Problem 16 - Chemistry The Molecular Science

Question

Balance these redox reactions, and identify the oxidizing agent and the reducing agent. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{g})\) (c) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Ti}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{TiO}_{2}(\mathrm{~s})\) in acidic solution (d) \(\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{MnO}_{2}(\mathrm{~s})\) in acidic solution (e) \(\mathrm{FeS}_{2}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{SO}_{2}(\mathrm{~g})\) (f) \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \longrightarrow \mathrm{O}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g})\) (g) \(\mathrm{Zn}(\mathrm{s})+\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s})+\operatorname{Hg}(\ell)\) in basic solution

Step-by-Step Solution

Answer

Reactions balanced. Identified oxidizing and reducing agents for all reactions.

Step 1: Write the Unbalanced Reaction for (a)

The given reaction is \( \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) \).

Step 2: Identify Oxidation and Reduction for (a)

- Carbon in \( \mathrm{CO} \) is oxidized to \( \mathrm{CO}_{2} \).- Oxygen in \( \mathrm{O}_{3} \) is reduced to \( \mathrm{O}_{2} \).- \( \mathrm{O}_{3} \) is the oxidizing agent and \( \mathrm{CO} \) is the reducing agent.

Step 3: Balance Atoms and Charge for (a)

The balanced reaction is already presented in the simplest form:\[ \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{3}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \]

Step 4: Write the Unbalanced Reaction for (b)

The given reaction is \( \mathrm{H}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g}) \).

Step 5: Identify Oxidation and Reduction for (b)

- Hydrogen is oxidized.- Chlorine is reduced.- \( \mathrm{Cl}_{2} \) is the oxidizing agent, and \( \mathrm{H}_{2} \) is the reducing agent.

Step 6: Balance Atoms and Charge for (b)

The balanced reaction is:\[ \mathrm{H}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2\mathrm{HCl}(\mathrm{g}) \]

Step 7: Write the Unbalanced Reaction for (c)

The given reaction is \( \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) + \mathrm{Ti}^{2+}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{TiO}_{2}(\mathrm{s}) \).

Step 8: Identify Oxidation and Reduction for (c)

- \( \mathrm{Ti}^{2+} \) is oxidized to \( \mathrm{TiO}_{2} \).- \( \mathrm{H}_{2} \mathrm{O}_{2} \) is reduced to \( \mathrm{H}_{2} \mathrm{O} \).- \( \mathrm{H}_{2} \mathrm{O}_{2} \) is the oxidizing agent, and \( \mathrm{Ti}^{2+} \) is the reducing agent.

Step 9: Balance Atoms and Charge for (c)

Assume acidic conditions:1. Balance oxygen using water: \[ \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) + \mathrm{Ti}^{2+}(\mathrm{aq}) \rightarrow \mathrm{TiO}_{2}(\mathrm{s}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \]2. Balance hydrogen using \( \mathrm{H}^{+} \): \[ 2\mathrm{H}^{+} + \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) + \mathrm{Ti}^{2+}(\mathrm{aq}) \rightarrow \mathrm{TiO}_{2}(\mathrm{s}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \]3. Balance charge with electrons: \[ \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) + \mathrm{Ti}^{2+}(\mathrm{aq}) \rightarrow \mathrm{TiO}_{2}(\mathrm{s}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) + 2\mathrm{e}^{-} \]

Step 10: Write the Unbalanced Reaction for (d)

The reaction is \( \mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g}) + \mathrm{MnO}_{2}(\mathrm{s}) \).

Step 11: Identify Oxidation and Reduction for (d)

- \( \mathrm{Cl}^{-} \) is oxidized to \( \mathrm{Cl}_{2} \).- \( \mathrm{MnO}_{4}^{-} \) is reduced to \( \mathrm{MnO}_{2} \).- \( \mathrm{MnO}_{4}^{-} \) is the oxidizing agent, and \( \mathrm{Cl}^{-} \) is the reducing agent.

Step 12: Balance Atoms and Charge for (d)

In acidic conditions:1. Balance \( \mathrm{Cl} \): \[ 2\mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g}) + \mathrm{MnO}_{2}(\mathrm{s}) \]2. Add \( \mathrm{H}^{+} \) and \( \mathrm{H}_{2}\mathrm{O} \): \[ 4\mathrm{H}^{+} + 2\mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Cl}_{2} + \mathrm{MnO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \]3. Balance charge: \[ 2\mathrm{Cl}^{-} + \mathrm{MnO}_{4}^{-} + 4\mathrm{H}^{+} + 2\mathrm{e}^{-} \rightarrow \mathrm{Cl}_{2} + \mathrm{MnO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \]

Step 13: Write the Unbalanced Reaction for (e)

The reaction is \( \mathrm{FeS}_{2}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}) + \mathrm{SO}_{2}(\mathrm{g}) \).

Step 14: Identify Oxidation and Reduction for (e)

- Sulfur in \( \mathrm{FeS}_{2} \) is oxidized to \( \mathrm{SO}_{2} \).- Oxygen is reduced.- \( \mathrm{O}_{2} \) is the oxidizing agent and \( \mathrm{FeS}_{2} \) is the reducing agent.

Step 15: Balance Atoms and Charge for (e)

The balanced reaction is:\[ 4\mathrm{FeS}_{2} + 11\mathrm{O}_{2} \rightarrow 2\mathrm{Fe}_{2} \mathrm{O}_{3} + 8\mathrm{SO}_{2} \]

Step 16: Write the Unbalanced Reaction for (f)

The reaction is \( \mathrm{O}_{3}(\mathrm{g}) + \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + \mathrm{NO}_{2}(\mathrm{g}) \).

Step 17: Identify Oxidation and Reduction for (f)

- \( \mathrm{NO} \) is oxidized to \( \mathrm{NO}_{2} \).- \( \mathrm{O}_{3} \) is reduced to \( \mathrm{O}_{2} \).- \( \mathrm{O}_{3} \) is the oxidizing agent, and \( \mathrm{NO} \) is the reducing agent.

Step 18: Balance Atoms and Charge for (f)

The balanced reaction:\[ \mathrm{O}_{3}(\mathrm{g}) + \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + \mathrm{NO}_{2}(\mathrm{g}) \]

Step 19: Write the Unbalanced Reaction for (g)

The reaction is \( \mathrm{Zn}(\mathrm{s}) + \mathrm{HgO}(\mathrm{s}) \rightarrow \mathrm{Zn(OH)}_{2}(\mathrm{s}) + \mathrm{Hg}(\ell) \).

Step 20: Identify Oxidation and Reduction for (g)

- \( \mathrm{Zn} \) is oxidized.- \( \mathrm{HgO} \) is reduced to \( \mathrm{Hg} \).- \( \mathrm{HgO} \) is the oxidizing agent and \( \mathrm{Zn} \) is the reducing agent.

Step 21: Balance Atoms and Charge for (g)

In basic conditions:1. Balance Zn and Hg: \[ \mathrm{Zn} + \mathrm{HgO} \rightarrow \mathrm{Zn(OH)}_{2} + \mathrm{Hg} \]2. Add \( \mathrm{OH}^{-} \) to balance O and H: \[ \mathrm{Zn} + \mathrm{HgO} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Zn(OH)}_{2} + \mathrm{Hg} \]

Key Concepts

Oxidizing Agent

In a redox reaction, an oxidizing agent is a substance that causes another substance to lose electrons, thereby becoming reduced itself. This means the oxidizing agent gains electrons. In reaction (a) from the exercise, \( \mathrm{O}_{3} \) acts as the oxidizing agent because it causes the oxidation of carbon monoxide (\( \mathrm{CO} \)) by accepting electrons and is reduced to oxygen (\( \mathrm{O}_{2} \)). Suspiciously, the name can imply that the oxidizing agent is itself oxidized, but it's important to remember that the oxidizing agent actually gets reduced.
To identify the oxidizing agent in a chemical reaction, you can:

• Look for the species that gains electrons.
• Identify which element's oxidation state decreases.
• Check the reactant species that accepts electrons in the reaction.

Understanding the role of an oxidizing agent helps clarify the electron transfer process crucial in redox reactions.

Reducing Agent

The reducing agent in a redox reaction is the substance that donates electrons to another substance, leading to its own oxidation. In simpler terms, it loses electrons while causing the other substance to gain electrons and become reduced. For example, in reaction (b) we see that \( \mathrm{H}_{2} \) serves as the reducing agent.
The hydrogen (\( \mathrm{H}_{2} \)) donates electrons to chlorine (\( \mathrm{Cl}_2 \)), forming hydrochloric acid (\( \mathrm{HCl} \)), while oxidizing itself. This identification helps understand which species shifts its electrons.

• Identifying the reducing agent involves finding the species that loses electrons.
• Check for the element whose oxidation state increases.
• Look for reactants that provide electrons in the reaction.

Remember, the ability to recognize a reducing agent is key in predicting electron flow and the direction of redox reactions.

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms for each element is equal on both sides of the reaction. This process obeys the law of conservation of mass. For example, in reaction (c), the reaction is comprehensively balanced using steps:1\. Balance oxygen first by adding water molecules.2\. Then balance hydrogen using hydrogen ions (\( \mathrm{H}^+ \)).3\. Finally, balance the charge by adding electrons.
During the process, it’s crucial to:

• Understand the oxidation state changes to recognize the oxidized and reduced species.
• Use \( \mathrm{H}^+ \) and water in acidic solutions or hydroxide (\( \mathrm{OH}^- \)) in basic solutions to balance hydrogen and oxygen.
• Finish by ensuring all electrons are balanced, contributing to charges to either side of the equation.

These steps give a clear, linear approach to obtaining a balanced equation, essential for accurate stoichiometric calculations and analysis.