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AP Chemistry

AP Chemistry Help: Oxidation Reduction Redox Reactions

Review real example questions for Oxidation Reduction Redox Reactions in AP Chemistry.

Question 1

Aluminum metal reacts with aqueous copper(II) chloride to produce aqueous aluminum chloride and copper metal, as shown:

2 Al(s)+3 CuCl2(aq)→2 AlCl3(aq)+3 Cu(s)\mathrm{2\,Al(s) + 3\,CuCl_2(aq) \rightarrow 2\,AlCl_3(aq) + 3\,Cu(s)}2Al(s)+3CuCl2​(aq)→2AlCl3​(aq)+3Cu(s)

How many electrons are transferred per aluminum atom that reacts?

  1. 2 electrons
  2. 4 electrons
  3. 1 electron
  4. 6 electrons
  5. 3 electrons
Explanation: This question assesses understanding of oxidation–reduction (redox) reactions. To determine electrons transferred per Al atom, assign oxidation numbers: Al(s) is 0, Cu in CuCl₂(aq) is +2 (Cl -1 each), Al in AlCl₃(aq) is +3, Cu(s) is 0. Each Al atom's oxidation number increases from 0 to +3, losing 3 electrons, while Cu decreases from +2 to 0, gaining 2 electrons; the balanced equation shows 2 Al and 3 Cu, so total 6 electrons transferred (3 per Al times 2). It's 3 electrons per aluminum atom. A tempting distractor is 2 electrons, but that's per Cu, not Al—a common error is confusing per atom with total transfer without balancing. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation number changes per atom and use coefficients for total electrons.

Question 2

Hydrogen peroxide can react with iodide ions in acidic solution to produce iodine and water:

H2O2(aq)+2I−(aq)+2H+(aq)→I2(aq)+2H2O(l)\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)H2​O2​(aq)+2I−(aq)+2H+(aq)→I2​(aq)+2H2​O(l)

Which species is reduced?

  1. I−(aq)
  2. H+(aq)
  3. I2(aq)
  4. H2O2(aq)
  5. H2O(l)
Explanation: 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.

Question 3

A student adds aluminum metal to an aqueous solution of silver nitrate, producing silver metal:

Al(s)+3AgNO3(aq)→Al(NO3)3(aq)+3Ag(s)\text{Al}(s) + 3\text{AgNO}_3(aq) \rightarrow \text{Al(NO}_3)_3(aq) + 3\text{Ag}(s)Al(s)+3AgNO3​(aq)→Al(NO3​)3​(aq)+3Ag(s)

How many electrons are transferred per aluminum atom in this reaction?

  1. 1
  2. 2
  3. 3
  4. 6
  5. 9
Explanation: This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: aluminum in Al(s) is 0, silver in AgNO₃ is +1, aluminum in Al(NO₃)₃ is +3, and silver in Ag(s) is 0. Aluminum's oxidation number increases from 0 to +3, indicating loss of 3 electrons per atom, while silver decreases from +1 to 0, showing gain of 1 electron per atom. Nitrate remains unchanged. A tempting distractor is 6, but that might come from doubling the electrons for the balanced equation; forgetting per-atom count is common. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers.

Question 4

In aqueous solution, dichromate ions oxidize sulfite ions to sulfate ions:

Cr2O72−(aq)+3SO32−(aq)+8H+(aq)→2Cr3+(aq)+3SO42−(aq)+4H2O(l)\text{Cr}_2\text{O}_7^{2-}(aq) + 3\text{SO}_3^{2-}(aq) + 8\text{H}^+(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{SO}_4^{2-}(aq) + 4\text{H}_2\text{O}(l)Cr2​O72−​(aq)+3SO32−​(aq)+8H+(aq)→2Cr3+(aq)+3SO42−​(aq)+4H2​O(l)

Which species is oxidized?

  1. Cr2O72−(aq)\text{Cr}_2\text{O}_7^{2-}(aq)Cr2​O72−​(aq)
  2. SO32−(aq)\text{SO}_3^{2-}(aq)SO32−​(aq)
  3. H+(aq)\text{H}^+(aq)H+(aq)
  4. Cr3+(aq)\text{Cr}^{3+}(aq)Cr3+(aq)
  5. SO42−(aq)\text{SO}_4^{2-}(aq)SO42−​(aq)
Explanation: This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: sulfur in SO32−\text{SO}_3^{2-}SO32−​ is +4, chromium in Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2​O72−​ is +6, sulfur in SO42−\text{SO}_4^{2-}SO42−​ is +6, and chromium in Cr3+\text{Cr}^{3+}Cr3+ is +3. Sulfur's oxidation number increases from +4 to +6, indicating loss of electrons and oxidation, while chromium decreases from +6 to +3, showing gain of electrons and reduction. Hydrogen and oxygen numbers balance out. A tempting distractor is Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2​O72−​, but it is reduced, not oxidized; misidentifying the oxidized species is common in complex ions. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers.

Question 5

In an acidic solution, iron(II) ions react with dichromate ions according to the overall reaction:

6Fe2+(aq)+Cr2O72−(aq)+14H+(aq)→6Fe3+(aq)+2Cr3+(aq)+7H2O(l)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)6Fe2+(aq)+Cr2​O72−​(aq)+14H+(aq)→6Fe3+(aq)+2Cr3+(aq)+7H2​O(l)

Which species is reduced?

  1. Fe^2+(aq)
  2. H+(aq)
  3. Cr2O7^2−(aq)
  4. Fe^3+(aq)
  5. H2O(l)
Explanation: 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.

Question 6

Aqueous dichromate reacts with iodide ions in acidic solution, producing chromium(III) ions and iodine:

Cr2O72−(aq)+14H+(aq)+6I−(aq)→2Cr3+(aq)+3I2(s)+7H2O(l)\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)Cr2​O72−​(aq)+14H+(aq)+6I−(aq)→2Cr3+(aq)+3I2​(s)+7H2​O(l)

Which species is oxidized?

  1. Cr2_22​O72−_7^{2-}72−​(aq)
  2. H+^++(aq)
  3. I−^-−(aq)
  4. Cr3+^{3+}3+(aq)
  5. H2_22​O(l)
Explanation: 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.

Question 7

Hydrogen peroxide decomposes according to the equation below:

2H2O2(aq)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(aq)\rightarrow 2\text{H}_2\text{O}(l)+\text{O}_2(g)2H2​O2​(aq)→2H2​O(l)+O2​(g)

Which statement correctly describes the redox changes of oxygen in this reaction?

  1. Oxygen is only oxidized: −1→0-1\rightarrow 0−1→0
  2. Oxygen is only reduced: −1→−2-1\rightarrow -2−1→−2
  3. Oxygen is both oxidized and reduced: −1→0-1\rightarrow 0−1→0 and −1→−2-1\rightarrow -2−1→−2
  4. Oxygen remains at oxidation number −1-1−1 throughout
  5. Oxygen is both oxidized and reduced: 0→−10\rightarrow -10→−1 and −2→−1-2\rightarrow -1−2→−1
Explanation: 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.

Question 8

Chlorine gas is bubbled into an aqueous solution of potassium iodide, producing iodine and chloride ions:

Cl2(g)+2KI(aq)→2KCl(aq)+I2(s)\text{Cl}_2(g)+2\text{KI}(aq)\rightarrow 2\text{KCl}(aq)+\text{I}_2(s)Cl2​(g)+2KI(aq)→2KCl(aq)+I2​(s)

Which species is reduced in the reaction?

  1. K+^++(aq)
  2. Cl−^-−(aq)
  3. Cl2_22​(g)
  4. I2_22​(s)
  5. I−^-−(aq)
Explanation: 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.

Question 9

Nitrogen monoxide reacts with oxygen to form nitrogen dioxide:

2NO(g)+O2(g)→2NO2(g)2\text{NO}(g)+\text{O}_2(g)\rightarrow 2\text{NO}_2(g)2NO(g)+O2​(g)→2NO2​(g)

Which statement correctly describes the oxidation-number change for nitrogen?

  1. Nitrogen changes from +4+4+4 in NO to +2+2+2 in NO2_22​
  2. Nitrogen changes from +2+2+2 in NO to +4+4+4 in NO2_22​
  3. Nitrogen changes from 000 in NO to +2+2+2 in NO2_22​
  4. Nitrogen changes from −2-2−2 in NO to 000 in NO2_22​
  5. Nitrogen remains +3+3+3 in both NO and NO2_22​
Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In 2NO(g) + O₂(g) → 2NO₂(g), we must determine nitrogen's oxidation numbers: in NO, N is +2 (since O is -2 and the compound is neutral), and in NO₂, N is +4 (since 2×(-2) = -4 from oxygen requires N to be +4). Nitrogen changes from +2 to +4, losing 2 electrons per atom (oxidized). A common error is assuming N is +4 in NO because students might incorrectly think all nitrogen oxides have similar oxidation states. Remember: calculate oxidation numbers systematically using the rule that oxygen is typically -2 in compounds.

Question 10

Nitrogen monoxide reacts with oxygen to form nitrogen dioxide, as shown:

2 NO(g)+O2(g)→2 NO2(g)\mathrm{2\,NO(g) + O_2(g) \rightarrow 2\,NO_2(g)}2NO(g)+O2​(g)→2NO2​(g)

How does the oxidation number of nitrogen change in this reaction?

  1. It remains +2
  2. It decreases from +2 to 0
  3. It increases from 0 to +2
  4. It increases from +2 to +4
  5. It decreases from +2 to +1
Explanation: This question assesses understanding of oxidation–reduction (redox) reactions. To track nitrogen's oxidation number change, assign them: in NO(g), O is -2 so N is +2; O in O₂(g) is 0; in NO₂(g), O is -2 (total -4) so N is +4. Nitrogen's oxidation number increases from +2 to +4, showing loss of electrons and oxidation, while oxygen is incorporated but its role supports the change. The increase is from +2 to +4, not a decrease or no change. A tempting distractor is 'it decreases from +2 to 0,' but N goes to +4, not 0—a common error is miscalculating NO₂ as N at 0 by ignoring oxygen's contribution. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers carefully for polyatomic molecules.