Question

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}-(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

Short answer

The balanced reactions are: a. \(3Cu(s) + 2NO_3^-(aq) + 12H^+(aq) \rightarrow 3Cu^{2+}(aq) + 2NO(g) + 8H_2O(l)\) b. \(7Cl^-(aq) + 2Cr_2O_7^{2-}(aq) + 28H^+(aq) \rightarrow 7Cl_2(g) + 4Cr^{3+}(aq) + 14H_2O(l)\) c. \(Pb(s) + PbO_2(s) + H_2SO_4(aq) \rightarrow PbSO_4(s) + 2H_2O(l)\) d. \(3Mn^{2+}(aq) + 15NaBiO_3(s) + 6H^+(aq) \rightarrow 3MnO_4^-(aq) + 5Bi^{3+}(aq) + 8H_2O(l) + 15Na^+(aq)\) e. \(3Zn(s) + 2H_3AsO_4(aq) + 6H^+(aq) \rightarrow 3Zn^{2+}(aq) + 2AsH_3(g) + 6H_2O(l)\)

Step-by-step solution

a. Balancing Cu(s) + NO₃⁻(aq) → Cu²⁺(aq) + NO(g) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cu → Cu²⁺ + 2e⁻ Reduction half-reaction: NO₃⁻ + e⁻ → NO Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 2H₂O to balance oxygen and 6H⁺ to balance hydrogen: NO₃⁻ + 2H₂O + 6H⁺ + 3e⁻ → NO + 6H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3Cu → 3Cu²⁺ + 6e⁻ 2(NO₃⁻ + 2H₂O + 6H⁺ + 3e⁻) → 2(NO + 6H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Cu + 2NO₃⁻ + 4H₂O + 12H⁺ → 3Cu²⁺ + 2NO + 12H₂O Step 5: Simplify the reaction, if possible: 3Cu(s) + 2NO₃⁻(aq) + 4H₂O(l) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 12H₂O(l) Removing 4H₂O from both sides: 3Cu(s) + 2NO₃⁻(aq) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 8H₂O(l) The balanced reaction is: 3Cu(s) + 2NO₃⁻(aq) + 12H⁺(aq) → 3Cu²⁺(aq) + 2NO(g) + 8H₂O(l)

b. Balancing Cr₂O₇²⁻(aq) + Cl⁻(aq) → Cr³⁺(aq) + Cl₂(g) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cl⁻ → Cl₂ + 2e⁻ Reduction half-reaction: Cr₂O₇²⁻ → 2Cr³⁺ Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 14H⁺ to balance the change and 7H₂O to balance oxygen: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 7 and the reduction half-reaction by 2: 7(Cl⁻ → Cl₂ + 2e⁻) 2(Cr₂O₇²⁻ + 14H⁺) → 2(2Cr³⁺ + 7H₂O) Step 4: Add the half-reactions back together and cancel electrons: 7Cl⁻ + 2Cr₂O₇²⁻ + 28H⁺ → 7Cl₂ + 4Cr³⁺ + 14H₂O The balanced reaction is: 7Cl⁻(aq) + 2Cr₂O₇²⁻(aq) + 28H⁺(aq) → 7Cl₂(g) + 4Cr³⁺(aq) + 14H₂O(l)

c. Balancing Pb(s) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Pb → Pb²⁺ + 2e⁻ Reduction half-reaction: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. Add one SO₄²⁻ to the reduction half-reaction: PbO₂ + 4H⁺ + SO₄²⁻ → PbSO₄ + 2H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 1 (no need to multiply): Step 4: Add half-reactions back together, and cancel electrons: Pb + PbO₂ + 4H⁺ + SO₄²⁻ → Pb²⁺ + PbSO₄ + 2H₂O Step 5: Simplify the reaction, if possible: Pb(s) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) + 2H₂O(l) The balanced reaction is: Pb(s) + PbO₂(s) + H₂SO₄(aq) → PbSO₄(s) + 2H₂O(l)

d. Balancing Mn²⁺(aq) + NaBiO₃(s) → Bi³⁺(aq) + MnO₄⁻(aq) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Mn²⁺ → MnO₄⁻ Reduction half-reaction: BiO₃⁻ → Bi³⁺ Step 2: Balance each half-reaction For the oxidation half-reaction, add 4H₂O to balance oxygen and 8H⁺ to balance hydrogen, and then subtract 5e⁻ to balance the charges: Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻ For the reduction half-reaction, add 2H₂O to balance oxygen and 6H⁺ to balance hydrogen, and then add 3e⁻ to balance the charges: BiO₃⁻ + 6H⁺ + 3e⁻ → Bi³⁺ + 3H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 5: 3(Mn²⁺ + 4H₂O → MnO₄⁻ + 8H⁺ + 5e⁻) 5(BiO₃⁻ + 6H⁺ + 3e⁻ → Bi³⁺ + 3H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Mn²⁺ + 12H₂O + 15BiO₃⁻ + 30H⁺ → 3MnO₄⁻ + 24H⁺ + 5Bi³⁺ + 15H₂O Step 5: Simplify the reaction: 3Mn²⁺(aq) + 12H₂O(l) + 15NaBiO₃(s) + 6H⁺(aq) → 3MnO₄⁻(aq) + 5Bi³⁺(aq) + 8H₂O(l) + 15Na⁺(aq) The balanced reaction is: 3Mn²⁺(aq) + 15NaBiO₃(s) + 6H⁺(aq) → 3MnO₄⁻(aq) + 5Bi³⁺(aq) + 8H₂O(l) + 15Na⁺(aq)

e. Balancing H₃AsO₄(aq) + Zn(s) → AsH₃(g) + Zn²⁺(aq) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Zn → Zn²⁺ + 2e⁻ Reduction half-reaction: H₃AsO₄ → AsH₃ Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 3H⁺ + 3e⁻ to balance hydrogen and charges: H₃AsO₄ + 3H⁺ + 3e⁻ → AsH₃ + 3H₂O Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3(Zn → Zn²⁺ + 2e⁻) 2(H₃AsO₄ + 3H⁺ + 3e⁻ → AsH₃ + 3H₂O) Step 4: Add the half-reactions back together and cancel electrons: 3Zn + 2H₃AsO₄ + 6H⁺ → 3Zn²⁺ + 2AsH₃ + 6H₂O The balanced reaction is: 3Zn(s) + 2H₃AsO₄(aq) + 6H⁺(aq) → 3Zn²⁺(aq) + 2AsH₃(g) + 6H₂O(l)

Key concepts

Half-Reaction Method

The half-reaction method is a systematic approach to balance oxidation-reduction reactions, especially in an acidic solution. It breaks down the redox reaction into two separate equations: the oxidation half-reaction and the reduction half-reaction.
The oxidation half-reaction describes the loss of electrons from a substance while the reduction half-reaction describes the gain of electrons by another substance.
Each half-reaction is balanced individually for mass and charge, and then combined to form the overall balanced equation. This method ensures that electrons lost in oxidation equal those gained in reduction, maintaining charge neutrality.

Acidic Solution

In chemistry, an acidic solution has a higher concentration of H⁺ ions. Balancing redox reactions in such solutions requires careful handling of hydrogen atoms and charges.
To achieve balance of redox reactions in an acidic solution using the half-reaction method, additional H⁺ ions are often introduced to the equation to balance hydrogen atoms. Water (H₂O) may also be added to balance oxygen atoms.
The presence of H⁺ ions affects the electron counts needed for balancing, as they help track and balance charges across the redox process. Understanding the role of these ions is crucial in ensuring the accurate balancing of reactions in these conditions.

Balancing Redox Reactions

Balancing redox reactions involves equalizing the number of electrons transferred during the oxidation and reduction processes. The process typically includes:

• Simplification of original reaction into two half-reactions.
• Balancing the atoms in each half-reaction.
• Ensuring the same number of electrons are present in both half-reactions by multiplying them appropriately.
• Combining the two half-reactions to cancel out common terms like electrons.

Only through precise balancing of both atoms and charges can the full redox reaction be accurately represented, demonstrating the actual chemical changes occurring during the reaction.

Electron Transfer

Electron transfer is the fundamental process in any redox reaction. It involves the movement of electrons from one chemical species (the reducing agent) to another (the oxidizing agent).
This electron movement is what characterizes a reaction as a redox process, driving the transformation of substances involved. The count of electrons lost in oxidation must directly match the electrons gained in reduction to preserve electron balance.
Monitoring and equalizing this electron transfer is the key objective of using the half-reaction method, ensuring minimal energy loss and correct chemical conversion according to the reaction's stoichiometry.