Problem 2 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An example of a disproportionation reaction is: [Main April 12, 2019 (I)] (a) \(2 \mathrm{MnO}_{4}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) (b) \(2 \mathrm{NaBr}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{NaCl}+\mathrm{Br}_{2}\) (c) \(2 \mathrm{KMnO}_{4} \rightarrow \mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{MnO}_{2}+\mathrm{O}_{2}\) (d) \(2 \mathrm{CuBr} \rightarrow \mathrm{CuBr}_{2}+\mathrm{Cu}\)

Step-by-Step Solution

Answer

Option (d) is a disproportionation reaction.

Step 1: Understanding Disproportionation Reactions

A disproportionation reaction is a type of redox reaction in which a single element simultaneously undergoes both oxidation and reduction. This means that one portion of the element will lose electrons (oxidation) while another portion of the same element will gain electrons (reduction).

Step 2: Analyze Each Reaction Option

Let's analyze each given reaction for evidence of disproportionation: (a) In this reaction, Mn changes from an oxidation state of +7 in MnO4⁻ to +2 in Mn²⁺ (reduction), and I⁻ changes from -1 to 0 in I₂ (oxidation). Mn undergoes only reduction, and I undergoes only oxidation, hence it is not disproportionation. (b) Here, Na stays at +1 and Br goes from -1 in NaBr to 0 in Br₂ (oxidation). Cl goes from 0 in Cl₂ to -1 in NaCl (reduction). Two different elements are involved, so it is not a disproportionation reaction. (c) In this reaction, Mn is initially in a +7 oxidation state in KMnO₄ and changes to +6 in K₂MnO₄ (reduction) and +4 in MnO₂ (oxidation). The same element is undergoing both oxidation and reduction, indicating disproportionation. (d) This reaction features Cu going from the +1 oxidation state in CuBr to +2 in CuBr₂ (oxidation) and 0 in Cu (reduction), confirming it as a disproportionation reaction.

Step 3: Identifying Disproportionation Reaction

From the analysis, reaction (d) is a disproportionation reaction because copper (Cu) simultaneously undergoes oxidation and reduction, showing changes from +1 to +2 and 0 oxidation states, respectively.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental chemical processes where the transfer of electrons between species takes place. These reactions entail two simultaneous processes: oxidation, where a species loses electrons, and reduction, where a species gains electrons.
Understanding redox reactions is essential, as they are the backbone of many natural and industrial chemical processes, ranging from cellular respiration to the rusting of iron.

• **Oxidation**: Generally involves the gain of oxygen or the loss of hydrogen or electrons. For example, in the equation for reaction (d) above, copper (Cu) is oxidized from a +1 to a +2 state in CuBr₂, meaning it has lost an electron.
• **Reduction**: Involves the loss of oxygen or the gain of hydrogen or electrons. In the same reaction, copper in CuBr goes from +1 to 0 as Cu, indicating it has gained an electron.

In redox reactions, the total number of electrons lost in oxidation equals the total number of electrons gained in reduction, ensuring the conservation of charge.

Oxidation and Reduction

Oxidation and reduction are key processes in any redox reaction, each symbolically represented with changes in oxidation states for the elements involved. To understand these concepts, it's vital to recognize how oxidation states indicate the degree of oxidation or reduction. This helps you determine the type of reaction occurring and identify a disproportionation reaction effectively.
For example, in step 1 of the solution, the oxygen states are reviewed to check for disproportionation. This is a crucial step in identifying how one element may undergo both processes.

• **Oxidation State Changes**: Symbols like + and - indicate whether an element is being oxidized or reduced. A rise in the oxidation number, as seen from Cu +1 to Cu +2, means oxidation. Conversely, a decrease indicates reduction.
• **Identifying Disproportionation**: In a disproportionation reaction, a single element will exhibit both an increase and a decrease in its oxidation state, such as in the CuBr breakdown where copper simultaneously acts as both oxidizer and reducer.

By practicing with different reactions and calculating potential changes in oxidation numbers, one can better understand these critical chemical processes.

Chemical Equilibrium

Chemical equilibrium is a state in which the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products. In redox chemistry, equilibrium has unique implications, especially when balancing and identifying reactions.
To exemplify, the reaction front shows a definite direction usually defined by changes in concentration or states. However, once equilibrium is reached, the system achieves a stable point where reactants are continuously being converted to products and vice versa at equal rates.

• **Dynamic Nature**: Equilibrium is not static; it involves continual and dynamic interchange between reactants and products. It's like a seesaw that finally balances at a point where neither side is dominant.
• **Influence on Redox Reactions**: In redox equilibria, parameters such as pressure, temperature, and concentration can shift the balance, dictated by Le Chatelier's Principle.
• **Achieving Balance**: To balance redox reactions, including disproportionation ones, understanding the equilibrium state helps predict the extent and feasibility of a reaction.

These principles are essential in predicting how different reactions behave under various conditions, including those involving disproportionation.