These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data: Ag 107.87, Cr 52.00, O 16.00 g mol⁻¹. Solubility: all sodium, potassium, ammonium and nitrate salts are soluble in water; any other solubility information you need is given in the question.
Question 1 (multiple choice · foundation)
Calcium chlorite has the formula Ca(ClO₂)₂. What is the oxidation number of chlorine in this compound?
- (A) −1
- (B) +3
- (C) +4
- (D) +5
Answer and explanation
Answer: (B). Calcium is a group 2 metal, so it is +2 and the compound contains two ClO₂⁻ ions. In ClO₂⁻: Cl + 2(−2) = −1, so Cl = +3.
- (A) is the oxidation number of chlorine in a chloride ion, Cl⁻. This compound has no chloride ions.
- (C) treats ClO₂ as a neutral unit (Cl + 2(−2) = 0), forgetting the ion’s 1− charge.
- (D) is chlorine in the chlorate ion, ClO₃⁻, which has one more oxygen atom.
Question 2 (multiple choice · core)
Which of the following is a redox reaction?
- (A) AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq)
- (B) H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
- (C) 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
- (D) CaO(s) + H₂O(l) → Ca(OH)₂(s)
Answer and explanation
Answer: (C). In hydrogen peroxide, oxygen is −1 (a peroxide). In the products it is −2 in H₂O and 0 in O₂. Some oxygen atoms are reduced (−1 → −2) and others are oxidised (−1 → 0), so oxidation numbers change and the reaction is redox.
- (A) is precipitation. Ag stays +1, Br stays −1, and the solid AgBr forms from aqueous ions.
- (B) is acid-base. H₂SO₄ gives protons to OH⁻ ions to make water; every oxidation number is unchanged (S stays +6).
- (D) has no oxidation number change: Ca +2, O −2, H +1 on both sides. (The oxide ion accepts a proton from water, so it is acid-base.) Forming a solid does not make it redox.
Question 3 (multiple choice · core)
A student mixes equal volumes of the two solutions in each pair. Only one pair can form a precipitate. Which one?
- (A) NaCl(aq) and KNO₃(aq)
- (B) NH₄Cl(aq) and NaNO₃(aq)
- (C) Pb(NO₃)₂(aq) and KI(aq)
- (D) K₂SO₄(aq) and NH₄NO₃(aq)
Answer and explanation
Answer: (C). Swap partners and test each new combination against the rule. In (C) the new combinations are KNO₃ (soluble: potassium and nitrate) and PbI₂, which is not a sodium, potassium, ammonium or nitrate salt. PbI₂ is the only combination in any option that the rule does not guarantee is soluble, so (C) is the only pair that can form a precipitate. (In fact, PbI₂ forms as a bright yellow solid.) Net ionic equation: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
- (A) The new combinations are NaNO₃ and KCl: both are on the always-soluble list.
- (B) The new combinations are NH₄NO₃ and NaCl: both soluble.
- (D) The new combinations are KNO₃ and (NH₄)₂SO₄: both soluble.
Question 4 (calculation · core)
Propene, C₃H₆, is a hydrocarbon fuel.
(a) Write a balanced equation for the complete combustion of propene. (b) Find the average oxidation number of carbon in propene and in the product that contains carbon. (c) Use your answers to show that the total number of electrons lost equals the total gained in your balanced equation.
Worked solution
(a) Products are CO₂ and H₂O. C₃H₆ + 4.5O₂ → 3CO₂ + 3H₂O; doubling gives 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Check: C 6 = 6; H 12 = 12; O 18 = 12 + 6.
(b) In C₃H₆: 3C + 6(+1) = 0, so the average oxidation number of C is −2. In CO₂, C is +4.
(c) Each C rises by 6: 6 C atoms × 6 = 36 electrons lost. Each O falls from 0 to −2: 18 O atoms × 2 = 36 electrons gained. The totals are equal.
Suggested mark points (3): 1 for the balanced equation (any correct whole-number multiple); 1 for −2 and +4; 1 for 36 = 36 with the counts shown.
Common error: counting electrons per molecule of propene but per O₂ for oxygen, which gives unequal totals. Use the same balanced equation for both.
Question 5 (constructed response · core)
Classify each reaction as acid-base, redox (stating if it is also combustion) or precipitation. For each, give one piece of evidence from the equation.
(i) HSO₄⁻(aq) + CO₃²⁻(aq) → SO₄²⁻(aq) + HCO₃⁻(aq) (ii) 2Al(s) + 3Cl₂(g) → 2AlCl₃(s) (iii) Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) (iv) 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)
Worked solution and suggested Marlbridge rubric
(i) Acid-base. HSO₄⁻ loses an H⁺ (becoming SO₄²⁻) and CO₃²⁻ gains it (becoming HCO₃⁻). S stays +6 and C stays +4, so it is not redox.
(ii) Redox (not combustion). Al goes from 0 to +3 (oxidised); Cl goes from 0 to −1 (reduced). 2 Al × 3 = 6 electrons lost; 6 Cl × 1 = 6 gained. It is not combustion because O₂ is not a reactant, even though a solid forms and heat is released.
(iii) Precipitation. Fe stays +3, O −2, H +1. Aqueous ions combine into an ionic solid. OH⁻ does not gain or lose a proton.
(iv) Redox, and combustion. S goes from −2 to +4 (oxidised); O goes from 0 to −2 (reduced). H₂S reacts with oxygen gas, so it is combustion. Because the fuel contains sulfur, SO₂ forms as well as water.
| Point | What earns it |
|---|---|
| 1 | (i) acid-base, with the proton transfer identified |
| 1 | (ii) redox, with an oxidation number change for Al or Cl |
| 1 | (iii) precipitation, with “no oxidation number change” or “ions form an ionic solid” |
| 1 | (iv) redox and combustion, with the S change (−2 → +4) or the O change (0 → −2) |
A correct type with no evidence earns no point.
Question 6 (constructed response · core)
A student adds 30.0 mL of 0.150 M silver nitrate, AgNO₃(aq), to an excess of potassium chromate solution, K₂CrO₄(aq). A brick-red solid forms.
(a) Identify the solid. Justify your answer using solubility. (b) Write the net ionic equation for the reaction. (c) Use oxidation numbers to show that this is not a redox reaction. (d) Calculate the maximum mass of solid that can form.
Worked solution and suggested Marlbridge rubric
(a) The ions present are Ag⁺, NO₃⁻, K⁺ and CrO₄²⁻. The new combinations are KNO₃ and Ag₂CrO₄. KNO₃ is a potassium (and nitrate) salt, so it is soluble. The solid must be silver chromate, Ag₂CrO₄.
(b) 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s). K⁺ and NO₃⁻ are spectator ions.
(c) Ag is +1 before and after. In CrO₄²⁻, Cr + 4(−2) = −2, so Cr = +6; in Ag₂CrO₄, 2(+1) + Cr + 4(−2) = 0, so Cr = +6 again. O is −2 throughout. No oxidation number changes, so it is not redox.
(d) n(Ag⁺) = 0.0300 L × 0.150 mol L⁻¹ = 4.50 × 10⁻³ mol. Chromate is in excess, so Ag⁺ limits the yield. Two Ag⁺ make one Ag₂CrO₄: n(Ag₂CrO₄) = 2.25 × 10⁻³ mol. M(Ag₂CrO₄) = 2(107.87) + 52.00 + 4(16.00) = 331.74 g mol⁻¹. Mass = 2.25 × 10⁻³ mol × 331.74 g mol⁻¹ = 0.746 g.
| Point | What earns it |
|---|---|
| 1 | Ag₂CrO₄, with KNO₃ ruled out because potassium/nitrate salts are soluble |
| 1 | Correct net ionic equation with state symbols and no spectator ions |
| 1 | Cr shown as +6 on both sides (and Ag +1 on both sides) |
| 1 | Uses the 2 : 1 ratio of Ag⁺ to Ag₂CrO₄ |
| 1 | 0.746 g, with the molar mass shown |
Common error: ignoring the 2 : 1 ratio gives 1.49 g, twice the correct mass.
Question 7 (explanation · stretch)
A student writes: “Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) is an acid-base reaction, because H⁺ comes from an acid.” Evaluate the claim. In your answer, identify the species oxidised and the species reduced, and describe the transfer that actually takes place.
Model answer and suggested Marlbridge rubric
Model answer. The claim is incorrect. Zn goes from 0 to +2, so zinc is oxidised: each Zn atom loses 2 electrons. Hydrogen goes from +1 in H⁺ to 0 in H₂, so H⁺ is reduced: each H⁺ gains 1 electron, and 2 H⁺ gain 2 electrons in total. The reaction transfers electrons from Zn to H⁺, so it is redox. In an acid-base reaction, an H⁺ is passed from an acid to a base and every oxidation number stays the same. Here no species accepts a proton; the protons are turned into H₂ molecules. The H⁺ does come from an acid solution, but the reaction type depends on what is transferred, not on where the reactant came from.
| Point | What earns it |
|---|---|
| 1 | States the claim is incorrect and the reaction is redox |
| 1 | Zn oxidised (0 → +2) and H⁺ reduced (+1 → 0) |
| 1 | Electrons transferred from Zn to H⁺, 2 lost = 2 gained |
| 1 | Explains why it is not acid-base: no proton passed to a base / oxidation numbers change |
How did you do?
- Q1 wrong: practise the oxidation-number rules in the study guide, especially rule 3 for polyatomic ions.
- Q2, Q5 or Q7 wrong: use Figure 1 of the study guide and always check oxidation numbers first.
- Q3 or Q6(a) wrong: list every new cation–anion combination before applying the always-soluble rule.
- Q4 or Q6(d) wrong: count electrons and moles from the balanced equation, including coefficients.
Then tick off the topic checklist.