Problem 69 - Chemistry Matter and Change

Question

Use the half-reaction method to balance these equations for redox reactions. Add water molecules and hydrogen ions (in acid solutions) or hydroxide ions (in basic solutions) as needed. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{NH}_{3}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)} \\ {\text { b. } \mathrm{Br}_{2} \rightarrow \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-} \text { (in basic solution) }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. \(4\mathrm{NH}_3 + 6\mathrm{NO}_2 + 6\mathrm{H}^+ \rightarrow 5\mathrm{N}_2 + 6\mathrm{H}_2\mathrm{O}\) b. \(2\mathrm{Br}_2 + 12\mathrm{OH}^- \rightarrow 2\mathrm{Br}^- + 2\mathrm{BrO}_3^- + 6\mathrm{H}_2\mathrm{O}\)

Step 1: Identify Oxidation and Reduction Half-Reactions (Equation a)

For the reaction \( \mathrm{NH}_3(\mathrm{g}) + \mathrm{NO}_2(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(l) \), identify oxidation and reduction. \( \mathrm{NH}_3 \) is oxidized to \( \mathrm{N}_2 \), and \( \mathrm{NO}_2 \) is reduced to \( \mathrm{N}_2 \).

Step 2: Balance Atoms and Charges (Equation a)

Write and balance each half-reaction for atoms and charges: 1. \( \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \) (Oxidation) - Add 1 electron (e-) to balance: \( \mathrm{NH}_3 + 1\mathrm{e}^- \rightarrow \mathrm{N}_2 \)2. \( \mathrm{NO}_2 \rightarrow \mathrm{N}_2 \) (Reduction) - Add 1 electron (e-) and 2 protons (H+): \( \mathrm{NO}_2 + 1\mathrm{e}^- + 2\mathrm{H}^+ \rightarrow \mathrm{N}_2 + \mathrm{H}_2\mathrm{O} \) Balance the equations to: 1. \( 2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3\mathrm{H}_2\mathrm{O} + 6\mathrm{e}^-\)2. \( 6\mathrm{e}^- + 3\mathrm{NO}_2 + 6\mathrm{H}^+ \rightarrow 3\mathrm{N}_2 + 3\mathrm{H}_2\mathrm{O}\)

Step 3: Combine the Half-Reactions (Equation a)

Combine the half-reactions by ensuring electrons cancel out:\[ 4\mathrm{NH}_3+ 6\mathrm{NO}_2 + 6\mathrm{H}^+ \rightarrow 5\mathrm{N}_2 + 6\mathrm{H}_2\mathrm{O} \]

Step 4: Identify Oxidation and Reduction Half-Reactions (Equation b)

For the reaction \( \mathrm{Br}_2 \rightarrow \mathrm{Br}^- + \mathrm{BrO}_3^- \) in basic solution, recognize \( \mathrm{Br}_2 \) is both oxidized (to \( \mathrm{BrO}_3^- \)) and reduced (to \( \mathrm{Br}^- \)).

Step 5: Balance Atoms and Charges (Equation b)

Write each half-reaction:1. \( \mathrm{Br}_2 \rightarrow \mathrm{Br}^- \) - Balance electrons: \( \mathrm{Br}_2 + 2\mathrm{e}^- \rightarrow 2\mathrm{Br}^- \)2. \( \mathrm{Br}_2 \rightarrow \mathrm{BrO}_3^- \) - Add 6 electrons, 6 water molecules (H2O) and 12 hydroxide ions (OH-): \( \mathrm{Br}_2 + 6\mathrm{\,H}_2\mathrm{O} + 6\mathrm{e}^- \rightarrow 2\mathrm{BrO}_3^- + 12\,\mathrm{OH}^- \)Balance the equations:1. \( 6\mathrm{e}^- + 2\mathrm{Br}_2 \rightarrow 4\mathrm{Br}^- \)2. \( \mathrm{Br}_2 + 6\mathrm{\,OH}^- \rightarrow 2\mathrm{BrO}_3^- + 6\mathrm{e}^- + 6\mathrm{H}_2\mathrm{O} \)

Step 6: Combine the Half-Reactions (Equation b)

Combine the balanced half-reactions:\[ 2\mathrm{Br}_2 + 12\mathrm{OH}^- \rightarrow 2\mathrm{Br}^- + 2\mathrm{BrO}_3^- + 6\mathrm{H}_2\mathrm{O} \]

Key Concepts

Oxidation Half-Reaction

Understanding oxidation half-reactions is essential when balancing redox reactions. In a redox process, oxidation refers to the loss of electrons by a molecule, atom, or ion.
During this process, the oxidation number of the element increases as it loses electrons.
For example, in Equation a of our exercise, ammonia (\( \mathrm{NH}_3 \)) is being oxidized to nitrogen gas (\( \mathrm{N}_2 \)). Ammonia loses electrons in this conversion.

• The balanced oxidation half-reaction: \( 2\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 6\mathrm{e}^- + 6\mathrm{H}^+ \)
• The electrons lost from ammonia must be accounted for when balancing the complete redox equation.

Through the oxidation half-reaction, you can see how the electrons are lost and plan to redistribute them to balance the complete reaction later.

Reduction Half-Reaction

The reduction half-reaction represents the part of a redox process where a substance gains electrons, leading to a decrease in its oxidation number.
Reduction is fundamental for balancing redox reactions, as it complements oxidation. In Equation a, nitrogen dioxide (\( \mathrm{NO}_2 \)) is reduced to nitrogen gas (\( \mathrm{N}_2 \)).
The nitrogen dioxide gains electrons, completing the reduction.

• The balanced reduction half-reaction: \( \mathrm{NO}_2 + 1\mathrm{e}^- + 2\mathrm{H}^+ \rightarrow \mathrm{N}_2 + \mathrm{H}_2\mathrm{O} \)
• The electron gain ensures the balance of the charge when combined with the oxidation half-reaction.

By clearly understanding the reduction half-reaction, one can ensure that both parts of the redox reactions are balanced before combining.

Acidic Solutions

Balancing redox reactions in acidic solutions involves specific steps to manage additional protons (\( \mathrm{H}^+ \)) present in the solution.
In these conditions, \( \mathrm{H}^+ \) ions are available for balancing charge and mass within the half-reactions.
For the reaction of ammonia and nitrogen dioxide, the solution is acidic.

• Extra \( \mathrm{H}^+ \) ions were added to the reduction half-reaction to achieve a balance in Equation a.
• This helps balance out the hydrogen atoms and charges in the half-reactions.

Incorporating \( \mathrm{H}^+ \) ions often also results in water being formed, further contributing to the completeness of the balanced reaction. This step ensures all elements and charges are evenly distributed.

Basic Solutions

Balancing redox reactions in basic solutions requires a different approach, notably the use of hydroxide ions (\( \mathrm{OH}^- \)).
Here, balancing involves both extra hydroxide ions and any water molecules needed to achieve complete balance.
For the reaction involving bromine (\( \mathrm{Br_2} \)), the solution is basic.

• Extra \( \mathrm{OH}^- \) ions are added to balance hydrogen and oxygen elements, as seen in Equation b.
• Specifically, hydroxide ions react with any excess hydrogen to form water molecules.

In basic solutions, the concentrations of hydroxide ions ensure that charges are neutralized across the reactions. Thus, electrons are balanced effectively, completing the redox process.