Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

In the redox reaction \(\mathrm{xKMnO}_{4}+\mathrm{yNH}_{3} \longrightarrow \mathrm{KNO}_{3}+\mathrm{MnO}_{2}+\mathrm{KOH}\) \(+\mathrm{H}_{2} \mathrm{O}\) (a) \(x=3, y=8\) (b) \(x=6, y=3\) (c) \(x=5, y=10\) (d) \(x=8, y=3\)

Step-by-Step Solution

Answer

Option (a) x=3, y=8 is the balanced scenario for this reaction.

Step 1: Analyze the Reaction Types

First, identify the type of reaction, which is a redox reaction. In redox reactions, oxidation and reduction occur simultaneously. Potassium permanganate (KMnO₄) is reduced, and ammonia (NH₃) is oxidized.

Step 2: Assign Oxidation Numbers

Assign oxidation numbers to each element in the reaction to track electron transfer. For KMnO₄, K is +1, Mn is +7, and O is -2. In NH₃, N is -3, and H is +1. In the products, MnO₂ has Mn at +4, and in KNO₃, N is +5.

Step 3: Determine Change in Oxidation States

Determine how the oxidation numbers change. Mn in KMnO₄ changes from +7 to +4, indicating reduction. N in NH₃ changes from -3 to +5, indicating oxidation.

Step 4: Balance Oxidation and Reduction Half-Reactions

Write and balance the two half-reactions: the reduction of MnO₄⁻ to MnO₂ and the oxidation of NH₃ to NO₃⁻. Ensure electrons lost in oxidation equal electrons gained in reduction.

Step 5: Balance the Electrons

Balance the electrons transferred in the half-reactions. Add appropriate coefficients to the chemicals in order to equate electron gain and loss.

Step 6: Combine the Balanced Half-Reactions

Combine the balanced half-reactions into a single balanced equation, verifying that the number of atoms and charge are conserved on both sides.

Step 7: Check Coefficient Options

Compare the balanced equation to the given options (a) to (d). Calculate expected values for coefficients x and y to determine which option is correctly balanced based on stoichiometry.

Step 8: Choose and Validate the Correct Answer

Choose the option where the coefficients x and y lead to a balanced chemical equation. Calculate based on stoichiometry which scenario balances all elements.

Key Concepts

Oxidation States

Understanding oxidation states is key to mastering redox reactions. An oxidation state tells us whether an element has lost or gained electrons during a chemical process. It helps us identify which species are oxidized and which are reduced. In the given reaction, let's break down the process:

• Potassium permanganate (KMnO₄) contains Mn with an oxidation state of +7. As it turns into MnO₂, the Mn oxidation state changes to +4. This change signifies that Mn gains electrons, indicating reduction.


• Ammonia (NH₃) has nitrogen with an oxidation state of -3. As it transforms into KNO₃, nitrogen's oxidation state becomes +5. Here, nitrogen loses electrons, a sign of oxidation.

Quantifying these changes allows us to see how electrons are transferred during the reaction, a critical first step in balancing it.

Balancing Chemical Equations

Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the reaction. In redox reactions, this balancing also needs to account for electron transfer. Here's how you tackle it:

• Write the unbalanced chemical equation first. For our problem, the equation involves KMnO₄ and NH₃ as reactants, and they must form KNO₃, MnO₂, KOH, and H₂O.


• Next, utilizing the oxidation states, determine which elements have changed oxidation numbers, thus identifying the ones that need balancing.
• Introduce coefficients to the equation so that atoms and charges are equivalent on both sides.

Balancing is a dynamic process, often requiring you to iterate your solution until all parts fit into place snugly, reflecting conservation of mass and charge.

Half-Reactions

Half-reactions are a convenient way to separately consider the oxidation and reduction processes within a redox reaction. By doing this, we can determine how electrons are exchanged:

• The oxidation half-reaction describes how NH₃ is transformed, with nitrogen losing electrons to become NO₃⁻.


• The reduction half-reaction handles the transformation of MnO₄⁻ in KMnO₄ to MnO₂, with manganese gaining electrons.


• By separately considering these half-reactions, you can balance the electron transfer before bringing them back together.

This division simplifies tracking and balancing electron flow, essential for correctly finalizing the chemical equation.

Electron Transfer

Electron transfer is at the heart of redox reactions, defining their very nature. In each reaction, one reactant gives up electrons (oxidation), while another gains them (reduction). Here's how to focus on electron flow:

• Identify the reducing agent, which is the species that donates electrons. In our case, NH₃ acts as a reducing agent by losing electrons.


• The oxidizing agent, here KMnO₄, accepts these electrons, thus becoming reduced. Mn in KMnO₄ accepts electrons and is reduced to MnO₂.


• The total number of electrons gained must equal the total lost to ensure the overall reaction is balanced.

By following the electrons' journey through a reaction, you gain a clearer understanding of the fundamental exchanges that drive chemical transformations.