Disproportionation Reactions – Disproportionation Reactions, Example, Practice Problems and FAQs

Everyone has that one friend that excels in the classroom, right? In addition to achieving well for themselves, he/she aids us in performing well on tests. As a result, we refer to these people as clever and smart because they are able to both gain and impart knowledge.

The same atom can go through oxidation and reduction at the same time in some chemical processes, just like these people. These types of reactions are called disproportionation reactions.

We will learn about disproportionation reactions and examples on this concept page.

TABLE OF CONTENTS

• Disproportionation Reactions
• Examples of Disproportionation Reactions
• n-factor of Disproportionation Reactions
• Practice Problems
• Frequently Asked Questions – FAQ

Disproportionation Reactions

Disproportionation reactions, also known as dismutation reactions, are essentially redox reactions in which the same atom is simultaneously reduced and oxidised from one oxidation state to two separate oxidation states.

In essence, a molecule with an intermediate oxidation state is converted into two new compounds with higher and lower oxidation states. Species undergo simultaneous reduction and oxidation to produce two distinct byproducts.

An element undergoing disproportionation must exhibit a minimum of three different oxidation states in order for the reaction to take place.

Examples of Disproportionation Reactions

Let us consider the given example of disproportionation of hydrogen peroxide (H₂O₂) to dioxygen (O₂) and water (H₂O).

In hydrogen peroxide (H₂O₂), the oxidation state of oxygen is -1. When hydrogen peroxide is converted into water (H₂O), the oxidation state of oxygen decreaases to -2. As the oxidation state of oxygen decreases, oxygen has undergone reduction in this case.

In hydrogen peroxide (H₂O₂), the oxidation state of oxygen is -1. When hydrogen peroxide is converted into dioxygen (O₂), the oxidation state of oxygen increases to 0. As the oxidation state of oxygen increases, oxygen has undergone oxidation in this case.

Here, we can see clearly that hydrogen peroxide (H₂O₂) undergoes simultaneous reduction and oxidation to produce two distinct byproducts namely, water (H₂O) and dioxygen (O₂).

n-Factor of Disproportionation Reaction

Consider the following reaction.

2H2O22H2O+O2

n-factor=

Number of atoms |O.S of the atom in the product - O.S of the atom in the reactant|

Reduction: H₂O₂H₂O

$n-factor =2\times |-2-(-1\left)\right|=2$

Oxidation: H2O2O2

$n-factor =2\times |0-(-1\left)\right|=2$

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Practice Problems

1. In the given reaction, which substance will behave as the oxidising as well as the reducing agent?

a. ${SO}_3\left(g\right)+H_{2}O\left(g\right)\to H_2{SO}_4\left(g\right)$
b. $Pb{SO}_4\left(aq\right)+Zn\left(s\right)\to Zn{SO}_4\left(aq\right)+Pb\left(s\right)$
c. ${Cl}_2\left(g\right)+2NaOH\left(aq\right)\to NaCl\left(aq\right)+NaClO\left(aq\right)+H_{2}O\left(l\right)$
d. All of these

Answer: C

1. ${SO}_3\left(g\right)+H_{2}O\left(g\right)\to H_2{SO}_4\left(g\right)$

In the reaction given in option A, the oxidation state of sulphur is the same (+6) in both the reactant (SO₃(g)) and the product (H₂SO₄(g)). The same is the case with the oxidation state of oxygen (-2) in the reactants (SO3(g) and H₂O(g)) and the product (H₂SO₄(g)). Since SO₃(g) does not undergo reduction and oxidation simultaneously, it cannot act as an oxidising as well as a reducing agent.

In other words, S has +6 oxidation state in both SO₃(g) and H₂SO₄(g). Hence, there is no gain or loss of electrons. O has -2 oxidation state in SO3(g),H₂O(g) and H₂SO₄(g). Hence, there is no gain or loss of electrons. H has +1 oxidation state in both H₂O(g) and H₂SO₄(g). Hence, there is no gain or loss of electrons. If there is no gain or loss of electrons in a reaction, neither reduction nor oxidation has taken place.

2. $Pb{SO}_4\left(aq\right)+Zn\left(s\right)\to Zn{SO}_4\left(aq\right)+Pb\left(s\right)$

In the reaction given in option B, the oxidation state of lead decreases from +2 in PbSO₄(aq) to 0 in Pb(s). Since the oxidation state of lead decreases, it has undergone reduction.

The oxidation state of zinc increases from o in Zn(s) to +2 in ZnSO₄(aq). Since the oxidation state of zinc increases, it has undergone oxidation.

The oxidation state of sulphur remains the same (+6) in the reactants (PbSO₄(aq)) and the products (ZnSO₄(aq)). Since there is no change in the oxidation states, neither reduction or oxidation takes place.

The oxidation state of oxygen remains the same (-2) in the reactants (PbSO₄(aq)) and the products (ZnSO₄(aq)). Since there is no change in the oxidation states, neither reduction or oxidation takes place.

In other words, the oxidation state of Pb in PbSO₄(aq) is +2, it accepts two electrons from Zn and becomes Pb(s) with the oxidation state (0). Therefore Pb²⁺ (PbSO₄) undergoes reduction, or in other words, it acts as an oxidising agent.

The oxidation state of Zn in Zn(s) is 0 and it donates two electrons to Pb²⁺ and becomes Zn²⁺(aq) with the oxidation state +2. Therefore, Zn undergoes oxidation, or in other words, it acts as a reducing agent. But neither species can act as both a reducing as well as an oxidising agent as neither undergoes oxidation and reduction simultaneously.

3. ${Cl}_2\left(g\right)+2NaOH\left(aq\right)\to NaCl\left(aq\right)+NaClO\left(aq\right)+H_{2}O\left(l\right)$

In the reaction given in option C, the oxidation state of chlorine changes from 0 in Cl₂(g) to -1 in NaCl(aq) and +1 in NaClO(aq). As the oxidation state of the same species increase as well as decreases in the reaction, it has undergone oxidation and reduction simultaneously. Therefore, it can act as both an oxidising as well as a reducing agent.

The oxidation state of the other species in this reaction remains the same. For instance, oxygen is in the -2 oxidation state in both the reactant (NaOH(aq)) and the products (NaClO(aq) and H₂O(l)). The same is the case with sodium (Oxidation state = +1 in both the reactant and the products) and hydrogen (Oxidation state = +1 in both the reactant and the products).

In other words, Cl has 0 oxidation state which decreases to give -1 oxidation state in NaCl(aq) Hence, reduction takes place. In Cl₂(g) , Cl has 0 oxidation state which increases to give +1 oxidation state in NaClO(aq). Hence, oxidation takes place. Here, we can see clearly that Cl₂(g) undergo simultaneous reduction and oxidation to produce two distinct byproducts. Hence, Cl₂(g) can act as both a reducing and oxidising agent.

So, option C is the correct answer.

2. Which of the following reactions is a disproportionation reaction?

a. $3H_3{PO}_2\to {PH}_3+2H_3{PO}_3$
b. $FeO\left(s\right)+H_2\left(g\right)\to Fe\left(s\right)+H_{2}O\left(aq\right)$
c. Both (A) and (B)
d. None of these

Answer: A

1. $3H_3{PO}_2\to {PH}_3+2H_3{PO}_3$

In the reaction given in option A, the oxidation state of phosphorus changes from +1 oxidation state in H₃PO₂ to -3 in PH₃ and +3 in H₃PO₃. As the oxidation state of the same species increases as well as decreases in the reaction, it has undergone oxidation and reduction simultaneously. Therefore, it can act as both an oxidising as well as a reducing agent. Therefore, it is a disproportionation reaction.

2. $FeO\left(s\right)+H_2\left(g\right)\to Fe\left(s\right)+H_{2}O\left(aq\right)$

In the reaction given in option B, the oxidation state of iron decreases from +2 in FeO(s) to 0 in Fe(s). Since the oxidation state of iron decreases, it has undergone reduction.

The oxidation state of hydrogen increases from o in H₂(g) to +1 in H₂O(aq). Since the oxidation state of hydrogen increases, it has undergone oxidation.

The oxidation state of oxygen remains the same (-2) in the reactant (FeO(s)) and the product (H₂O). Thus, it neither undergoes oxidation nor reduction.

Though there are reduction and oxidation going on in the given reaction, the same species is not undergoing oxidation and reduction simultaneously. So, it is not a disproportionation reaction.

So, option B is the correct answer.

3. Which of the following reactions exhibits disproportionation?

a. $H_{2}O\left(g\right)+F_2\left(g\right)\to 2HF\left(g\right)+O_2\left(g\right)$
b. $2CuCl\to Cu+CuC{l}_2$
c. Both of these
d. None of these

Answer: B

1. $H_{2}O\left(g\right)+F_2\left(g\right)\to 2HF\left(g\right)+O_2\left(g\right)$

In the reaction given in option A, the oxidation state of oxygen increases from -2 in H₂O(g) to 0 in O₂(g). Since the oxidation state increases, it has undergone oxidation.

The oxidation state of fluorine decreases from 0 in F₂(g) to -1 in HF(g). Since the oxidation state decreases, it has undergone reduction.

Though there are reduction and oxidation going on in the given reaction, the same species is not undergoing oxidation and reduction simultaneously. So, it is not a disproportionation reaction.

2. $2CuCl\to Cu+CuC{l}_2$

In the reaction given in option B, the oxidation state of copper changes from +1 in CuCl to 0 in Cu and +2 in CuCl₂. Since the oxidation state of the same species increases and decreases in the same reaction, it has undergone reduction and oxidation simultaneously. Therefore, it is a disproportionation reaction.

So, option B is the correct answer.

4. Which of the following reactions is not a disproportionation reaction?

a. $2CuBr\to Cu+Cu{Br}_2$
b. ${ClO}_3^{-}\to {Cl}^{-}+{ClO}_4^{-}$
c. ${MnO}_4^{2-}\to {MnO}_2+{MnO}_4^{-}$
d. None of these

Answer: D

1. $2CuBr\to Cu+Cu{Br}_2$

In the reaction given in option A, the oxidation state of copper changes from +1 in CuBr to 0 in Cu and +2 in CuBr₂. Since the oxidation state of the same species increases and decreases in the same reaction, it has undergone reduction and oxidation simultaneously. Therefore, it is a disproportionation reaction.

2. ${ClO}_3^{-}\to {Cl}^{-}+{ClO}_4^{-}$

In the reaction given in option B, the oxidation state of chlorine changes from +5 in ClO₃^- to -1 in Cl- and +7 in ClO₄^-. Since the oxidation state of the same species increases and decreases in the same reaction, it has undergone reduction and oxidation simultaneously. Therefore, it is a disproportionation reaction.

3. ${MnO}_4^{2-}\to {MnO}_2+{MnO}_4^{-}$

In the reaction given in option C, the oxidation state of manganese changes from +6 in MnO₄^2- to +4 in MnO₂ and +7 in MnO₄^-. Since the oxidation state of the same species increases and decreases in the same reaction, it has undergone reduction and oxidation simultaneously. Therefore, it is a disproportionation reaction.

Thus, the reactions given in options A, B and C are disproportionation reactions.

So, option D is the correct answer.

Frequently Asked Questions – FAQ

1. What are the different types of redox reactions?
Answer: Redox reactions can be generally classified into five types. They are

1. Combination reactions
2. Decomposition reactions
3. Displacement reactions
4. Disproportionation reactions
5. Conproportionation reactions

2. How do chemical reactions play an important role in our life?
Answer: Bonds between atoms in molecules break and reform in different manners. Chemical changes are at the heart of virtually every biological process in the universe. Chemical reactions cause stars to form or chemical changes in our sun's core to ignite it. Chemical changes gave rise to life on Earth. Hence, Chemical reactions play an important role in our lives.

3. What is the opposite of a disproportionation reaction?
Answer: Conproportionation often referred to as synproportionation, is the opposite of disproportionation and occurs when a chemical in an intermediate oxidation state is created from components in lower and higher oxidation levels.

4. Can we say that a disproportionation reaction is a redox reaction?
Answer: Yes, the disproportionation reaction is a kind of redox reaction as the same species undergoes oxidation and reduction simultaneously.