Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

For the equation \(\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{NO} .\) The number of electrons in its balanced form would be (a) 6 (b) 4 (c) 3 (d) 9

Step-by-Step Solution

Answer

The balanced equation requires 3 electrons, so the answer is (c) 3.

Step 1: Identify the Reaction

The given chemical equation is \( ext{NO}_3^- + 4 ext{H}^+ + ext{e}^- \longrightarrow 2 ext{H}_2 ext{O} + ext{NO}\). At a glance, it's not clear if the equation is balanced with respect to both atoms and charges.

Step 2: Determine Oxidation States

Identify the change in oxidation states for the elements involved in the reaction. Here, nitrogen changes from +5 (in \( ext{NO}_3^-\)) to +2 (in \( ext{NO}\)). This implies a reduction process where electrons are gained.

Step 3: Calculate Electron Requirement

The reduction of nitrogen from +5 to +2 requires three electrons per nitrogen atom. Since there is one nitrate ion (\( ext{NO}_3^-\)) involved, it will demand 3 electrons to balance.

Step 4: Balance Electrons

Add electrons to the left side of the equation to indicate that three electrons are required for the reduction process. Since option (c) suggests 3 as an answer and matches this calculation, this indicates that the correct answer is 3.

Key Concepts

Oxidation States

Understanding oxidation states is crucial when working with redox (reduction-oxidation) reactions. An oxidation state is a number assigned to an element in a compound or ion to represent the number of electrons lost or gained by an atom of that element. It helps determine how electrons are transferred in a chemical reaction.
In the given equation, observe that nitrogen in the nitrate ion ( ext{NO}_3^- ) starts with an oxidation state of +5. This state changes to +2 in nitrogen monoxide ( ext{NO} ), indicating nitrogen has gained electrons. By tracking these states for each element, one can identify whether an atom was oxidized or reduced.

• Oxidation involves an increase in oxidation state (loss of electrons).
• Reduction involves a decrease in oxidation state (gain of electrons).

Recognizing these changes lays the foundation for balancing redox reactions.

Electron Transfer

Electron transfer is at the heart of redox reactions. In any chemical reaction involving electron transfer, one species will donate electrons (oxidation), and another will accept them (reduction). By observing how electrons shift between reactants and products, you can determine how to balance the reaction.
In the given example, we note that each nitrogen atom accepts three electrons when it transitions from a +5 state to a +2 state. Therefore, the nitrate ion ( ext{NO}_3^- ) captures electrons to become nitrogen monoxide ( ext{NO} ), highlighting a classic reduction process.

• Identify how many electrons are involved in each oxidation and reduction.
• Count the electrons transferred to ensure the reaction maintains equality in its balanced state.

This understanding helps ensure that the reaction equation remains both charge and mass balanced.

Chemical Equation Balancing

Balancing chemical equations, especially in redox reactions, ensures that the number of atoms for each element and the total charge are the same on both sides of the equation. To balance redox reactions, we often split them into half-reactions - one for oxidation and one for reduction.
In the analysis provided, balancing involves ensuring that the electrons lost in the reduction part matches those required by reactants. The calculation determined three electrons were needed. Hence, adding three electrons to the equation helped harmonize it.
Steps to balance include:

• Identifying and separating the half-reactions.
• Balancing elements and charges for each half-reaction.
• Ensuring the total electrons gained and lost between the half-reactions are equal.
• Adding any necessary terms, like coefficients, to both sides to reach a balanced state.

By following these steps, you guarantee mass conservation and charge neutrality in the final equation.