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

(b) M/5, M/3, M/2

Step 1: Understanding Equivalent Weight

The equivalent weight of a compound in a reaction is determined by the amount of the substance that gains or donates one mole of electrons. In essence, it is the molecular weight divided by the number of electrons transferred in the reaction.

Step 2: Determine Reaction in Acidic Medium

In an acidic medium, the half-reaction for \(abla KMnO_4 \) involves \(abla Mn^{7+} \) getting reduced to \(abla Mn^{2+} \). This accounts for a change of \(abla +5\) in oxidation states, thus 5 electrons are involved.

Step 3: Calculate Equivalent Weight in Acidic Medium

For the acidic medium, the equivalent weight is calculated as \(abla M/5 \) because 5 electrons are transferred during the reaction.

Step 4: Determine Reaction in Neutral Medium

In a neutral medium, the half-reaction involves \(abla Mn^{7+} \) changing to \(abla MnO_2 \) (in \(abla Mn^{4+} \) state). The change involves a transfer of 3 electrons.

Step 5: Calculate Equivalent Weight in Neutral Medium

For the neutral medium, the equivalent weight is calculated as \(abla M/3 \) due to the transfer of 3 electrons in the reaction.

Step 6: Determine Reaction in Alkaline Medium

In a strong alkaline medium, \(abla Mn^{7+} \) gets reduced to \(abla MnO_4^{2-} \) where it remains in \(abla Mn^{6+} \) oxidation state whereby only 1 electron is transferred.

Step 7: Calculate Equivalent Weight in Alkaline Medium

In an alkaline medium, the equivalent weight corresponds to \(abla M/2 \) as 2 electrons are transferred.

Key Concepts

Equivalent Weight

When we talk about equivalent weight in chemistry, it refers to the idea of how a specific substance interacts in a chemical reaction. It helps us understand how much of a substance participates in electron exchange during reactions. Specifically, the equivalent weight is calculated as the molecular weight divided by the number of electrons exchanged.

Here’s how you can find it:

• Identify the substance involved in the reaction.
• Determine the molecular weight of the substance.
• Find out how many electrons are gained or lost by the substance in the reaction.
• Divide the molecular weight by the number of electrons transferred.

Understanding equivalent weight is vital, especially when dealing with redox reactions, where electron transfer is central.

Oxidation Reduction Reactions

Oxidation-reduction, or redox reactions, are chemical processes where one substance gains electrons (reduction) while another loses electrons (oxidation). This fascinating electron exchange is crucial for many chemical reactions and processes.

In redox reactions:

• "Oxidation" refers to the loss of electrons, leading to an increase in oxidation state.
• "Reduction" means gaining electrons, causing a decrease in oxidation state.
• The substance that gets oxidized is known as the reducing agent because it reduces another substance by donating electrons.
• The substance that gets reduced is called the oxidizing agent because it oxidizes the other by accepting electrons.

A common example of a redox reaction is rusting of iron, where iron reacts with oxygen resulting in the formation of iron oxide.

KMnO4 Reactions in Different Media

Potassium permanganate ( KMnO_4 ) is a strong oxidizing agent used in various chemical reactions. Its behavior varies significantly across different media, affecting its equivalent weight.

In an acidic medium, KMnO_4 is reduced from Mn^{7+} to Mn^{2+} , involving a change of five electrons. Thus, the equivalent weight is M/5 .

In a neutral medium, the half-reaction involving Mn^{7+} reduces down to MnO_2 ( Mn^{4+} ), involving three electrons. As a result, the equivalent weight is M/3 .

In an alkaline medium, KMnO_4 reduces from Mn^{7+} to MnO_4^{2-} , retaining the Mn^{6+} state, indicating a transfer of only one electron, giving an equivalent weight of M/2 . Understanding these changes is critical for applying KMnO_4 in analytical chemistry effectively.