount and mass of oxygen gas

From the balanced equation:

\[
2\ce{H2O2} \rightarrow \ce{O2}
\]

Two moles of hydrogen peroxide produce one mole of oxygen.

Therefore:

\[
n(\ce{O2})=\frac{0.200}{2}
\]

\[
\boxed{n(\ce{O2})=0.100\ \mathrm{mol}}
\]

Calculate the mass:

\[
m=nM
\]

\[
m(\ce{O2})=0.100\times32.0
\]

\[
\boxed{m(\ce{O2})=3.20\ \mathrm{g}}
\]

c) Electron transfer

Oxygen in hydrogen peroxide has an oxidation state of −1.

In water, oxygen has an oxidation state of −2.

In oxygen gas, it has an oxidation state of 0.

For every reaction represented by:

\[
\ce{2H2O2 -> 2H2O + O2}
\]

Two oxygen atoms each gain one electron.

Two oxygen atoms each lose one electron.

Therefore, two moles of hydrogen peroxide involve two moles of electrons gained and two moles lost.

The reaction extent is:

\[
\frac{0.200}{2}=0.100\ \mathrm{mol}
\]

Hence:

\[
n(e^-)=0.100\times2
\]

\[
\boxed{n(e^-)=0.200\ \mathrm{mol}}
\]

There are 0.200 mol of electrons gained and 0.200 mol lost.

The catalyst increases the reaction rate but does not change the balanced equation or the stoichiometric ratios.

Hydrogen peroxide is both oxidised and reduced because its oxygen atoms follow two different oxidation-state pathways.

Question 3

Consider the following two reactions involving copper(I) ions.

Reaction A

\[
\ce{2Cu^+(aq) -> Cu^2+(aq) + Cu(s)}
\]

Reaction B

\[
\ce{Zn(s) + 2Cu^+(aq) -> Zn^2+(aq) + 2Cu(s)}
\]

A student claims that both reactions are disproportionation because both produce solid copper from copper(I) ions.

Evaluate this claim.

Identify which reaction is disproportionation and explain why the other reaction is not.

Solution 3

Only Reaction A is disproportionation. In Reaction B, all copper(I) ions are reduced, while zinc is oxidised.

Reaction A

Copper starts with an oxidation state of +1.

In the products:

• \(\mathrm{Cu}^{2+}\) has an oxidation state of +2.
• \(\ce{Cu(s)}\) has an oxidation state of 0.

Therefore:

\[
+1\rightarrow+2
\]

One copper(I) ion loses an electron and is oxidised.

Meanwhile:

\[
+1\rightarrow0
\]

Another copper(I) ion gains an electron and is reduced.

The electron changes balance:

• One electron is lost.
• One electron is gained.

The same starting species, \(\mathrm{Cu}^{+}\), undergoes both processes.

Reaction A is therefore disproportionation.

Reaction B

Now consider the zinc reaction.

Zinc changes from:

\[
0\rightarrow+2
\]

It loses two electrons and is oxidised.

Copper changes from:

\[
+1\rightarrow0
\]

Each copper(I) ion gains one electron.

Since there are two copper(I) ions, they gain two electrons altogether.

Therefore:

• Zinc is oxidised.
• Copper(I) is reduced.
• No copper(I) ions undergo oxidation.

Reaction B is an ordinary redox displacement reaction, not disproportionation.

The student’s mistake is assuming that the appearance of a product containing the same element automatically proves disproportionation.

You must check whether the element undergoes both oxidation and reduction, not just whether it appears in multiple species.

Question 4

Two separate samples each contain 0.0600 mol of chlorine.

Sample A reacts completely with cold, dilute sodium hydroxide.

Sample B reacts completely with hot, concentrated sodium hydroxide.

The reactions are:

Sample A:

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

Sample B:

\[
\ce{3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O}
\]

Assume sodium hydroxide is present in sufficient quantity.

For each sample, calculate:

a) The amount of chloride ions produced.

b) The amount of the oxidised chlorine product produced.

c) The amount of electrons transferred.

d) Explain why the electron transfer differs between the two reactions.

Solution 4

Sample A produces 0.0600 mol each of chloride and hypochlorite, with 0.0600 mol of electrons transferred. Sample B produces 0.100 mol of chloride and 0.0200 mol of chlorate, with 0.100 mol of electrons transferred.

a) Amount of chloride ions

For Sample A, the equation gives a 1:1 ratio between chlorine and chloride.

\[
n(\mathrm{Cl}^{-})=0.0600\times1
\]

\[
\boxed{n(\mathrm{Cl}^{-})=0.0600\ \mathrm{mol}}
\]

For Sample B, three moles of chlorine produce five moles of chloride.

\[
n(\mathrm{Cl}^{-})=0.0600\times\frac{5}{3}
\]

\[
\boxed{n(\mathrm{Cl}^{-})=0.100\ \mathrm{mol}}
\]

b) Amount of oxidised chlorine product

For Sample A, the chlorine-to-hypochlorite ratio is 1:1.

\[
n(\mathrm{ClO}^{-})=0.0600\ \mathrm{mol}
\]

For Sample B, three moles of chlorine produce one mole of chlorate.

\[
n(\mathrm{ClO_{3}}^{-})=0.0600\times\frac{1}{3}
\]

\[
\boxed{n(\mathrm{ClO_{3}}^{-})=0.0200\ \mathrm{mol}}
\]

c) Amount of electrons transferred

For Sample A, chlorine changes from 0 to −1 in chloride and from 0 to +1 in hypochlorite.

One electron is gained for every chlorine molecule reacting.

Therefore:

\[
\boxed{n(e^-)=0.0600\ \mathrm{mol}}
\]

For Sample B, five chlorine atoms each gain one electron, while one chlorine atom loses five electrons.

The balanced equation represents five moles of electrons transferred for every three moles of chlorine.

\[
n(e^-)=0.0600\times\frac{5}{3}
\]

\[
\boxed{n(e^-)=0.100\ \mathrm{mol}}
\]

d) Explain the difference

The difference comes from the oxidation state of the oxidised chlorine product.

In Sample A, chlorine is oxidised from 0 to +1.

In Sample B, chlorine is oxidised from 0 to +5.

The second reaction therefore involves a larger oxidation-state increase for the chlorine atoms forming the oxidised product.

More electrons must be transferred to balance the corresponding reduction.

Both reactions are disproportionation, but the reaction conditions change the products and the electron ratios.

Question 5

This final problem combines disproportionation, reaction conditions, stoichiometry, and the difference between an overall reaction and a reaction pathway.

A sample contains 0.0600 mol of chlorine.

The chlorine first reacts completely with cold, dilute sodium hydroxide:

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

The resulting solution is then heated until all the hypochlorite ions decompose according to:

\[
\ce{3ClO^- -> 2Cl^- + ClO3^-}
\]

Assume:

• All the original chlorine reacts.
• Sufficient hydroxide ions are available for the first reaction.
• Chloride ions produced in the first reaction remain unchanged during heating.
• No other chlorine-containing products form.

Answer the following:

a) Calculate the amounts of chloride and hypochlorite produced in the first reaction.

b) Calculate the final amounts of chloride and chlorate after heating.

c) Combine the two reactions to obtain one overall net ionic equation.

d) Calculate the electron transfer for the overall reaction.

e) A student observes the final products and claims that chlorine must have reacted directly under hot, concentrated alkaline conditions from the beginning. Explain why the final products cannot establish this conclusion.

Solution 5

The final solution contains 0.100 mol of chloride and 0.0200 mol of chlorate. The overall reaction is a disproportionation reaction, but the final products do not prove that chlorine followed a single reaction pathway.

This problem requires us to track the chlorine through both stages, rather than treating the final products as if they all formed at once.

a) Calculate the products of the first reaction

The first equation is:

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

The mole ratio is:

\[
1:1:1
\]

Therefore, 0.0600 mol of chlorine produces:

\[
\boxed{n(\mathrm{Cl}^{-})=0.0600\ \mathrm{mol}}
\]

and:

\[
\boxed{n(\mathrm{ClO}^{-})=0.0600\ \mathrm{mol}}
\]

Both products must be carried forward into the second stage.

This is important: the chloride ions already produced do not disappear when the solution is heated.

b) Calculate the final products after heating

The second reaction is:

\[
\ce{3ClO^- -> 2Cl^- + ClO3^-}
\]

We initially have 0.0600 mol of hypochlorite.

The amount of reaction, based on the stoichiometric coefficient of three, is:

\[
\frac{0.0600}{3}=0.0200\ \mathrm{mol}
\]

Therefore, the amount of additional chloride produced is:

\[
n(\mathrm{Cl}^{-})=0.0200\times2
\]

\[
=0.0400\ \mathrm{mol}
\]

The amount of chlorate produced is:

\[
n(\mathrm{ClO_{3}}^{-})=0.0200\times1
\]

\[
=0.0200\ \mathrm{mol}
\]

However, remember that 0.0600 mol of chloride was already present from the first reaction.

The final chloride amount is:

\[
n(\mathrm{Cl}^{-})=0.0600+0.0400
\]

\[
\boxed{n(\mathrm{Cl}^{-})=0.100\ \mathrm{mol}}
\]

The final chlorate amount is:

\[
\boxed{n(\mathrm{ClO_{3}}^{-})=0.0200\ \mathrm{mol}}
\]

c) Combine the two reactions

Start with the first reaction:

\[
\ce{Cl2 + 2OH^- -> Cl^- + ClO^- + H2O}
\]

Multiply every coefficient by three:

\[
\ce{3Cl2 + 6OH^- ->}
\]

\[
\ce{3Cl^- + 3ClO^- + 3H2O}
\]

Now add the second reaction:

\[
\ce{3ClO^- -> 2Cl^- + ClO3^-}
\]

The three hypochlorite ions cancel because they are produced in the first stage and consumed in the second.

The combined equation is:

\[
\ce{3Cl2 + 6OH^- ->}
\]

\[
\ce{5Cl^- + ClO3^- + 3H2O}
\]

Therefore:

\[
\boxed{
\ce{3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O}
}
\]

This is exactly the same overall net ionic equation obtained for chlorine reacting under hot, concentrated alkaline conditions.

d) Calculate the overall electron transfer

We must compare the original oxidation states with the final oxidation states.

Initially, all chlorine atoms in \(\ce{Cl2}\) have oxidation state 0.

In the final products:

• Chlorine in \(\mathrm{Cl}^{-}\) has oxidation state −1.
• Chlorine in \(\mathrm{ClO_{3}}^{-}\) has oxidation state +5.

The 0.100 mol of chlorine atoms ending as chloride each gain one electron.

Therefore:

\[
n(e^-\text{ gained})=0.100\times1
\]

\[
=0.100\ \mathrm{mol}
\]

The 0.0200 mol of chlorine atoms ending as chlorate each lose five electrons.

Therefore:

\[
n(e^-\text{ lost})=0.0200\times5
\]

\[
=0.100\ \mathrm{mol}
\]

Hence:

\[
\boxed{
n(e^-\text{ gained})=
n(e^-\text{ lost})=
0.100\ \mathrm{mol}
}
\]

The overall electron balance is conserved.

There is an additional subtlety here.

If you calculate the electron changes separately for the two stages, you obtain:

• First stage: 0.0600 mol of electrons transferred.
• Second stage: 0.0800 mol of electrons transferred.

You should not simply add these values to obtain the overall electron transfer.

Why?

Because hypochlorite is an intermediate. Its chlorine atoms have already changed from oxidation state 0 to +1 in the first stage.

In the second stage, some of those atoms are reduced from +1 to −1, while others are oxidised from +1 to +5.

For the atoms ending as chloride, the first stage’s increase from 0 to +1 is partly reversed by the subsequent decrease from +1 to −1.

The correct overall electron balance comes from comparing the original reactant with the final products.

This is a useful reminder that an intermediate oxidation-state change can be reversed or extended during a later reaction.

e) Evaluate the student’s conclusion

The student’s conclusion is not justified.

The final products are consistent with the overall equation:

\[
\ce{3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O}
\]

However, we have just demonstrated that the same overall equation can arise through two successive reactions:

1. Chlorine first disproportionates into chloride and hypochlorite in cold, dilute alkali.
2. Hypochlorite subsequently disproportionates into additional chloride and chlorate when heated.

The overall equation describes the net change between the initial reactants and final products.

It does not, by itself, reveal every intermediate or establish the detailed reaction mechanism.

This is the crucial distinction: a balanced overall equation tells you what changes overall, not necessarily how the reaction happens.

08What to check next

Once you can identify disproportionation, the next useful skill is balancing unfamiliar redox reactions using oxidation states and half-equations.

The same electron-accounting method applies to galvanic cells, but there is an important difference.

In disproportionation, oxidation and reduction occur within the same overall reaction mixture. In a galvanic cell, the oxidation and reduction half-reactions occur at separate electrodes, with electrons travelling through an external circuit.

Being able to recognise which species is oxidised, which is reduced, and how many electrons are transferred gives you the foundation for understanding both.