Disproportionation: Definition & Examples

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  • 0:04 Disproportionation
  • 1:39 Examples
  • 5:19 Lesson Summary
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Lesson Transcript
Hemnath (Vikash) Seeboo

Taught Science (mainly Chemistry, Physics and Math) at high school level and has a Master's Degree in Education.

Expert Contributor
Will Welch

Will has a doctorate in chemistry from the University of Wyoming and has experience in a broad selection of chemical disciplines and college-level teaching.

In this lesson, you will learn about disproportionation redox reactions. You will also analyze various redox reactions and be able to identify that disproportionation is taking place.


Have you ever worked with a touch screen device? Perhaps you're studying this lesson using one. A touch screen serves two purposes: it can be used to both input data while simultaneously obtaining output displays. How does that apply to disproportionation?

Redox reactions involve the reduction of one reactant and the oxidation of another reactant; so, one substance is reduced while another substance is oxidized. However, just like touch screens that have dual purposes, there are reactions where a molecule, atom, or ion can at the same time be simultaneously oxidized and reduced. The substance acts as both an oxidizing agent and as a reducing agent. This type of reaction is known as a disproportionation reaction.

More specifically, a disproportionation reaction is a special type of redox reaction in which an element simultaneously gives electrons and accepts electrons to form different products. This means that oxidation and reduction occur simultaneously with an atom, element, molecule, or ion acting both as a reducing and an oxidizing agent.

It's really important to recall that any change in oxidation state is due to either a loss or gain of electrons. Any increase in oxidation state indicates loss of electrons and hence oxidation (Lose Electrons Oxidation). A decrease in oxidation state implies a gain of electrons and hence reduction (Gain Electrons Reduction). You'll notice in disproportionation reactions that there's an element that must have at least three different oxidation states.


There are quite a few reactions that actually undergo disproportionation. We'll study three such types of redox reactions.

Let's first take a look at disproportionation of copper (I) chloride.

What happens when a solution of copper (I) chloride is left in a beaker for a period of time? Well, the solution becomes blue with a pink deposit in the beaker. Analysis of the products reveals the formation of copper (II) chloride and copper (s) metal, which you can see play out below:

Disproportionation of copper (I) chloride

The oxidation states of copper in copper (I) chloride, copper (II) chloride, and copper (s) produced are +1, +2 and 0, respectively.

From copper (I) chloride to copper (II) chloride, the oxidation number of copper increases from +1 to +2, indicating that there is a loss of an electron. This loss of electron indicates that copper (I) chloride is being oxidized and is acting as a reducing agent.

You'll also notice that the oxidation number of copper decreases from +1 to 0 when copper (I) chloride is converted to copper (s). This is associated with the gain of one electron. We can conclude that copper (I) chloride is being reduced and therefore acting as an oxidizing agent.

Since copper (I) chloride is acting both as a reducing and an oxidizing agent, the reaction is described as disproportionation.

Now let's take a look at disproportionation of hydrogen peroxide.

The decomposition of hydrogen peroxide is a commonly used reaction to prepare oxygen in the laboratory. Did you also know that the decomposition of hydrogen peroxide is also used in the paper industry to bleach wood pulp? Let's review this redox reaction.

Hydrogen peroxide decomposes into water and oxygen gas in the presence of a catalyst, which you can see playing out below:

Decomposition of hydrogen peroxide

The oxidation states of oxygen atom in hydrogen peroxide, water, and oxygen gas are respectively -1, -2, and 0.

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Additional Activities

Disproportionation Problems

Formal Redox Problems

The following scenarios are examples of classic disproportionation reactions that involve formal redox. Identify the disproportionated element in each scenario. Explain why you know it is that element, and provide oxidation states where pertinent.

  1. The conversion of mercury(II) chloride to mercury(I) chloride and mercury metal.
  2. The reaction between dissolved chlorine gas and hydroxide ion to form chloride ion, chlorate ion and water.
  3. Bromine fluoride gas is converted into BrF3 and bromine gas.
  4. The conversion of trihydrogenphosphite to phosphate and phosphine.

Organic Disproportionation Problems

Organic chemistry often does not define disproportioination in terms of formal oxidation numbers, but atoms that start in one state and end up in two opposing protonation states or states of electron richness are considered disproportionation reactions. Identify the disporportionated species.

  1. 2 water molecules forming hydronium and hydroxide.
  2. 2 bicarbonate ions converting to a carbonate and a carbonic acid.
  3. 2 carbon monoxide molecules forming elemental carbon and carbon dioxide.


Formal Redox Answers

  1. This is the same as the copper chloride example in the lesson. mercury is disproportionated, going from oxidation state +2 to +1 and 0.
  2. Chlorine is disproportionated going from elemental chlorine with oxidation state 0 to chloride (-1) and chlorate (+5)
  3. Bromine is disproportionated. Fluoride is electronegative enough that in BrF, the bromine can be considered Br+1. In the products, it is Br+3 and elemental Br (0).
  4. Phosphorus is disproportionated. It goes from an oxidation state of +3 in the phosphite species to +5 in phosphoric acid and -3 in phosphine.

Organic Disproportionation Answers

  1. Oxygen is the disproportionated species. It is bonded to two hydrogens in water but then 3 in hydronium and 1 in hydroxide. This is the prototypical example of how an neutral molecule can disproportionate to form an acid and a base.
  2. Similar to the above, a proton is being transferred from one oxygen to another so that of the three oxygens to start with in bicarbonate, one is protonated and the result is carbonic acid with two protons and carbonate with zero protons. Bicarbonate is a neutral species, and the products are an acid and a base.
  3. This is similar to a redox disproportionation reaction only because elemental carbon is in oxidation state 0 by definition. In CO, it can be considered to be in a -1 oxidation state, and in carbon dioxide, in a +4 oxidation state.

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