Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the $\mathrm{O}$ atoms in hydrogen peroxide, ${\mathrm{H}}_2{\mathrm{O}}_2$, have an atypical oxidation state. (a) ${\mathrm{NO}}_2^{-}(aq)+{\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}(aq)-\cdots$ ${\mathrm{Cr}}^{3+}(aq)+{\mathrm{NO}}_3^{-}(aq)$ (acidic solution) (b) $\mathrm{S}(\mathrm{s})+{\mathrm{HNO}}_3(aq)⟶{\mathrm{H}}_2{\mathrm{SO}}_3(aq)+{\mathrm{N}}_2\mathrm{O}(g)$ (acidic solution) (c) ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}(aq)+{\mathrm{CH}}_3\mathrm{OH}(aq)⟶$ ${\mathrm{HCO}}_2\mathrm{H}(aq)+{\mathrm{Cr}}^{3+}(aq)$ (acidic solution) (d) ${\mathrm{MnO}}_4^{-(aq)}+{\mathrm{Cl}}^{-}(aq)⟶{\mathrm{Mn}}^{2+}(aq)+{\mathrm{Cl}}_2(aq)$ (acidic solution) (e) ${\mathrm{NO}}_2^{-}(aq)+\mathrm{Al}(s)⟶{\mathrm{NH}}_4{}^{+}(aq)+{\mathrm{AlO}}_2^{-}(aq)$ (basic solution) (f) ${\mathrm{H}}_2{\mathrm{O}}_2(aq)+{\mathrm{ClO}}_2(aq)⟶{\mathrm{ClO}}_2^{-}(aq)+{\mathrm{O}}_2(g)$ (basic solution)

Short answer

(a) Short answer: Balanced equation: $2C{r}_2{O}_7^{2-}+6N{O}_2^{-}+14H^{+}\to 2C{r}^{3+}+6N{O}_3^{-}+7H_{2}O$. Oxidizing agent: $C{r}_2{O}_7^{2-}$, Reducing agent: $N{O}_2^{-}$ (b) Short answer: Balanced equation: $3S+2HN{O}_3\to 3H_{2}S{O}_3+N_{2}O$. Oxidizing agent: $HN{O}_3$, Reducing agent: $S$ (c) Short answer: Balanced equation: $2C{r}_2{O}_7^{2-}+6C{H}_{3}OH+14H^{+}\to 2C{r}^{3+}+6HC{O}_{2}H+7H_{2}O$. Oxidizing agent: $C{r}_2{O}_7^{2-}$, Reducing agent: $C{H}_{3}OH$ (d) Short answer: Balanced equation: $Mn{O}_4^{-}+5C{l}^{-}+8H^{+}\to M{n}^{2+}+2.5C{l}_2+4H_{2}O$. Oxidizing agent: $Mn{O}_4^{-}$, Reducing agent: $C{l}^{-}$ (e) Short answer: Balanced equation: $2Al+3N{O}_2^{-}\to 2Al{O}_2^{-}+3N{H}_4^{+}$. Oxidizing agent: $N{O}_2^{-}$, Reducing agent: $Al$ (f) Short answer: Balanced equation: $2H_2{O}_2+4Cl{O}_2\to 6Cl{O}_2^{-}+4H_{2}O$. Oxidizing agent: $Cl{O}_2$, Reducing agent: $H_2{O}_2$

Step-by-step solution

Identify the change in oxidation states

For NO₂⁻, the nitrogen has an oxidation state of +3, and for NO₃⁻, it has an oxidation state of +5. For Cr₂O₇²⁻, chromium has an oxidation state of +6, and for Cr³⁺, it has an oxidation state of +3.

Balance half-reactions

Write two half-reactions: 1) Balance the nitrogen atoms: NO₂⁻ → NO₃⁻ 2) Balance the chromium atoms: Cr₂O₇²⁻ → 2Cr³⁺ Now, balance the electrons: 1) NO₂⁻ → NO₃⁻ + 2e⁻ 2) Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺

Combine and balance the final equation

To combine these half-reactions, we need to balance the electrons. We'll multiply the first half-reaction by 3 and then add the half-reactions: 3(NO₂⁻ → NO₃⁻ + 2e⁻) Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ ------------------------ 2Cr₂O₇²⁻ + 6NO₂⁻ → 2Cr³⁺ + 6NO₃⁻ Finally, balance the solutions charge by adding H⁺ ions: 2Cr₂O₇²⁻ + 6NO₂⁻ + 14H⁺ → 2Cr³⁺ + 6NO₃⁻ + 7H₂O

Identify the oxidizing and reducing agents

The oxidizing agent is the species that is being reduced and causing the oxidation. In this case, it's Cr₂O₇²⁻. The reducing agent is the species that is being oxidized and causing the reduction. In this case, it's NO₂⁻. For the other reactions, we'll follow the same steps: (b) S(s) + HNO₃(aq)→ H₂SO₃(aq) + N₂O(g) (acidic solution) Half-reactions: S + 2e⁻ → S²⁻ 2HNO₃ + 6e⁻ → N₂O + 3H₂O Balanced equation: 3S + 2HNO₃ → 3H₂SO₃ + N₂O Oxidizing agent: HNO₃ Reducing agent: S (c) Cr₂O₇²⁻(aq) + CH₃OH(aq) → HCO₂H(aq) + Cr³⁺(aq) (acidic solution) Half-reactions: Cr₂O₇²⁻ + 6e⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O CH₃OH → HCO₂H + 2e⁻ Balanced equation: 2Cr₂O₇²⁻ + 6CH₃OH + 14H⁺ → 2Cr³⁺ + 6HCO₂H + 7H₂O Oxidizing agent: Cr₂O₇²⁻ Reducing agent: CH₃OH (d) MnO₄⁻(aq) + Cl⁻(aq) → Mn²⁺(aq) + Cl₂(aq) (acidic solution) Half-reactions: MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O 2Cl⁻ → Cl₂ + 2e⁻ Balanced equation: MnO₄⁻ + 5Cl⁻ + 8H⁺ → Mn²⁺ + 2.5Cl₂ + 4H₂O Oxidizing agent: MnO₄⁻ Reducing agent: Cl⁻ (e) NO₂⁻(aq) + Al(s) → NH₄⁺(aq) + AlO₂⁻(aq) (basic solution) Balance under basic conditions with OH⁻ and H₂O: Half-reactions: NO₂⁻ + 3H₂O → NH₄⁺ + 4OH⁻ Al → AlO₂⁻ + 3H₂O + 4e⁻ Balanced equation: 2Al + 3NO₂⁻ → 2AlO₂⁻ + 3NH₄⁺ Oxidizing agent: NO₂⁻ Reducing agent: Al (f) H₂O₂(aq) + ClO₂(aq) → ClO₂⁻(aq) + O₂(g) (basic solution) Balance under basic conditions with OH⁻ and H₂O: Half-reactions: H₂O₂ + 2OH⁻ → 2H₂O + O₂ 2ClO₂ + 2OH⁻ → 2ClO₂⁻ + H₂O + O₂ Balanced equation: 2H₂O₂ + 4ClO₂ → 6ClO₂⁻ + 4H₂O Oxidizing agent: ClO₂ Reducing agent: H₂O₂

Key concepts

Oxidation States

Understanding oxidation states is crucial for balancing redox reactions. An oxidation state, or oxidation number, is a signed integer representing the number of electrons an atom either gains or loses to form a chemical bond. When an atom loses electrons, we say it has been oxidized, which increases its oxidation state. Conversely, when an atom gains electrons, it is reduced, and its oxidation state decreases.

In the given exercise, identifying the change in oxidation states is the first step toward solving redox reactions. For instance, in reaction (a), nitrogen's oxidation state changes from +3 in NO₂⁻ to +5 in NO₃⁻, indicating an oxidation process. Contrastingly, chromium's oxidation state decreases from +6 in Cr₂O₇²⁻ to +3 in Cr³⁺, signifying a reduction process.

Remember that some atoms can have an 'atypical' oxidation state, such as oxygen in hydrogen peroxide (H₂O₂). Normally, oxygen has an oxidation state of -2, but in H₂O₂, it is -1. Recognizing atypical states is necessary for accurately balancing equations and determining the oxidizing and reducing agents in redox reactions.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. It involves separating the overall chemical equation into two half-reactions: one for oxidation and another for reduction. Each half-reaction represents the fate of one of the species involved in the redox process.

As seen in the second step of solving the exercise, each half-reaction must be balanced separately. This involves not just balancing the elements involved, but also ensuring that the charges on both sides are equal, primarily by adding electrons. Balancing half-reactions for reaction (a) goes like this:

1. NO₂⁻ → NO₃⁻ (oxidation)
2. Cr₂O₇²⁻ → 2Cr³⁺ (reduction)

After balancing for atoms and charge, we then add the half-reactions back together and balance the overall equation for mass and charge, ensuring the number of electrons lost in oxidation equals those gained in reduction. The result is a stoichiometrically and charge consistent balanced redox equation.

Oxidizing and Reducing Agents

Finalizing the redox balancing process includes identifying oxidizing and reducing agents. The oxidizing agent is a substance that takes electrons from another substance during the redox reaction, thus being reduced itself. In contrast, the reducing agent donates electrons and becomes oxidized.

In part (a) of the exercise, Cr₂O₇²⁻ acts as the oxidizing agent as its oxidation state is reduced from +6 to +3. NO₂⁻ is the reducing agent, its oxidation state increases from +3 to +5. Identifying these agents is essential in understanding the redox process, as it provides insights into the reaction's direction and the roles each species plays.

The oxidizing and reducing agents drive the chemical changes in redox reactions. Their identification not only enables us to balance reactions but also to predict the products and reactants in various chemical environments, be it acidic or basic.