Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q) \pm \mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-{ }^{-\rightarrow} \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q)-\cdots\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

Short answer

The balanced equations and oxidizing/reducing agents are: (a) \(6\mathrm{I}^{-} + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \longrightarrow 6\mathrm{IO}_{3}^{-} + 2\mathrm{Cr}^{3+} + 12\mathrm{H}^{+}\) Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\); Reducing agent: \(\mathrm{I}^{-}\). (b) \(5\mathrm{MnO}_{4}^{-} + 16\mathrm{CH}_3\mathrm{OH} + 22\mathrm{H}^{+} \rightarrow 5\mathrm{Mn}^{2+} + 16\mathrm{HCO}_2\mathrm{H} + 11\mathrm{H}_2\mathrm{O}\) (c) \(2\mathrm{I}_2 + 2\mathrm{H}_2\mathrm{O} + 5\mathrm{OCl}^{-} \rightarrow 10\mathrm{IO}_{3}^{-} + 10\mathrm{Cl}^{-} + 10\mathrm{H}^{+}\) (d) \(2\mathrm{As}_{2} \mathrm{O}_{3} + 12\mathrm{H}_{2}\mathrm{O} + 10\mathrm{NO}_{3}^{-} \rightarrow 8\mathrm{H}_{3} \mathrm{AsO}_{4} + 5\mathrm{N}_{2} \mathrm{O}_{3} + 10\mathrm{H}^{+}\) (e) \(2\mathrm{MnO}_{4}^{-} + 3\mathrm{Br}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{MnO}_{2} + 3\mathrm{BrO}_{3}^{-} + 6\mathrm{OH}^{-}\) (f) \(4\mathrm{Pb}(\mathrm{OH})_{4}^{2-} + 8\mathrm{ClO}^{-} \rightarrow 4\mathrm{PbO}_2 + 8\mathrm{Cl}^{-} + 12\mathrm{OH}^{-}\)

Step-by-step solution

1. Assign oxidation states and identify redox half-reactions.

Using the rules of oxidation numbers, we can determine the oxidation states: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): Cr = +6, O = -2; \(\mathrm{I^{-}}\): I = -1; \(\mathrm{Cr}^{3+}\): Cr = +3; \(\mathrm{IO}_{3}^{-}\): I = +5, O = -2. The half-reactions are: Oxidation: \(2\mathrm{I^{-}} \longrightarrow \mathrm{I}_{2} + 2e^-\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6e^- \longrightarrow 2\mathrm{Cr}^{3+}\)

2. Balance half-reactions and charges.

For the oxidation half-reaction, balance atoms and charges: \(2\mathrm{I^{-}} \longrightarrow \mathrm{I}_{2} + 2e^-\) For the reduction half-reaction, balance atoms, charges, and add H2O for the oxygens: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

3. Combine half-reactions and simplify.

Multiply the oxidation half-reaction by 3 and combine both half-reactions, and then simplify: \(6\mathrm{I}^{-} \longrightarrow 3\mathrm{I}_{2} + 6e^-\) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6e^- \longrightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\) Combine: \(6\mathrm{I}^{-} + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} +\allowbreak 14\mathrm{H}^{+} \longrightarrow 3\mathrm{I}_2 + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\) Now, we can see that \(\mathrm{I}_{2}\) reacts further with water: \( 3\mathrm{I}_{2} + 6\mathrm{H}_2\mathrm{O} \longrightarrow 6\mathrm{IO}_{3}^{-} + 12\mathrm{H}^{+}\) Adding the two balanced reactions above, we get the final balanced equation: \(6\mathrm{I}^{-} + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \longrightarrow 6\mathrm{IO}_{3}^{-} + 2\mathrm{Cr}^{3+} + 12\mathrm{H}^{+}\)

4. Identify oxidizing and reducing agents.

Oxidizing agent: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (reduction takes place); Reducing agent: \(\mathrm{I}^{-}\) (oxidation takes place). The other equations follow the same procedure, and you can complete them as a practice. Here are the balanced equations for reference: (b) \(5\mathrm{MnO}_{4}^{-} + 16\mathrm{CH}_3\mathrm{OH} + 22\mathrm{H}^{+} \rightarrow 5\mathrm{Mn}^{2+} + 16\mathrm{HCO}_2\mathrm{H} + 11\mathrm{H}_2\mathrm{O}\) (c) \(2\mathrm{I}_2 + 2\mathrm{H}_2\mathrm{O} + 5\mathrm{OCl}^{-} \rightarrow 10\mathrm{IO}_{3}^{-} + 10\mathrm{Cl}^{-} + 10\mathrm{H}^{+}\) (d) \(2\mathrm{As}_{2} \mathrm{O}_{3} + 12\mathrm{H}_{2}\mathrm{O} + 10\mathrm{NO}_{3}^{-} \rightarrow 8\mathrm{H}_{3} \mathrm{AsO}_{4} + 5\mathrm{N}_{2} \mathrm{O}_{3} + 10\mathrm{H}^{+}\) (e) \(2\mathrm{MnO}_{4}^{-} + 3\mathrm{Br}^{-} + 2\mathrm{H}_{2}\mathrm{O} \rightarrow 4\mathrm{MnO}_{2} + 3\mathrm{BrO}_{3}^{-} + 6\mathrm{OH}^{-}\) (f) \(4\mathrm{Pb}(\mathrm{OH})_{4}^{2-} + 8\mathrm{ClO}^{-} \rightarrow 4\mathrm{PbO}_2 + 8\mathrm{Cl}^{-} + 12\mathrm{OH}^{-}\)

Key concepts

Oxidation-Reduction

Oxidation-reduction, often called redox reactions, are chemical processes where electrons are transferred from one molecule to another. In any redox reaction, there are two main components: oxidation and reduction. Oxidation refers to the loss of electrons, while reduction involves the gain of electrons. Not only do these reactions form the basis for many chemical processes in nature and industries, but they also play a crucial role in biological systems. Understanding these reactions helps in appreciating how batteries work or how our bodies use energy.
Redox reactions involve electron transfer, which is why identifying the changes in oxidation states of involved elements is crucial. These changes tell us which species gained electrons and which lost them, defining which is reduced and which is oxidized.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the same number of each type of atom is present on both sides of the equation. This is crucial when dealing with redox reactions, as it requires balancing not just atoms, but also charges.
To balance redox equations, especially in acidic or basic solutions, follow these steps:

• Separate the equation into two half-reactions - one for oxidation, the other for reduction.
• Balance all atoms except oxygen and hydrogen first.
• Balance oxygen atoms by adding water ( H_2O ) molecules.
• Balance hydrogen atoms by adding hydrogen ions ( H^+ in acidic solutions or hydroxide ions ( OH^- ) in basic solutions).
• Finally, balance the charges by adding electrons ( e^- ).
• Adjust the two half-reactions so that the electrons cancel each other out, then combine them to form a balanced equation.

This careful balance ensures that mass and charge are conserved, adhering to the fundamental law of conservation in chemical reactions.

Oxidizing Agents

An oxidizing agent is a substance that gains electrons in a redox reaction and is reduced in the process. It plays a vital role in driving the chemical reaction forward by accepting electrons from another species.
Some common oxidizing agents include:

• Oxygen ( O_2 ) itself
• Potassium permanganate ( KMnO_4 )
• Hydrogen peroxide ( H_2O_2 )
• Dichromate ions ( Cr_2O_7^{2-} )

In the equation Cr_2O_7^{2-} + I^- ightarrow Cr^{3+} + IO_3^- , the dichromate ion ( Cr_2O_7^{2-} ) acts as the oxidizing agent because it gains electrons (is reduced) to become Cr^{3+} .
Understanding the concept of oxidizing agents helps us identify the substances that are crucial in reactions like combustion, respiration, and industrial processes.

Reducing Agents

Reducing agents, also known as reductants, are molecules that lose electrons in a redox reaction, becoming oxidized. They donate electrons to another substance, effectively reducing that substance while they themselves are oxidized.
Some common reducing agents include:

• Hydrogen gas ( H_2 )
• Zinc ( Zn )
• Hydrochloric acid ( HCl )
• Iodide ions ( I^- )

For example, in the redox equation 6I^- + Cr_2O_7^{2-} + 14H^+ ightarrow 3I_2 + 2Cr^{3+} + 7H_2O , the iodide ion ( I^- ) serves as the reducing agent. It donates electrons to the dichromate ion, thus getting oxidized to I_2 .
Recognizing reducing agents is essential for understanding reactions related to metallurgy, photosynthesis, and even some types of fuel cells.

Half-Reactions

Half-reactions focus on either the oxidation or reduction process occurring in a redox equation. They are separated out to simplify the balancing of complex chemical equations by dealing with one chemical transformation at a time.
To write half-reactions:

• Identify the species that are oxidized and those that are reduced by examining changes in oxidation states.
• Write two unbalanced equations: one for oxidation and one for reduction. Each equation will show only the reactants and products that change oxidation states.
• Balance each half-reaction individually as described in balancing chemical equations, ensuring both mass and charge are balanced by adding electrons as necessary.

For example, in the reaction involving Cr_2O_7^{2-} and I^- , the half-reactions are:

• Oxidation: 2I^- ightarrow I_2 + 2e^-
• Reduction: Cr_2O_7^{2-} + 6e^- + 14H^+ ightarrow 2Cr^{3+} + 7H_2O

These half-reactions provide clear insight into electron flow, essential for enhancing our understanding of redox mechanisms.