Oxidation-Reduction (Redox) Reactions - AP Chemistry

1. A sample of magnesium metal burns in oxygen gas to form magnesium oxide: $2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$ Which statement correctly describes the oxidation-number changes for Mg and O?
   Answer: Mg: $0 \rightarrow +2$ and O: $0 \rightarrow -2$ This question tests understanding of oxidation-reduction (redox) reactions. In 2Mg(s) + O₂(g) → 2MgO(s), we track oxidation numbers: Mg starts at 0 (elemental) and becomes +2 in MgO, while O starts at 0 (in O₂) and becomes -2 in MgO. Magnesium loses 2 electrons per atom (0 → +2, oxidation), and oxygen gains 2 electrons per atom (0 → -2, reduction). This is a classic combustion reaction where the metal is oxidized and oxygen is reduced. A common mistake is thinking Mg goes to -2 because it combines with oxygen, but metals typically form positive ions. Remember: in redox reactions, track oxidation numbers from reactants to products to identify electron transfer.
2. When magnesium is burned in carbon dioxide, magnesium oxide and carbon form: $2\,\text{Mg}(s)+\text{CO}_2(g)\rightarrow 2\,\text{MgO}(s)+\text{C}(s)$ Which species is oxidized in the reaction?
   Answer: Mg(s) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction 2Mg(s) + CO₂(g) → 2MgO(s) + C(s), we assign oxidation numbers: Mg starts at 0 (elemental) and becomes +2 in MgO, while C in CO₂ is +4 and becomes 0 in C(s). Since Mg goes from 0 to +2, it loses electrons and is oxidized; since C goes from +4 to 0, it gains electrons and is reduced. The species that is oxidized is Mg(s). A common error would be choosing C in CO₂ because students might think the species that forms an element is oxidized, but C actually gains electrons (reduction). Remember: track oxidation numbers systematically—metals typically lose electrons (oxidation) while nonmetals often gain electrons (reduction).
3. In an acidic solution, iron(II) ions react with dichromate ions according to the overall reaction: $6\text{Fe}^{2+}(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)$ Which species is reduced?
   Answer: $Cr2O7^2$−(aq) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction, Fe²⁺ changes to Fe³⁺ (oxidation number +2 to +3, loses electrons, is oxidized), while Cr in Cr₂O₇²⁻ has oxidation number +6 and becomes Cr³⁺ with oxidation number +3 (gains electrons, is reduced). To find Cr's oxidation number in Cr₂O₇²⁻: 2(Cr) + 7(-2) = -2, so Cr = +6. The species being reduced is Cr₂O₇²⁻ (dichromate ion). A common error is thinking H⁺ is reduced because it appears to form water, but H remains +1 throughout. Remember: reduction is gain of electrons (decrease in oxidation number), and track oxidation numbers systematically.
4. Ammonia is oxidized by copper(II) oxide when heated, producing nitrogen gas, copper metal, and water vapor: $\mathrm{2\,NH_3(g) + 3\,CuO(s) \rightarrow N_2(g) + 3\,Cu(s) + 3\,H_2O(g)}$ Which species acts as the oxidizing agent?
   Answer: CuO(s) This question assesses understanding of oxidation–reduction (redox) reactions. To find the oxidizing agent, assign oxidation numbers: N in NH₃(g) is -3 (H +1 each), Cu in CuO(s) is +2 (O -2), N in N₂(g) is 0, Cu(s) is 0, O in H₂O(g) is -2. Nitrogen increases from -3 to 0, losing electrons and being oxidized, while copper decreases from +2 to 0, gaining electrons and being reduced; thus, CuO(s) is the oxidizing agent. Ammonia is reduced to nitrogen, but CuO causes the oxidation. A tempting distractor is NH₃(g), but it is the reducing agent—a common error is assigning N in ammonia as 0 instead of -3. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers in solids and gases alike.
5. Hydrogen peroxide can react with iodide ions in acidic solution to produce iodine and water: $\text{H}_2\text{O}_2(aq) + 2\text{I}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{I}_2(aq) + 2\text{H}_2\text{O}(l)$ Which species is reduced?
   Answer: H2O2(aq) This question tests understanding of oxidation-reduction (redox) reactions. In H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l), iodide (I⁻) goes from -1 to 0 in I₂ (loses electrons, is oxidized), while oxygen in H₂O₂ has oxidation number -1 (unusual for oxygen) and becomes -2 in H₂O (gains electrons, is reduced). To verify O in H₂O₂: 2(+1) + 2(O) = 0, so O = -1. The species being reduced is H₂O₂. A common error is thinking I⁻ is reduced because it forms I₂, but forming a diatomic molecule from ions involves electron loss. Remember: reduction is gain of electrons (decrease in oxidation number), and peroxides have oxygen at -1.
6. In aqueous solution, bromine reacts with sulfide ions: $\text{Br}_2(aq) + \text{S}^{2-}(aq) \rightarrow 2\text{Br}^-(aq) + \text{S}(s)$ Which statement correctly describes the oxidation-number changes?
   Answer: Br changes from $0$ to $-1$, and S changes from $-2$ to $0$ This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: bromine in $\text{Br}_2$ is $0$, sulfur in $S^{2-}$ is $-2$, bromine in $\text{Br}^{-}$ is $-1$, and sulfur in $\text{S}(s)$ is $0$. Bromine's oxidation number decreases from $0$ to $-1$, and sulfur's increases from $-2$ to $0$. This tracks electron gain for bromine (reduction) and loss for sulfur (oxidation). A tempting distractor is A, but it incorrectly states Br changes from $-1$ to $0$; assuming initial Br is $-1$ like halides is common. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers.
7. A piece of aluminum metal is placed into a solution of silver nitrate, and the reaction proceeds as shown: $\text{Al}(s) + 3\text{AgNO}_3(aq) \rightarrow \text{Al(NO}_3)_3(aq) + 3\text{Ag}(s)$ How many electrons are transferred per aluminum atom as the reaction occurs?
   Answer: 3 This question tests understanding of oxidation-reduction (redox) reactions. In Al(s) + 3AgNO₃(aq) → Al(NO₃)₃(aq) + 3Ag(s), aluminum goes from 0 to +3 (loses 3 electrons per atom), while silver goes from +1 in AgNO₃ to 0 in Ag(s) (each Ag⁺ gains 1 electron). Since one Al atom loses 3 electrons and three Ag⁺ ions each gain 1 electron, the electron transfer balances. The answer is 3 electrons per aluminum atom. A common mistake is multiplying by the coefficient 3 in front of AgNO₃ to get 9, but the question asks per aluminum atom, not per reaction. Remember: track electrons transferred per atom, not per mole or per equation as written.
8. Hydrogen sulfide gas is oxidized by oxygen gas to form sulfur dioxide and liquid water, as represented by the balanced equation: $\mathrm{2\,H_2S(g) + 3\,O_2(g) \rightarrow 2\,SO_2(g) + 2\,H_2O(l)}$ Which element undergoes an increase in oxidation number?
   Answer: Sulfur This question assesses understanding of oxidation–reduction (redox) reactions. To identify the element with increasing oxidation number, assign them: in H₂S(g), H is +1 so S is -2; O in O₂(g) is 0; in SO₂(g), O is -2 so S is +4; in H₂O(l), H is +1 and O is -2. Sulfur's oxidation number increases from -2 to +4, indicating oxidation, while oxygen's decreases from 0 to -2, showing reduction; hydrogen remains +1. No change occurs for hydrogen or all elements, as only sulfur increases. A tempting distractor is oxygen, but its oxidation number decreases, not increases—a common error is forgetting that elemental oxygen is 0, not -2. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers for each element to see changes.
9. Hydrogen peroxide reacts with iodide ions in acidic solution to produce iodine and water: $\text{H}_2\text{O}_2(aq)+2\,\text{I}^-(aq)+2\,\text{H}^+(aq)\rightarrow \text{I}_2(aq)+2\,\text{H}_2\text{O}(l)$ Which species is the reducing agent?
   Answer: I$^-$(aq) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction H₂O₂(aq) + 2I⁻(aq) + 2H⁺(aq) → I₂(aq) + 2H₂O(l), we track oxidation numbers: I⁻ starts at -1 and becomes 0 in I₂, while O in H₂O₂ is -1 (peroxide) and becomes -2 in H₂O. Since I⁻ goes from -1 to 0, it loses electrons and is oxidized; since O goes from -1 to -2, it gains electrons and is reduced. The reducing agent is the species that gets oxidized, which is I⁻(aq). A common error would be choosing H₂O₂ because students might think the compound that decomposes is the reducing agent, but H₂O₂ is actually the oxidizing agent being reduced. Remember: the reducing agent donates electrons and is oxidized in the process.
10. Aqueous dichromate reacts with iodide ions in acidic solution, producing chromium(III) ions and iodine: $\text{Cr}_2\text{O}_7^{2-}(aq)+14\text{H}^+(aq)+6\text{I}^-(aq)\rightarrow 2\text{Cr}^{3+}(aq)+3\text{I}_2(s)+7\text{H}_2\text{O}(l)$ Which species is oxidized?
    Answer: I$^-$(aq) This question tests understanding of oxidation-reduction (redox) reactions. In this reaction, we track oxidation numbers: Cr in Cr₂O₇²⁻ is +6 and becomes +3 in Cr³⁺ (gains electrons, reduced), while I⁻ at -1 becomes I₂ at 0 (loses electrons, oxidized). Since I⁻ loses electrons (goes from -1 to 0), it is the species being oxidized. A common error is thinking Cr₂O₇²⁻ is oxidized because it's a complex ion, but we must track individual element oxidation states, not the overall charge. Remember: oxidation means loss of electrons (increase in oxidation number); the species being oxidized is the reducing agent.
11. Tin(II) chloride reacts with iron(III) chloride in aqueous solution: $\text{SnCl}_2(aq) + 2\text{FeCl}_3(aq) \rightarrow \text{SnCl}_4(aq) + 2\text{FeCl}_2(aq)$ Which species is the reducing agent?
    Answer: SnCl$_2$(aq) This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: tin in SnCl₂ is +2, iron in FeCl₃ is +3, tin in SnCl₄ is +4, and iron in FeCl₂ is +2. Tin's oxidation number increases from +2 to +4, indicating loss of electrons and oxidation, making SnCl₂ the reducing agent, while iron decreases from +3 to +2, showing gain of electrons and reduction. Chlorine remains -1. A tempting distractor is FeCl₃, but it is reduced, not the reducing agent; confusing the agents is common. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers.
12. Chlorine gas is bubbled into an aqueous solution of potassium iodide, producing iodine and chloride ions: $\text{Cl}_2(g)+2\text{KI}(aq)\rightarrow 2\text{KCl}(aq)+\text{I}_2(s)$ Which species is reduced in the reaction?
    Answer: Cl$_2$(g) This question tests understanding of oxidation-reduction (redox) reactions. In Cl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(s), we track oxidation numbers: Cl₂ starts at 0 (elemental) and becomes Cl⁻ with oxidation number -1, while I⁻ (oxidation number -1) becomes I₂ at 0. Since Cl₂ gains electrons (0 to -1), it is reduced; I⁻ loses electrons (-1 to 0) and is oxidized. A common error is thinking I⁻ is reduced because it forms a solid, but physical state doesn't determine oxidation/reduction—only electron transfer does. Remember: reduction is gain of electrons (decrease in oxidation number); track the change in oxidation states, not physical states.
13. A piece of magnesium metal is placed into aqueous hydrochloric acid, producing hydrogen gas: $\text{Mg}(s)+2\text{HCl}(aq)\rightarrow \text{MgCl}_2(aq)+\text{H}_2(g)$ How many electrons are transferred per Mg atom in the reaction?
    Answer: 2 This question tests understanding of oxidation-reduction (redox) reactions. In Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g), we track oxidation numbers: Mg goes from 0 (elemental) to +2 in MgCl₂ (loses 2 electrons, oxidized), while H⁺ in HCl (oxidation number +1) becomes H₂ at 0 (gains electrons, reduced). Since each Mg atom goes from 0 to +2, it transfers 2 electrons. A common error is thinking 4 electrons are transferred because there are 2 HCl molecules, but we count electrons per Mg atom, not per equation. Remember: count electron transfer per atom of the species specified in the question, using oxidation number changes.
14. In aqueous solution, hydrogen peroxide reacts with iodide ions in acidic solution to form iodine: $\text{H}_2\text{O}_2(aq) + 2\text{I}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{I}_2(aq) + 2\text{H}_2\text{O}(l)$ Which species is reduced?
    Answer: H$_2$O$_2$(aq) This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: oxygen in H₂O₂ is -1, iodine in I⁻ is -1, oxygen in H₂O is -2, and iodine in I₂ is 0. Oxygen's oxidation number decreases from -1 to -2, indicating gain of electrons and reduction for H₂O₂, while iodine increases from -1 to 0, showing loss of electrons and oxidation. Hydrogen remains +1. A tempting distractor is I⁻, but it is oxidized, not reduced; reversing the processes is common. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers.
15. Aqueous hydrogen peroxide reacts with iodide ions in acidic solution to produce iodine and water, as shown: $\mathrm{H_2O_2(aq) + 2\,I^-(aq) + 2\,H^+(aq) \rightarrow I_2(aq) + 2\,H_2O(l)}$ Which statement correctly describes the redox changes?
    Answer: I$^-$(aq) is oxidized and H$_2$O$_2$(aq) is reduced This question assesses understanding of oxidation–reduction (redox) reactions. To describe the redox changes, assign oxidation numbers: O in H₂O₂(aq) is -1 each (H +1, total balanced), I in I⁻(aq) is -1, H in H⁺(aq) is +1, I in I₂(aq) is 0, O in H₂O(l) is -2. Iodine increases from -1 to 0, losing electrons and being oxidized, while oxygen in H₂O₂ decreases from -1 to -2, gaining electrons and being reduced; H remains +1. Thus, I⁻(aq) is oxidized and H₂O₂(aq) is reduced. A tempting distractor is 'I⁻(aq) is reduced and H₂O₂(aq) is oxidized,' but that's reversed—a common error is assigning O in peroxide as -2 instead of -1. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers, especially for exceptions like peroxides.
16. A student adds solid zinc to an aqueous solution of copper(II) sulfate, and the following reaction occurs: $\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$ Which species is oxidized in the reaction?
    Answer: Zn(s) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), we need to track oxidation numbers: Zn starts at 0 (elemental form) and becomes Zn²⁺ in ZnSO₄ (oxidation number +2), while Cu²⁺ in CuSO₄ (oxidation number +2) becomes Cu(s) with oxidation number 0. Since Zn loses electrons (0 → +2), it is oxidized, and Cu²⁺ gains electrons (+2 → 0), so it is reduced. A common error is thinking CuSO₄ is oxidized because it disappears, but we must track individual elements, not compounds. Remember: oxidation is loss of electrons (increase in oxidation number), and the species that loses electrons is the one being oxidized.
17. A reaction between iron(III) oxide and carbon monoxide in a blast furnace is represented by the balanced equation: $\text{Fe}_2\text{O}_3(s)+3\,\text{CO}(g)\rightarrow 2\,\text{Fe}(s)+3\,\text{CO}_2(g)$ Which species acts as the reducing agent?
    Answer: CO(g) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), we track oxidation numbers: Fe in Fe₂O₃ is +3 and becomes 0 in Fe(s), while C in CO is +2 and becomes +4 in CO₂. Since Fe goes from +3 to 0, it gains electrons and is reduced; since C goes from +2 to +4, it loses electrons and is oxidized. The reducing agent is the species that gets oxidized, which is CO(g). A common error would be choosing Fe₂O₃(s) because students might think the compound containing the reduced element is the reducing agent, but the reducing agent is the species that causes reduction by donating electrons. Remember: the reducing agent is oxidized (loses electrons), while the oxidizing agent is reduced (gains electrons).
18. Hydrogen peroxide decomposes according to the equation below: $2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g)$ Which statement correctly describes the redox changes of oxygen in this reaction?
    Answer: Oxygen is both oxidized and reduced: $-1\rightarrow 0$ and $-1\rightarrow -2$ This question tests understanding of oxidation-reduction (redox) reactions. In 2H₂O₂(aq) → 2H₂O(l) + O₂(g), oxygen in H₂O₂ has oxidation number -1 (peroxide), and it changes to -2 in H₂O (gains 1 electron, reduced) and to 0 in O₂ (loses 1 electron, oxidized). This is a disproportionation reaction where the same element (oxygen) is both oxidized and reduced: -1 → 0 (oxidation) and -1 → -2 (reduction). A common error is thinking oxygen remains at -1 or only undergoes one type of change. Remember: in disproportionation reactions, the same element in one oxidation state produces two different oxidation states.
19. In acidic solution, permanganate ions react with iron(II) ions according to the equation below: $\text{MnO}_4^{-}(aq)+5\text{Fe}^{2+}(aq)+8\text{H}^+(aq)\rightarrow \text{Mn}^{2+}(aq)+5\text{Fe}^{3+}(aq)+4\text{H}_2\text{O}(l)$ Which species is the oxidizing agent?
    Answer: MnO$_4^-$(aq) This question tests understanding of oxidation-reduction (redox) reactions. In this reaction, we track oxidation numbers: Mn in MnO₄⁻ goes from +7 to +2 in Mn²⁺ (gains 5 electrons, reduced), while Fe²⁺ goes to Fe³⁺ (loses 1 electron per ion, oxidized). The oxidizing agent is the species that gets reduced while causing another species to be oxidized—that's MnO₄⁻. A common error is thinking H⁺ is the oxidizing agent because it appears in the reaction, but H⁺ maintains its +1 oxidation state throughout. Remember: the oxidizing agent accepts electrons and gets reduced in the process.
20. A student adds solid zinc to an aqueous solution of copper(II) sulfate, and the following reaction occurs: $\text{Zn}(s)+\text{CuSO}_4(aq)\rightarrow \text{ZnSO}_4(aq)+\text{Cu}(s)$ Which species is oxidized in the reaction?
    Answer: Zn(s) This question tests understanding of oxidation-reduction (redox) reactions. In the reaction Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s), we need to track oxidation numbers: Zn starts at 0 (elemental form) and becomes Zn²⁺ in ZnSO₄ (oxidation number +2), while Cu²⁺ in CuSO₄ (oxidation number +2) becomes Cu at 0. Since Zn goes from 0 to +2, it loses 2 electrons and is oxidized; Cu²⁺ gains 2 electrons (from +2 to 0) and is reduced. A common error is thinking CuSO₄ is oxidized because it's a reactant, but we must track individual elements, not compounds. Remember: oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number).

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