2.1.3 Disproportionation Reaction

In most redox reactions, one element is oxidized while a different element is reduced. However, there is a special class of reactions where a single element plays both roles at the same time. This phenomenon is known as disproportionation and is an important concept in redox chemistry.

Definition

A disproportionation reaction (also known as a self-oxidation-reduction reaction) is a redox reaction in which the same element is simultaneously oxidized and reduced.

In simple words, a single substance reacts to form two different products, and the element that undergoes the change ends up in a higher oxidation state in one product and a lower oxidation state in the other.

Core Concept

The core idea is that one type of atom starts at a single oxidation state in the reactant and then splits into two different oxidation states in the products. For this to happen, the element must have at least three accessible oxidation states. The intermediate oxidation state acts as a "middle ground" from which the element can either lose electrons (go higher) or gain electrons (go lower). The total number of electrons lost by the oxidized atoms must equal the total number of electrons gained by the reduced atoms.

Detailed Explanation

The textbook provides an excellent example to explain this concept: the decomposition of hydrogen peroxide (H_2O_2).

The chemical equation for this reaction is: 2H_2O_2_{(aq)} ⟶ 2H_2O_{(l)} + O_2_{(g)}

Let's analyze this reaction step-by-step using oxidation numbers.

Step 1: Assign Oxidation Numbers to the Reactant

In hydrogen peroxide (H_2O_2), the oxidation number of hydrogen is +1. Let the oxidation number of oxygen be x. Since the molecule is neutral, the sum of the oxidation numbers must be zero: 2(+1) + 2(x) = 0 +2 + 2x = 0 2x = -2 x = -1 So, the oxidation number of oxygen in H_2O_2 is -1. Note that this is an exception to the general rule that oxygen is -2 in most compounds.

Step 2: Assign Oxidation Numbers to the Products

The products are water (H_2O) and oxygen gas (O_2).

• In H_2O: The oxidation number of hydrogen is +1 and oxygen is -2.
• In O_2: Oxygen is in its elemental form, so its oxidation number is 0.

Step 3: Analyze the Changes

Now let's compare the oxidation number of oxygen in the reactant to its values in the products.

• Path 1: Oxygen in H_2O
• Oxidation number changes from -1 (in H_2O_2) to -2 (in H_2O).
• This is a decrease in oxidation number, which means the oxygen has gained electrons and is reduced.

• Path 2: Oxygen in O_2
• Oxidation number changes from -1 (in H_2O_2) to 0 (in O_2).
• This is an increase in oxidation number, which means the oxygen has lost electrons and is oxidized.

Conclusion

Because the same element (oxygen) undergoes both oxidation and reduction in this single reaction, the decomposition of hydrogen peroxide is a disproportionation reaction.

• Reduction Half-Reaction: H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O
• Oxidation Half-Reaction: H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-

The textbook summarizes this with an image showing the element oxygen starting at -1 and splitting into -2 (reduction) and 0 (oxidation).

Intermediate Level

Disproportionation is a special type of redox reaction. For it to occur, the element must be in an intermediate oxidation state. This means it's not at its highest possible oxidation state (so it can go higher by being oxidized) and not at its lowest possible state (so it can go lower by being reduced). In the case of H_2O_2, oxygen is at -1, but its common states are 0 (in O_2) and -2 (in oxides). So it can go either way. This "instability" of the intermediate state is what drives the disproportionation reaction.

Important Points

• - Disproportionation: A reaction where the same element is simultaneously oxidized and reduced.
• - It is also called a self-oxidation-reduction reaction.
• - The element must be in an intermediate oxidation state.
• - The reaction results in the element appearing in two different oxidation states in the products—one higher and one lower.
• - The textbook example is the decomposition of hydrogen peroxide (H_2O_2).

Remember

• - Disproportionation = Same element, both oxidized and reduced.
• - It requires an element to have at least three oxidation states (e.g., -1, 0, and -2 for oxygen).
• - The element acts as both the oxidizing agent and the reducing agent.
• - The total electrons lost must equal the total electrons gained.

Common Mistakes

• - Confusing it with a regular redox reaction. In a normal redox reaction, different elements are oxidized and reduced. In disproportionation, it's the same element.
• - Thinking any unstable compound disproportionates. Many compounds are unstable but undergo decomposition (e.g., heating calcium carbonate to give calcium oxide and carbon dioxide), which is not a redox reaction at all. Disproportionation specifically involves a change in oxidation state.
• - Missing the intermediate state requirement. An element at its highest oxidation state (e.g., +7 for Mn) cannot be oxidized further and, therefore, cannot disproportionate. It can only be reduced.
• - Incorrectly assigning oxidation numbers. If you get the oxidation numbers wrong, you won't see the change. Make sure to carefully apply the rules, especially for exceptions like H_2O_2 (where O is -1).

Quick Revision

A disproportionation reaction is a special type of redox reaction where a single element is simultaneously oxidized and reduced. The textbook example is the decomposition of hydrogen peroxide (H_2O_2), where oxygen in the -1 state is both reduced (to -2 in water) and oxidized (to 0 in oxygen gas). This type of reaction requires the element to be in an intermediate oxidation state and acts as both the oxidizing and reducing agent.

60-Second Revision

A disproportionation reaction is a redox reaction where the same element is both oxidized and reduced. For example, in the decomposition of H_2O_2, oxygen (-1) is reduced to -2 in H_2O and oxidized to 0 in O_2. It acts as both the oxidizing and reducing agent. This requires the element to have an intermediate oxidation state.

Frequently Asked Questions (FAQs)

Q1. What is a disproportionation reaction? Answer: A disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced. This means a single element in a reactant goes to two different oxidation states in the products—one higher and one lower.

Q2. Give the textbook example of a disproportionation reaction. Answer: The decomposition of hydrogen peroxide (2H_2O_2 → 2H_2O + O_2). In this reaction, the oxygen in H_2O_2 has an oxidation number of -1. It is both reduced to -2 in water (H_2O) and oxidized to 0 in oxygen gas (O_2).

Q3. Why is a disproportionation reaction also called a self-oxidation-reduction reaction? Answer: It is called a self-oxidation-reduction reaction because a single substance (or element within it) acts as both the oxidizing agent and the reducing agent. The element oxidizes itself and reduces itself simultaneously.

Q4. What condition must an element satisfy to undergo disproportionation? Answer: An element must be in an intermediate oxidation state. It cannot be at its highest possible state (because it would have no electrons to lose, so it cannot be oxidized) or at its lowest possible state (because it would have no electrons to gain, so it cannot be reduced).

Q5. In the disproportionation of H_2O_2, which product is formed by oxidation and which by reduction? Answer: The oxygen gas (O_2) is formed by oxidation, as the oxidation number of oxygen increases from -1 to 0. The water (H_2O) is formed by reduction, as the oxidation number of oxygen decreases from -1 to -2.

Q6. Can an element at its highest oxidation state undergo disproportionation? Answer: No. An element at its highest oxidation state (e.g., +7 for manganese in permanganate, MnO_4^-) cannot be oxidized further. Therefore, it cannot undergo disproportionation; it can only be reduced.

Q7. Is the decomposition of calcium carbonate (CaCO_3 \rightarrow CaO + CO_2) a disproportionation reaction? Answer: No. The decomposition of calcium carbonate is not a redox reaction at all, because the oxidation numbers of all the elements remain the same before and after the reaction. Disproportionation specifically requires a change in oxidation states.

Q8. What is the reverse of a disproportionation reaction called? Answer: The reverse is called comproportionation, a reaction where two different oxidation states of the same element react to form a single, intermediate oxidation state.

Multiple Choice Questions (MCQs)

• Q1. What is a disproportionation reaction?
• A. A reaction where two different elements are oxidized and reduced.
• B. A reaction where the same element is both oxidized and reduced.
• C. A reaction where an element is only oxidized.
• D. A reaction where an element is only reduced.
• Correct Answer: B
• Explanation: By definition, a disproportionation reaction is a redox reaction in which a single element is simultaneously oxidized and reduced.

Q2. In the decomposition of hydrogen peroxide (2H_2O_2 → 2H_2O + O_2), which product contains the oxidized form of oxygen? A. H_2O B. O_2 C. Both D. None Correct Answer: B Explanation: The oxygen in O_2 has an oxidation number of 0, which is higher than its starting state of -1 in H_2O_2. An increase in oxidation number means oxidation.

• Q3. What is the oxidation number of oxygen in hydrogen peroxide (H_2O_2)?
• A. -2
• B. -1
• C. 0
• D. +1
• Correct Answer: B
• Explanation: In peroxides, the oxidation number of oxygen is -1. This is a common exception to the rule that oxygen is -2.

• Q4. For an element to undergo disproportionation, it must be in:
• A. Its highest oxidation state.
• B. Its lowest oxidation state.
• C. An intermediate oxidation state.
• D. The zero oxidation state.
• Correct Answer: C
• Explanation: The element must be in an intermediate state so that it can either increase its oxidation number (be oxidized) or decrease it (be reduced).

• Q5. In a disproportionation reaction, the element that undergoes both oxidation and reduction acts as:
• A. Only an oxidizing agent.
• B. Only a reducing agent.
• C. Both an oxidizing and a reducing agent.
• D. Neither.
• Correct Answer: C
• Explanation: The same element is oxidized (acts as a reducing agent) and reduced (acts as an oxidizing agent) in the same reaction.

• Q6. Which of the following is a characteristic of a disproportionation reaction?
• A. Two different elements change their oxidation states.
• B. The total number of electrons lost is not equal to the number of electrons gained.
• C. A single element changes to two different oxidation states.
• D. The oxidation state of all elements remains the same.
• Correct Answer: C
• Explanation: The key feature is that a single element in one oxidation state splits into two different oxidation states in the products.

• Q7. In the reaction 2H_2O_2 → 2H_2O + O_2, what is the change in the oxidation number of oxygen that is reduced?
• A. -1 to 0
• B. -1 to -2
• C. 0 to -1
• D. -2 to -1
• Correct Answer: B
• Explanation: Oxygen is reduced when its oxidation number decreases. In H_2O_2 it is -1, and in H_2O it becomes -2.

• Q8. If an element is at its highest possible oxidation state, can it undergo disproportionation?
• A. Yes, always.
• B. Yes, but only with a catalyst.
• C. No, it cannot be oxidized further.
• D. No, it cannot be reduced further.
• Correct Answer: C
• Explanation: For disproportionation, the element must be able to be both oxidized and reduced. If it's at its highest state, it cannot be oxidized further.

• Q9. The reaction 2Cu^+ → Cu + Cu^{2+} is an example of:
• A. Comproportionation
• B. Disproportionation
• C. A simple redox reaction
• D. A precipitation reaction
• Correct Answer: B
• Explanation: Copper(I) (Cu^+) is in an intermediate oxidation state (+1). It is both reduced to Cu(0) and oxidized to Cu^{2+}(+2).

• Q10. In a disproportionation reaction, the total number of electrons lost:
• A. Is greater than the number of electrons gained.
• B. Is less than the number of electrons gained.
• C. Is exactly equal to the number of electrons gained.
• D. Depends on the specific reaction.
• Correct Answer: C
• Explanation: This is a fundamental law of redox reactions. The total increase in oxidation numbers must equal the total decrease in oxidation numbers.

Multiple Choice Questions (MCQs)

• Q1. In the disproportionation of hydrogen peroxide, what is the oxidation number of oxygen in the reactant?
• A. 0
• B. -1
• C. -2
• D. +1
• Correct Answer: B
• Explanation: In hydrogen peroxide (H_2O_2), oxygen has an oxidation number of -1.

• Q2. Which of the following best defines a disproportionation reaction?
• A. A reaction where an element is reduced and another is oxidized.
• B. A reaction where the same element is both oxidized and reduced.
• C. A reaction where an element is only oxidized.
• D. A reaction where an element is only reduced.
• Correct Answer: B
• Explanation: Disproportionation is when a single element undergoes both oxidation and reduction simultaneously.

• Q3. In the reaction 2H_2O_2 → 2H_2O + O_2, the oxygen in H_2O_2 is:
• A. Only oxidized
• B. Only reduced
• C. Both oxidized and reduced
• D. Neither oxidized nor reduced
• Correct Answer: C
• Explanation: The oxygen in H_2O_2 (-1) is both reduced to -2 in H_2O and oxidized to 0 in O_2.

• Q4. Which of the following is NOT a requirement for a disproportionation reaction?
• A. The element must be in an intermediate oxidation state.
• B. The element must have at least three accessible oxidation states.
• C. Two different elements must change oxidation numbers.
• D. The same element must be both oxidized and reduced.
• Correct Answer: C
• Explanation: Disproportionation involves only one element changing its oxidation state in two different ways, not two different elements.

Q5. In the disproportionation of H_2O_2, which product is formed by reduction? A. O_2 B. H_2O C. Both are formed by reduction. D. Neither is formed by reduction. Correct Answer: B Explanation: The oxygen in H_2O has an oxidation number of -2, which is lower than its starting state of -1. A decrease in oxidation number is reduction.

• Q6. The reaction 2Cu^+ → Cu + Cu^{2+} is an example of:
• A. Oxidation
• B. Reduction
• C. Disproportionation
• D. Comproportionation
• Correct Answer: C
• Explanation: Copper(I) is simultaneously oxidized to Cu^{2+} and reduced to Cu, making it a disproportionation reaction.

• Q7. What is the term for the reverse of a disproportionation reaction?
• A. Oxidation
• B. Reduction
• C. Comproportionation
• D. Decomposition
• Correct Answer: C
• Explanation: Comproportionation is when two different oxidation states of the same element react to form a single intermediate oxidation state.

• Q8. Which statement is true for a disproportionation reaction?
• A. The total electrons lost is always greater than electrons gained.
• B. The total electrons lost is always less than electrons gained.
• C. The total electrons lost equals the total electrons gained.
• D. Electrons are not transferred.
• Correct Answer: C
• Explanation: In any redox reaction, including disproportionation, the number of electrons lost must equal the number of electrons gained.

• Q9. In the decomposition of hydrogen peroxide, the element that is disproportionated is:
• A. Hydrogen
• B. Oxygen
• C. Both hydrogen and oxygen
• D. Neither
• Correct Answer: B
• Explanation: Only oxygen changes its oxidation number, from -1 to both -2 and 0. Hydrogen remains at +1 throughout.

Q10. Why can't an element at its lowest oxidation state undergo disproportionation? A. It cannot be reduced further. B. It cannot be oxidized further. C. It cannot be both oxidized and reduced. D. It cannot exist in that state. Correct Answer: A Explanation: At its lowest oxidation state, an element has already gained as many electrons as it can. It cannot be reduced further, so it cannot undergo disproportionation.

Multiple Choice Questions (MCQs)

• Q1. What is a disproportionation reaction?
• A. A reaction where one element is oxidized and another is reduced.
• B. A reaction where the same element is both oxidized and reduced.
• C. A reaction where an element is only oxidized.
• D. A reaction where an element is only reduced.
• Correct Answer: B
• Explanation: Disproportionation is defined as a redox reaction where a single element undergoes both oxidation and reduction simultaneously.

Q2. In the decomposition of hydrogen peroxide (2H_2O_2 → 2H_2O + O_2), what is the oxidation number of oxygen in H_2O_2? A. -2 B. -1 C. 0 D. +1 Correct Answer: B Explanation: In hydrogen peroxide, oxygen is in the -1 oxidation state, which is an exception to the general rule.

• Q3. In the same reaction, which product contains the oxidized form of oxygen?
• A. H_2O
• B. O_2
• C. Both
• D. None
• Correct Answer: B
• Explanation: The oxygen in O_2 has an oxidation number of 0, which is higher than -1, so it has been oxidized.

• Q4. In the disproportionation of H_2O_2, the oxygen in H_2O_2 acts as:
• A. Only an oxidizing agent.
• B. Only a reducing agent.
• C. Both an oxidizing and a reducing agent.
• D. Neither.
• Correct Answer: C
• Explanation: Since it is both oxidized and reduced, it acts as both a reducing agent (when oxidized) and an oxidizing agent (when reduced).

• Q5. For an element to undergo disproportionation, it must be in:
• A. Its highest oxidation state.
• B. Its lowest oxidation state.
• C. An intermediate oxidation state.
• D. The zero oxidation state.
• Correct Answer: C
• Explanation: The element must be able to go to a higher state (oxidation) and a lower state (reduction), so it must be intermediate.

• Q6. Which of the following is a disproportionation reaction?
• A. 2H_2 + O_2 → 2H_2O
• B. Zn + 2HCl → ZnCl_2 + H_2
• C. 2H_2O_2 → 2H_2O + O_2
• D. NaOH + HCl → NaCl + H_2O
• Correct Answer: C
• Explanation: The decomposition of hydrogen peroxide is a classic example of disproportionation.

• Q7. In the reaction 2Cu^+ → Cu + Cu^{2+}, what is the change in oxidation number for the copper that is oxidized?
• A. 0 to +1
• B. +1 to 0
• C. +1 to +2
• D. +2 to +1
• Correct Answer: C
• Explanation: The copper that is oxidized goes from +1 to +2, an increase in oxidation number.

• Q8. What is the reverse of a disproportionation reaction called?
• A. Oxidation
• B. Reduction
• C. Comproportionation
• D. Decomposition
• Correct Answer: C
• Explanation: Comproportionation is the reaction where two different oxidation states of the same element combine to form an intermediate state.

• Q9. In the disproportionation of H_2O_2, the total number of electrons lost is:
• A. Greater than the number of electrons gained.
• B. Less than the number of electrons gained.
• C. Equal to the number of electrons gained.
• D. Cannot be determined.
• Correct Answer: C
• Explanation: This is a fundamental principle: in any balanced redox reaction, the electrons lost equal the electrons gained.

• Q10. Can an element at its highest oxidation state undergo disproportionation?
• A. Yes, always.
• B. Yes, but only with a catalyst.
• C. No, it cannot be oxidized further.
• D. No, it cannot be reduced further.
• Correct Answer: C
• Explanation: To undergo disproportionation, the element must be able to be both oxidized and reduced. At its highest state, it cannot be oxidized further.