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AP Chemistry Practice Quiz: Oxidation-Reduction (Redox) Reactions

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

When constructing a balanced chemical equation for a redox reaction, what is the primary purpose of separating the overall reaction into half-reactions?

All Questions (7)

When constructing a balanced chemical equation for a redox reaction, what is the primary purpose of separating the overall reaction into half-reactions?

A) To isolate the spectator ions from the reacting species.

B) To ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction.

C) To determine the physical states (s, l, g, aq) of the reactants and products.

D) To calculate the overall enthalpy change of the reaction.

Correct Answer: B

The fundamental principle of balancing redox reactions using the half-reaction method is to balance mass and charge. A key step in this process is to ensure that the transfer of electrons is balanced. By separating the reaction into oxidation and reduction half-reactions, we can explicitly track the electrons and ensure that the number of electrons lost by the oxidized species is equal to the number of electrons gained by the reduced species before combining them into the final balanced equation.

Which of the following best represents a balanced reduction half-reaction?

A) Zn(s) → Zn²⁺(aq) + 2e⁻

B) Cl₂(g) + 2e⁻ → 2Cl⁻(aq)

C) H₂(g) → 2H⁺(aq) + 2e⁻

D) Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g)

Correct Answer: B

A reduction half-reaction represents a species gaining electrons, which are shown as reactants. Option B correctly shows chlorine gas (Cl₂) gaining two electrons to form two chloride ions (2Cl⁻), with both mass and charge being balanced. Options A and C represent oxidation half-reactions because electrons are products (lost). Option D represents a complete, balanced redox reaction, not a half-reaction.

A redox reaction is constructed from the following two half-reactions: Oxidation: Fe(s) → Fe³⁺(aq) + 3e⁻ Reduction: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) What is the correctly balanced overall chemical equation for this redox reaction?

A) Fe(s) + O₂(g) + 4H⁺(aq) → Fe³⁺(aq) + 2H₂O(l)

B) 3Fe(s) + 4O₂(g) + 4H⁺(aq) → 3Fe³⁺(aq) + 2H₂O(l)

C) 4Fe(s) + 3O₂(g) + 12H⁺(aq) → 4Fe³⁺(aq) + 6H₂O(l)

D) Fe³⁺(aq) + 2H₂O(l) → Fe(s) + O₂(g) + 4H⁺(aq)

Correct Answer: C

To construct the balanced overall equation, the number of electrons transferred must be equal in both half-reactions. The least common multiple of 3 electrons (from the oxidation half-reaction) and 4 electrons (from the reduction half-reaction) is 12. Therefore, the oxidation half-reaction is multiplied by 4, and the reduction half-reaction is multiplied by 3. This gives: 4Fe(s) → 4Fe³⁺(aq) + 12e⁻ 3O₂(g) + 12H⁺(aq) + 12e⁻ → 6H₂O(l) Combining these and canceling the 12 electrons yields the overall balanced equation: 4Fe(s) + 3O₂(g) + 12H⁺(aq) → 4Fe³⁺(aq) + 6H₂O(l).

Consider the overall balanced redox equation: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s) Which of the following correctly represents the oxidation half-reaction that was used to construct this equation?

A) Al(s) → Al³⁺(aq) + 3e⁻

B) Cu²⁺(aq) + 2e⁻ → Cu(s)

C) 2Al(s) → 2Al³⁺(aq) + 6e⁻

D) 3Cu²⁺(aq) + 6e⁻ → 3Cu(s)

Correct Answer: A

The overall equation shows solid aluminum, Al(s), being converted to aluminum ions, Al³⁺(aq). This process involves a loss of electrons, which is oxidation. The half-reaction for one aluminum atom losing three electrons to become an aluminum ion is Al(s) → Al³⁺(aq) + 3e⁻. Option B is the reduction half-reaction. Options C and D represent the half-reactions after they have been multiplied by coefficients to balance the electron transfer for the overall equation, but the fundamental half-reaction is represented in its simplest whole-number ratio, which is A.

When the redox equation below is balanced using half-reactions in an acidic solution, what is the total number of electrons transferred? MnO₄⁻(aq) + C₂O₄²⁻(aq) → Mn²⁺(aq) + CO₂(g)

A) 2

B) 5

C) 7

D) 10

Correct Answer: D

First, the two half-reactions must be written and balanced: Reduction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) Oxidation: C₂O₄²⁻(aq) → 2CO₂(g) + 2e⁻ To balance the electrons, the reduction half-reaction must be multiplied by 2, and the oxidation half-reaction must be multiplied by 5. This makes the number of electrons in each half-reaction equal to 10 (the least common multiple of 5 and 2). Therefore, a total of 10 electrons are transferred from the oxalate ions to the permanganate ions.

When constructing a balanced redox equation from its half-reactions, which of the following statements is always true?

A) The oxidation half-reaction must be multiplied by a larger integer than the reduction half-reaction.

B) Water molecules must be added to the product side of the overall equation.

C) The total positive charge on the reactant side must equal the total positive charge on the product side.

D) Electrons must not appear as a reactant or product in the final overall equation.

Correct Answer: D

The process of balancing half-reactions involves adjusting stoichiometric coefficients so that the number of electrons lost in oxidation is identical to the number of electrons gained in reduction. When the half-reactions are added together to form the overall equation, these electrons cancel out completely. Therefore, a correctly constructed and balanced redox equation will never show free electrons.

The following half-reactions are used to construct a balanced redox equation: Half-reaction 1: Au³⁺(aq) + 3e⁻ → Au(s) Half-reaction 2: Ni(s) → Ni²⁺(aq) + 2e⁻ To represent the final balanced equation, what coefficients must be placed in front of Au(s) and Ni(s) respectively?

A) 1 and 1

B) 3 and 2

C) 2 and 3

D) 1 and 3

Correct Answer: C

To balance the electron transfer, the number of electrons in both half-reactions must be equal. The least common multiple of 3 electrons (from the Au half-reaction) and 2 electrons (from the Ni half-reaction) is 6. To achieve 6 electrons in each, Half-reaction 1 must be multiplied by 2, and Half-reaction 2 must be multiplied by 3. This results in the balanced half-reactions: 2Au³⁺(aq) + 6e⁻ → 2Au(s) 3Ni(s) → 3Ni²⁺(aq) + 6e⁻ Therefore, the coefficient for Au(s) is 2, and the coefficient for Ni(s) is 3.