Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

In the redox reaction \(\mathrm{xMnO}+\mathrm{yPbO}_{2}+\mathrm{zHNO}_{3} \longrightarrow \mathrm{HMnO}_{4}\) \(+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) \(\mathrm{x}=2, \mathrm{y}=5, \mathrm{z}=10\) (b) \(x=10, y=2, z=5\) (c) \(x=3, y=7, z=6\) (d) \(x=5, y=2, z=10\)

Step-by-Step Solution

Answer

The correct choice is (a) \( x=2, y=5, z=10 \).

Step 1: Assign Oxidation States

Determine the oxidation states of each element in the reactants and products. \( \text{Mn in } \mathrm{MnO} \) typically has an oxidation state of +2, Pb in \( \mathrm{PbO}_{2} \) has +4, and H typically is +1, O is -2, and N in \( \mathrm{HNO}_{3} \) is +5. In the products, \( \text{Mn in } \mathrm{HMnO}_{4} \) has an oxidation state of +7, Pb in \( \mathrm{Pb}(\mathrm{NO}_{3})_{2} \) is +2.

Step 2: Identify Changes in Oxidation States

The oxidation state of Mn increases from +2 in \( \mathrm{MnO} \) to +7 in \( \mathrm{HMnO}_{4} \), indicating oxidation. The oxidation state of Pb decreases from +4 in \( \mathrm{PbO}_{2} \) to +2 in \( \mathrm{Pb(NO}_{3})_{2} \), indicating reduction.

Step 3: Balance the Redox Reaction

Ensure the changes in oxidation states are balanced by equalizing the total increase and decrease in oxidation states. Mn changes from +2 to +7, a change of +5, and Pb changes from +4 to +2, a change of -2 per atom. The least common multiple of 5 and 2 is 10, so we use this to balance.

Step 4: Determine Coefficients

To balance using the least common multiple method, set: \( 2 \times \mathrm{Mn} : 5 \times \mathrm{PbO}_{2} \). Thus, the coefficients needed are (a) 2 Mn, as the change is \( 2 \times 5 \), and \( 5 \times \left(-2\right)\). This implies \( (a) \text{ correct choice is } \; (a) \, \mathbf{x=2,\, y=5,\, z=10} \) as it matches needed balance.

Key Concepts

Oxidation States

In chemistry, oxidation states, also known as oxidation numbers, help us understand how electrons are distributed in compounds. They give us insight into the electron transfer that occurs during redox reactions. For example, in the provided redox reaction, each atom's oxidation state was examined. The oxidation state of manganese (Mn) in \(\mathrm{MnO}\) is typically +2. In contrast, lead (Pb) in \(\mathrm{PbO}_2\) is +4. The oxidation states for hydrogen (H), oxygen (O), and nitrogen (N) are generally +1, -2, and +5 in \(\mathrm{HNO}_3\) respectively. However, when we look at the products, the oxidation state changes; manganese becomes +7 in \(\mathrm{HMnO}_4\) and lead becomes +2 in \(\mathrm{Pb(NO}_3)_2\). By analyzing these shifts in oxidation states, we can identify where oxidation and reduction occur.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures mass conservation. The challenge is to maintain equal numbers of each type of atom on both sides of the reaction. In this redox reaction, we start by examining the total change in oxidation states to balance them.
The process involved using the least common multiple (LCM) to balance the shifts in oxidation states of Mn and Pb.
Here’s how:

• The oxidation state of Mn increases by 5 (from +2 to +7), requiring 2 Mn atoms for a total increase of 10.
• The reduction of Pb involves a decrease of 2 per atom, needing 5 Pb atoms for a total decrease of 10.

This approach ensures that the equation is balanced in terms of atoms and charge, which is critical when dealing with reactions that include changes in oxidation states.

Oxidation and Reduction

Oxidation and reduction are simultaneous processes that involve electron transfer. Oxidation refers to the loss of electrons, while reduction is the gain of electrons. These processes are evident when we examine oxidation states:

• In this reaction, Manganese (Mn) is oxidized as its oxidation state goes from +2 in \(\mathrm{MnO}\) to +7 in \(\mathrm{HMnO}_4\). This indicates a loss of electrons.
• Lead (Pb) is reduced as it changes from +4 in \(\mathrm{PbO}_2\) to +2 in \(\mathrm{Pb(NO}_3)_2\), showing a gain of electrons.

Recognizing these changes helps in understanding how electron flow occurs and what agents are involved as oxidizing and reducing counterparts.

Stoichiometry

Stoichiometry is the quantitative relationship among compounds involved in chemical reactions. This concept is key to determining the correct proportions of reactants and products.
In the context of the redox reaction discussed:

• First, we consider the stoichiometric coefficients that balance both mass and charge. Specifically, these coefficients ensure the total electrons lost equals the total electrons gained.
• For each Mn atom oxidized, two Pb atoms are needed for the reduction to match the electrons transferred, dictating coefficients of \( x=2\) and \( y=5\)
• Water \(\mathrm{H}_2\mathrm{O}\) and nitric acid \(\mathrm{HNO}_3\) are also balanced to meet the stoichiometric needs of hydrogen and oxygen in the reaction, resulting in the coefficient \( z=10\).

Mastering stoichiometry is crucial because it serves as the foundational tool that chemists use to predict the outcomes of chemical reactions accurately.