Question

Number of electrons transferred in each case when \(\mathrm{KMnO}_{4}\) acts as an oxidizing agent to give \(\mathrm{MnO}_{2}\), \(\mathrm{Mn}^{2+}, \mathrm{Mn}(\mathrm{OH})_{2}\) and \(\mathrm{MnO}_{4}^{2-}\) are respectively \(\quad[\mathbf{2 0 0 2}]\) (a) \(1,3,4\) and 5 (b) \(4,3,1\) and 5 (c) \(5,4,3\) and 1 (d) \(3,5,4\) and 1

Short answer

Option (d): (3,5,4) and 1.

Step-by-step solution

Determine Oxidation States of Mn

To find the number of electrons transferred, we first need to determine the oxidation state of Mn in each compound. In \(\mathrm{KMnO}_{4}\), Mn has an oxidation state of \(+7\). For each product:- In \(\mathrm{MnO}_{2}\), Mn has an oxidation state of \(+4\).- In \(\mathrm{Mn}^{2+}\), Mn has an oxidation state of \(+2\).- In \(\mathrm{Mn(OH)}_{2}\), Mn has an oxidation state of \(+2\) as well.- In \(\mathrm{MnO}_{4}^{2-}\), Mn has an oxidation state of \(+6\).

Calculate Electron Transfer for Each Reaction

Now, calculate the electron transfer for each reaction:1. From \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnO}_{2}\): Mn changes from \(+7\) to \(+4\). This is a change of \(7 - 4 = 3\) electrons.2. From \(\mathrm{KMnO}_{4}\) to \(\mathrm{Mn}^{2+}\): Mn changes from \(+7\) to \(+2\). This is a change of \(7 - 2 = 5\) electrons.3. From \(\mathrm{KMnO}_{4}\) to \(\mathrm{Mn(OH)}_{2}\): Same as \(\mathrm{Mn}^{2+}\), the change is also \(7 - 2 = 5\) electrons.4. From \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnO}_{4}^{2-}\): Mn changes from \(+7\) to \(+6\). This is a change of \(7 - 6 = 1\) electron.

Match Changes with Provided Options

Match the calculated electron changes with the given options:- For \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnO}_{2}\), 3 electrons are transferred.- For \(\mathrm{KMnO}_{4}\) to \(\mathrm{Mn}^{2+}\), 5 electrons are transferred.- For \(\mathrm{KMnO}_{4}\) to \(\mathrm{Mn(OH)}_{2}\), 5 electrons are transferred.- For \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnO}_{4}^{2-}\), 1 electron is transferred. These correspond to option (d): \((3,5,4)\) and 1.

Key concepts

Electron Transfer

In redox reactions, understanding electron transfer is fundamental. It represents the movement of electrons from one substance to another. When a substance gains electrons, it gets reduced, and when it loses electrons, it gets oxidized.
This movement is key in reactions involving oxidizing agents like potassium permanganate (\(\mathrm{KMnO}_{4}\)).

• Every chemical element has an oxidation state, indicating the number of electrons lost or gained.
• Electrons transferring result in a change in these states.

Through calculating these changes, students can determine exactly how many electrons are involved.
For instance, when \(\mathrm{KMnO}_{4}\) acts as an oxidizing agent:

• Reduction of Mn from +7 in \(\mathrm{KMnO}_{4}\) to +4 in \(\mathrm{MnO}_{2}\) involves 3 electrons.
• Reduction to +2 in \(\mathrm{Mn}^{2+}\) and \(\mathrm{Mn(OH)}_{2}\) involves 5 electrons each.
• Reduction to +6 in \(\mathrm{MnO}_{4}^{2-}\) involves 1 electron.

By tracking these atom-level adjustments, students get a clear picture of how electrons define the path and outcome of the reaction.

Oxidizing Agent

An oxidizing agent, like potassium permanganate, is a substance that can accept electrons. This agent helps another compound get oxidized by itself being reduced.The oxidizing agent is crucial in redox reactions because:

• It draws electrons away from other molecules in the reaction.
• By gaining electrons, it changes the overall charge dynamically, influencing other participants.

For potassium permanganate (\(\mathrm{KMnO}_{4}\)):
As it interacts with different compounds, it accepts electrons, facilitating significant changes.

• In scenarios where \(\mathrm{KMnO}_{4}\) oxidizes another compound, it potentially changes Mn's oxidation state from +7 to lower states.
• The acceptance of different numbers of electrons creates different manganese compounds with various oxidation states: +4, +2, or +6.

Remember, the strength of an oxidizing agent is often linked to its affinity for electrons. The higher the electron affinity, the stronger the oxidizing capability.

Redox Reactions

Redox reactions describe the simultaneous reduction and oxidation processes occurring in chemical reactions. Representing the changing of oxidation states, redox reactions require one compound to lose electrons while another gains.

• "Redox" comes from "reduction" and "oxidation," highlighting the dual nature of these processes.
• The oxidized compound sees its oxidation state increase, while the reduced compounds see theirs decrease.

In the case of \(\mathrm{KMnO}_{4}\):

• As the compound gives away electrons, \(\mathrm{KMnO}_{4}\) experiences a reduction in the oxidation state of Mn.
• This contrasts with the substance interacting with it, as that substance undergoes oxidation.

Equations showcasing these reactions affirm how central and applicable these processes are in chemistry, proving that understanding them equips students to tackle a wide array of chemical interactions efficiently. These dual shifts personify the complex dance of atoms and electrons, forming the basis of many biological and industrial chemical processes.