AP Chemistry Quiz: Oxidation Reduction Redox Reactions

What this quiz covers

This quiz focuses on Oxidation Reduction Redox Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 2 electrons
2. 4 electrons
3. 1 electron
4. 6 electrons
5. 3 electrons (correct answer)

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

1. I−(aq)
2. H+(aq)
3. I2(aq)
4. H2O2(aq) (correct answer)
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

1. 1
2. 2
3. 3 (correct answer)
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

1. $\text{Cr}_2\text{O}_7^{2-}(aq)$
2. $\text{SO}_3^{2-}(aq)$ (correct answer)
3. $\text{H}^+(aq)$
4. $\text{Cr}^{3+}(aq)$
5. $\text{SO}_4^{2-}(aq)$

Explanation: This question tests understanding of oxidation–reduction (redox) reactions. Assign oxidation numbers: sulfur in $\text{SO}_3^{2-}$ is +4, chromium in $\text{Cr}_2\text{O}_7^{2-}$ is +6, sulfur in $\text{SO}_4^{2-}$ is +6, and chromium in $\text{Cr}^{3+}$ 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 $\text{Cr}_2\text{O}_7^{2-}$, 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

1. Fe^2+(aq)
2. H+(aq)
3. Cr2O7^2−(aq) (correct answer)
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

1. Cr$_2$O$_7^{2-}$(aq)
2. H$^+$(aq)
3. I$^-$(aq) (correct answer)
4. Cr$^{3+}$(aq)
5. H$_2$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

1. Oxygen is only oxidized: $-1\rightarrow 0$
2. Oxygen is only reduced: $-1\rightarrow -2$
3. Oxygen is both oxidized and reduced: $-1\rightarrow 0$ and $-1\rightarrow -2$ (correct answer)
4. Oxygen remains at oxidation number $-1$ throughout
5. Oxygen is both oxidized and reduced: $0\rightarrow -1$ and $-2\rightarrow -1$

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

1. K$^+$(aq)
2. Cl$^-$(aq)
3. Cl$_2$(g) (correct answer)
4. I$_2$(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

1. Nitrogen changes from $+4$ in NO to $+2$ in NO$_2$
2. Nitrogen changes from $+2$ in NO to $+4$ in NO$_2$ (correct answer)
3. Nitrogen changes from $0$ in NO to $+2$ in NO$_2$
4. Nitrogen changes from $-2$ in NO to $0$ in NO$_2$
5. Nitrogen remains $+3$ in both NO and NO$_2$

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

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 (correct answer)
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.

Question 11

1. MnO$_4^-$(aq)
2. H$^+$(aq)
3. CO$_2$(g)
4. Mn$^{2+}$(aq)
5. C$_2$O$_4^{2-}$(aq) (correct answer)

Explanation: This question assesses understanding of oxidation–reduction (redox) reactions. To find the reducing agent, assign oxidation numbers: Mn in MnO₄⁻(aq) is +7 (O is -2, total -8, ion -1), C in C₂O₄²⁻(aq) is +3 each (O -2 total -8, ion -2 so two C +6), and in products, Mn²⁺ is +2, C in CO₂(g) is +4 (O -2 total -4). Carbon's oxidation number increases from +3 to +4, losing electrons and being oxidized, while Mn's decreases from +7 to +2, gaining electrons and being reduced; thus, C₂O₄²⁻(aq) is the reducing agent. A tempting distractor is MnO₄⁻(aq), but it is the oxidizing agent—a common error is assigning C in oxalate as +4 instead of +3 by miscounting oxygen. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers to identify the agent causing reduction.

Question 12

1. N is reduced from $+2$ to $+4$, and O is oxidized from $0$ to $-2$
2. N is oxidized from $+2$ to $+4$, and O is reduced from $0$ to $-2$ (correct answer)
3. N is reduced from $+4$ to $+2$, and O is reduced from $0$ to $-2$
4. N is oxidized from $0$ to $+4$, and O is reduced from $-2$ to $0$
5. No oxidation numbers change because oxygen is present on both sides

Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In the reaction 2NO(g) + O₂(g) → 2NO₂(g), we assign oxidation numbers: in NO, N is +2 (since O is -2); in NO₂, N is +4 (since 2×(-2) = -4 requires N to be +4 for neutrality); O₂ starts at 0 (elemental form) and becomes -2 in NO₂. Since N goes from +2 to +4, it loses electrons and is oxidized; since O goes from 0 to -2, it gains electrons and is reduced. Therefore, N is oxidized from +2 to +4, and O is reduced from 0 to -2. A common error would be thinking N is reduced because NO₂ has more oxygen atoms, but we must track the actual oxidation number changes. Remember: oxidation means an increase in oxidation number (loss of electrons), while reduction means a decrease in oxidation number (gain of electrons).

Question 13

1. C in CO$_2$(g)
2. O in CO$_2$(g)
3. Mg(s) (correct answer)
4. C(s)
5. O in MgO(s)

Explanation: 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).

Question 14

1. I$^-$(aq) is reduced and H$_2$O$_2$(aq) is oxidized
2. H$^+$(aq) is reduced and I$^-$(aq) is oxidized
3. I$^-$(aq) is oxidized and H$_2$O$_2$(aq) is reduced (correct answer)
4. I$_2$(aq) is oxidized and H$_2$O(l) is reduced
5. No species changes oxidation number because oxygen is present on both sides

Explanation: 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.

Question 15

1. $Fe^{2+}(aq)$
2. $H_2O(l)$
3. $H^{+}(aq)$
4. $Cr_2O_7^{2-}(aq)$ (correct answer)
5. $Cr^{3+}(aq)$

Explanation: This question assesses understanding of oxidation–reduction (redox) reactions. To find the oxidizing agent, assign oxidation numbers: iron in $Fe^{2+}(aq)$ is +2, chromium in $Cr_2O_7^{2-}(aq)$ is +6 (oxygen is -2, total -14 for seven oxygens, so two Cr are +12 total), and in products, $Fe^{3+}$ is +3, $Cr^{3+}$ is +3. Iron's oxidation number increases from +2 to +3, losing electrons and being oxidized, while chromium's decreases from +6 to +3, gaining electrons and being reduced. Thus, $Cr_2O_7^{2-}(aq)$ is the oxidizing agent as it causes oxidation by accepting electrons. A tempting distractor is $Fe^{2+}(aq)$, but it is oxidized, not the oxidizing agent—a common error is confusing the species that loses electrons with the one that accepts them. Oxidation is loss of electrons; reduction is gain of electrons—track oxidation numbers to identify agents in redox reactions.

Question 16

1. O^2− (in SO3)
2. S(s)
3. SO2(g) (correct answer)
4. O2(g)
5. SO3(g)

Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In 2SO₂(g) + O₂(g) → 2SO₃(g), we track oxidation numbers: S in SO₂ has +4 (since O is -2, and the molecule is neutral: S + 2(-2) = 0, so S = +4), and S in SO₃ has +6 (S + 3(-2) = 0, so S = +6). Oxygen in O₂ starts at 0 and becomes -2 in SO₃. Since sulfur increases from +4 to +6 (loses electrons), SO₂ is oxidized. A common mistake is thinking O₂ is oxidized because it "disappears," but O₂ actually gains electrons (0 → -2) and is reduced. Remember: oxidation involves loss of electrons and an increase in oxidation number.

Question 17

1. $\text{MnO}_4^{-}(aq)$
2. $\text{H}^{+}(aq)$
3. $\text{CO}_2(g)$
4. $\text{C}_2\text{O}_4^{2-}(aq)$ (correct answer)
5. $\text{H}_2\text{O}(l)$

Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In the reaction, Mn in $MnO_4^{-}$ has oxidation number +7 ($(Mn + 4(-2) = -1$, so $Mn = +7$) and becomes $Mn^{2+}$ (+2), gaining electrons and being reduced. Carbon in $C_2O_4^{2-}$ has oxidation number +3 ($(2C + 4(-2) = -2$, so $C = +3$) and becomes +4 in $CO_2$, losing electrons and being oxidized. The reducing agent is the species that gets oxidized, which is $C_2O_4^{2-}$ (oxalate ion). A common mistake is thinking $MnO_4^{-}$ is the reducing agent because it's reduced, but the reducing agent causes reduction while itself being oxidized. Remember: the reducing agent loses electrons and increases in oxidation number.

Question 18

1. Fe^3+(aq)
2. K+(aq)
3. I−(aq) (correct answer)
4. Cl−(aq)
5. I2(s)

Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In the reaction, Fe³⁺ in FeCl₃ becomes Fe²⁺ in FeCl₂ (gains electrons, is reduced), while I⁻ becomes I₂ (oxidation number -1 to 0, loses electrons, is oxidized). The reducing agent is the species that gets oxidized, which is I⁻ (iodide ion). K⁺ and Cl⁻ are spectator ions with unchanged oxidation numbers. A common error is thinking Fe³⁺ is the reducing agent because it appears to be "reduced," but the reducing agent causes reduction while itself being oxidized. Remember: the reducing agent loses electrons and gets oxidized in the process.

Question 19

1. H2S(g)
2. HCl(aq)
3. S(s)
4. Cl2(aq) (correct answer)
5. H+(aq)

Explanation: This question tests understanding of oxidation-reduction (redox) reactions. In H₂S(g) + Cl₂(aq) → 2HCl(aq) + S(s), we assign oxidation numbers: S in H₂S has -2, S in elemental form has 0, Cl in Cl₂ has 0, and Cl in HCl has -1. Sulfur goes from -2 to 0 (loses electrons, is oxidized), while chlorine goes from 0 to -1 (gains electrons, is reduced). The oxidizing agent is the species that gets reduced, which is Cl₂. A common mistake is thinking H₂S is the oxidizing agent because it contains the element being oxidized, but the oxidizing agent causes oxidation while itself being reduced. Remember: the oxidizing agent gains electrons and gets reduced in the process.

Question 20

1. Mg: $0 \rightarrow +2$ and O: $0 \rightarrow -2$ (correct answer)
2. Mg: $0 \rightarrow -2$ and O: $0 \rightarrow +2$
3. Mg: $+2 \rightarrow 0$ and O: $-2 \rightarrow 0$
4. Mg: $0 \rightarrow +1$ and O: $0 \rightarrow -1$
5. Mg: $+2 \rightarrow +4$ and O: $-2 \rightarrow -1$

Explanation: 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.