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.

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 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.

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.

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.

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.

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. 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.

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. 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.