What is Disproportionation Reaction?

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Grade Level:

Class 12

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Definition

What is it?

A disproportionation reaction is a special type of redox reaction where a single element in a reactant is both oxidized and reduced simultaneously. This means the same element gains electrons (reduction) and loses electrons (oxidation) in the same chemical process, forming products with different oxidation states.

Simple Example

Quick Example

Imagine a student, Rahul, who is excellent at both scoring runs and taking wickets in a cricket match. In a single match, he scores 50 runs (showing one skill) and also takes 3 wickets (showing another skill). Similarly, in a disproportionation reaction, one element acts in two opposite ways – both gaining and losing electrons.

Worked Example

Step-by-Step

Let's look at the decomposition of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2).

Step 1: Write the unbalanced equation: H2O2 --> H2O + O2
---Step 2: Assign oxidation states to the elements in H2O2. Hydrogen is +1, so for H2O2, (2 * +1) + (2 * x) = 0, which means 2x = -2, so x = -1. Oxygen in H2O2 has an oxidation state of -1.
---Step 3: Assign oxidation states to the elements in the products. In H2O, Hydrogen is +1, Oxygen is -2. In O2, Oxygen is 0 (elemental state).
---Step 4: Identify the element whose oxidation state changed. Oxygen changed from -1 (in H2O2) to -2 (in H2O) and also from -1 (in H2O2) to 0 (in O2).
---Step 5: Determine oxidation and reduction. Oxygen going from -1 to -2 means gaining electrons, which is reduction. Oxygen going from -1 to 0 means losing electrons, which is oxidation.
---Step 6: Since the same element (Oxygen) is both oxidized and reduced, this is a disproportionation reaction.

Why It Matters

Understanding disproportionation reactions is crucial in fields like medicine for drug synthesis and in environmental science for water treatment processes. Chemical engineers use this knowledge to design efficient industrial processes, and even in developing new battery technologies for EVs, these reactions play a role.

Common Mistakes

MISTAKE: Thinking that any redox reaction where one reactant forms two products is disproportionation. | CORRECTION: The key is that a *single element* within that reactant must change its oxidation state in *two opposite ways* (both increase and decrease).

MISTAKE: Confusing disproportionation with comproportionation reactions. | CORRECTION: In disproportionation, one element in one reactant forms products with higher and lower oxidation states. In comproportionation, two different oxidation states of the *same element* react to form a single product with an intermediate oxidation state.

MISTAKE: Incorrectly assigning oxidation states, especially for peroxides or elements in their elemental form. | CORRECTION: Always remember standard rules: H is +1 (except in metal hydrides), O is -2 (except in peroxides where it's -1, and superoxides where it's -1/2), and elements in their free state have an oxidation state of 0.

Practice Questions

Try It Yourself

QUESTION: Is the reaction 2Cu+ --> Cu2+ + Cu a disproportionation reaction? | ANSWER: Yes, because copper (Cu) in Cu+ has an oxidation state of +1. In the products, Cu2+ has an oxidation state of +2 (oxidation), and Cu has an oxidation state of 0 (reduction). The same element, Cu, is both oxidized and reduced.

QUESTION: Chlorine gas (Cl2) reacts with dilute NaOH to form NaCl, NaClO, and H2O. Identify the element undergoing disproportionation and its initial and final oxidation states. | ANSWER: Chlorine (Cl) undergoes disproportionation. Initial oxidation state in Cl2 is 0. Final oxidation states are -1 in NaCl (reduction) and +1 in NaClO (oxidation).

QUESTION: In the reaction 3MnO4(2-) + 4H+ --> 2MnO4(-) + MnO2 + 2H2O, identify the element that disproportionates. Show its initial and final oxidation states. | ANSWER: Manganese (Mn) disproportionates. Initial oxidation state in MnO4(2-) is +6. Final oxidation states are +7 in MnO4(-) (oxidation) and +4 in MnO2 (reduction).

MCQ

Quick Quiz

Which of the following reactions is an example of disproportionation?

2Na + Cl2 --> 2NaCl

Zn + CuSO4 --> ZnSO4 + Cu

3S + 6NaOH --> 2Na2S + Na2SO3 + 3H2O

Fe2O3 + 3CO --> 2Fe + 3CO2

The Correct Answer Is:

In option C, sulfur (S) has an oxidation state of 0 in elemental S. In Na2S, S is -2 (reduction). In Na2SO3, S is +4 (oxidation). Thus, sulfur is both oxidized and reduced.

Real World Connection

In the Real World

Disproportionation reactions are crucial in how bleaching agents like sodium hypochlorite (NaOCl) work. When you use a household bleach to clean clothes or surfaces, the hypochlorite ion disproportionates, releasing oxygen that helps remove stains and kill germs. This makes our homes cleaner and safer, just like how a water purifier ensures clean drinking water for every Indian household.

Key Vocabulary

Key Terms

REDOX REACTION: A chemical reaction involving both oxidation (loss of electrons) and reduction (gain of electrons) | OXIDATION STATE: A number assigned to an element in a compound that represents the number of electrons gained or lost by that element | OXIDATION: Loss of electrons, resulting in an increase in oxidation state | REDUCTION: Gain of electrons, resulting in a decrease in oxidation state | SIMULTANEOUSLY: Happening at the same time

What's Next

What to Learn Next

Now that you understand disproportionation, you can explore comproportionation reactions, which are the reverse process. Then, you can dive deeper into balancing redox reactions using the oxidation state method or the half-reaction method, which are super important for higher chemistry.