Problem 17 - A Complete Resource Book in Chemistry for JEE Main

Question

The equivalent weights of \(\mathrm{KMnO}_{4}\) in an acidic, a neutral and a strong alkaline medium respectively are \((\mathrm{M}=\) molecular weight \()\) (a) \(\mathrm{M} / 5, \mathrm{M} / 2, \mathrm{M}\) (b) \(\mathrm{M} / 5, \mathrm{M} / 3, \mathrm{M} / 2\) (c) \(\mathrm{M} / 5, \mathrm{M} / 3, \mathrm{M}\) (d) \(\mathrm{M} / 3, \mathrm{M}, \mathrm{M} / 5\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\mathrm{M}/5, \mathrm{M}/3, \mathrm{M}\).

Step 1: Determine the reaction in acidic medium

In acidic medium, \( \mathrm{KMnO}_{4} \) acts as a strong oxidizing agent and gets reduced from MnO\(_4^-\) to Mn\(^{2+}\). The change in oxidation state of manganese is from +7 to +2. So, the number of moles of electrons exchanged per mole of \( \mathrm{KMnO}_{4} \) is 5. Therefore, the equivalent weight in acidic medium is given by the formula \( \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{5} \).

Step 2: Determine the reaction in neutral medium

In a neutral medium, \( \mathrm{KMnO}_{4} \) is reduced from MnO\(_4^-\) to MnO\(_2\). The change in oxidation state is from +7 to +4. Thus, for the neutral medium, the equivalent weight is \( \frac{\text{Molecular Weight}}{3} \), since the number of electrons transferred is 3.

Step 3: Determine the reaction in alkaline medium

In a strong alkaline medium, \( \mathrm{KMnO}_{4} \) is reduced from MnO\(_4^-\) to MnO\(^{4-}_2 \), changing the oxidation state from +7 to +6. Therefore, the number of electrons transferred is 1. The equivalent weight in a strong alkaline medium is \( \text{Molecular Weight} \) as the transfer involves only one electron.

Step 4: Match the Equivalent Weights with the Options

We have calculated the equivalent weights as \( M/5 \) for acidic, \( M/3 \) for neutral, and \( M \) for alkaline medium. Compare these with the given options to find the correct set.

Key Concepts

Equivalent weight

Equivalent weight is a concept used in chemistry for judging the equivalency of substances in a chemical reaction. It represents the mass of a substance that will react with or displace a fixed quantity of another substance. Specifically, it is the mass of a substance that supplies or consumes one mole of electrons during a redox reaction.
In practical terms, the equivalent weight is calculated by dividing the molecular weight of a compound with the number of electrons transferred in a reaction. For instance, when dealing with potassium permanganate (\(\mathrm{KMnO}_4\)) in different mediums, understanding its equivalent weight helps predict how it will perform in varying chemical environments.- In acidic medium: Equivalent weight = \(\frac{Molecular\ Weight}{5}\)
- In neutral medium: Equivalent weight = \(\frac{Molecular\ Weight}{3}\)
- In alkaline medium: Equivalent weight = \(Molecular\ Weight\) itself

Oxidation state

Oxidation state is a concept that defines the degree of oxidation (loss of electrons) of an atom in a compound. It's represented as an integer and helps in determining how electrons are distributed in chemical reactions.
In the case of KMnO\(_4\), manganese has a variable oxidation state which changes depending upon the reaction medium. It's important to comprehend how these changes correlate with how a substance interacts in a chemical reaction.- In acidic conditions: Manganese in KMnO\(_4\) reduces from +7 to +2, leading to the exchange of 5 electrons.
- In neutral conditions: Manganese goes from +7 to +4, exchanging 3 electrons.
- In alkaline conditions: The reduction goes from +7 to +6, transferring just 1 electron.These transitions in oxidation states are crucial for determining equivalent weights.

KMnO4 reactions

Potassium permanganate, or KMnO\(_4\), is a versatile oxidation agent used widely in chemistry. Its role as an oxidizing agent means that it can accept electrons from other substances. The nature of these reactions varies with the medium in which they occur.- **Acidic Medium**: Here, KMnO\(_4\) reduces to Mn\(^{2+}\), significantly lowering its oxidation state from +7 to +2, demonstrating strong oxidation capabilities.
- **Neutral Medium**: In a neutral setup, KMnO\(_4\) reduces to MnO\(_2\), changing its oxidation state to +4.
- **Alkaline Medium**: Under alkaline conditions, KMnO\(_4\) reduces to a form with Mn in the oxidation state of +6. The changes in these reactions are key to understanding their varied chemical behaviors.

Redox reactions

Redox (reduction-oxidation) reactions are fundamental in chemistry, illustrating the transfer of electrons between substances. These reactions are characterized by one substance gaining electrons (reduction) while another loses electrons (oxidation).Potassium permanganate often acts as an oxidizing agent in these redox reactions, especially significant in its reactions through different mediums:
- **Acidic conditions** involve a significant electron exchange of five electrons as KMnO\(_4\) reduces to Mn\(^{2+}\).
- **Neutral conditions** engage in a moderate electron exchange of three electrons as KMnO\(_4\) reduces to MnO\(_2\).
- **Alkaline conditions** involve a minimal electron transfer of one as KMnO\(_4\) reduces to a +6 state.Redox reactions are crucial for energy transfer and are part of many biochemical and industrial processes.

Acidic, neutral, alkaline mediums

The behavior of chemical substances can vary significantly depending on whether they are in acidic, neutral, or alkaline mediums. This is especially true for agents like KMnO\(_4\). Understanding the characteristics of each environment helps in predicting the compound's behavior during chemical reactions:- **Acidic Medium**: Dominated by the presence of H\(^+\) ions, this environment enables KMnO\(_4\) to act as a strong oxidizing agent, transferring more electrons.
- **Neutral Medium**: Mostly water with equal H\(^+\) and OH\(^-\) levels. Reactions here involve less electron transfer than in acidic conditions.
- **Alkaline Medium**: Characterized by excess OH\(^-\) ions, affecting the extent of reduction in KMnO\(_4\), transferring the fewest number of electrons.Each medium uniquely influences reaction mechanisms and component interaction based on its ionic composition.