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.

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.

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.

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.

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.

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.

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.

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.

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.

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.