Problem 4 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of \(\mathrm{CO}_{2}\) is: (a) 1 (b) 10 (c) 2 (d) 5

Step-by-Step Solution

Answer

(a) 1

Step 1: Identify the Half-Reactions

First, identify the half-reactions for both the reduction of permanganate ion \((\mathrm{MnO}_4^-)\) and the oxidation of oxalate \((\mathrm{C}_2\mathrm{O}_4^{2-})\). The reduction of \(\mathrm{MnO}_4^-\) typically produces \(\mathrm{Mn}^{2+}\) in acidic medium.

Step 2: Write the Oxidation Reaction

Write down the oxidation half-reaction of oxalate: \[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-} \] This half-reaction shows that each mole of oxalate loses 2 electrons to form \(\text{CO}_2\).

Step 3: Analyze the Electron Exchange in Oxidation

From the oxidation half-reaction \(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-}\), note that the formation of two molecules of \(\text{CO}_2\) involves 2 electrons. Thus, forming one molecule of \(\text{CO}_2\) involves transferring 1 electron.

Key Concepts

Half-Reactions

In redox chemistry, a half-reaction is an equation that shows either the oxidation or reduction process separately. Each half-reaction involves the transfer of electrons, pivotal to understanding how the overall redox reaction progresses.
When analyzing any redox reaction, it's essential to identify the half-reactions first. This separation helps determine which species is being oxidized and which is being reduced. For example, in the reaction between oxalate and permanganate, we split the process into two half-reactions:

• The oxidation half, where oxalate ( \(\mathrm{C}_2\mathrm{O}_4^{2-}\)) loses electrons.
• The reduction half, where permanganate ( \(\mathrm{MnO}_4^-\)) gains electrons.

Understanding these half-reactions allows for a detailed analysis of electron movement and the stoichiometry of the reaction.

Oxidation

Oxidation is a fundamental concept in chemistry that involves the loss of electrons from a substance. As electrons are negatively charged, losing them means the oxidation state of the substance increases. This step is crucial in determining how substances chemically alter during a reaction.
In the context of the given exercise, the oxidation element is the oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)), which transforms into carbon dioxide (\(\text{CO}_2\)) by losing electrons. The balanced oxidation half-reaction is given by:

• \[\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-}\]

This equation shows that each oxalate ion releases two electrons as it forms two molecules of \(\text{CO}_2\). Therefore, oxidation is vital as it highlights the electron donation in redox reactions.

Electron Exchange

Electron exchange is the core of any redox process, determining how electrons transfer between atoms or ions during a reaction. In every redox reaction, one substance's oxidation results in another's reduction, effectively "exchanging" electrons between the two.
For our example involving oxalate and permanganate, the electron exchange occurs distinctly during the oxidation and reduction half-reactions:

• Oxidation of oxalate ion (\(\text{C}_2\text{O}_4^{2-}\)) involves losing electrons to form \(\text{CO}_2\).
• Reduction of permanganate ion (\(\text{MnO}_4^{-}\)) involves gaining electrons to form \(\text{Mn}^{2+}\).

Understanding the electron exchange helps in balancing redox reactions, ensuring the number of electrons lost equals those gained. This balance is crucial for any redox equation and establishes the foundation for predicting reaction outcomes. By breaking down the process into half-reactions, it becomes straightforward to track electron movement and appreciate their pivotal role in driving chemical transformations.