HSC Exam Practice

Sample response. Oxidation is the loss of electrons by a species during a chemical reaction.

Marking notes. 1 mark for “loss of electrons” (OIL RIG). Do not accept “reaction with oxygen” alone.

Sample response. An oxidant (oxidising agent) is the species that gains electrons in a redox reaction; it is itself reduced. A reductant (reducing agent) is the species that loses electrons; it is itself oxidised. The oxidant causes the other species to be oxidised by accepting its electrons.

Marking notes. 1 mark for oxidant = gains electrons / is reduced. 1 mark for reductant = loses electrons / is oxidised. Must be phrased with reference to electron gain/loss.

Sample response. Oxygen is usually assigned an oxidation state of −2 in compounds. Exception: in peroxides (e.g. H₂O₂), oxygen is −1. Hydrogen is usually assigned +1 in compounds. Exception: in metal hydrides (e.g. NaH), hydrogen is −1.

Marking notes. 1 mark for O = −2 in compounds. 1 mark for the O exception (peroxides, O = −1). 1 mark for H = +1 in compounds with the H exception (metal hydrides, H = −1). Award 2 marks if only one exception given alongside both standard rules.

Sample response. Mg: oxidation state changes from 0 (Mg metal) to +2 (in MgCl₂) — an increase → Mg is oxidised → Mg is the reductant. [1 mark for Mg identified as oxidised with ON change 0 → +2; 1 mark for correctly naming Mg as reductant.] H (in HCl, H = +1): changes to 0 (in H₂) — a decrease → H⁺ is reduced → H⁺ (from HCl) is the oxidant. [1 mark for H⁺ identified as reduced with ON change +1 → 0; 1 mark for correctly naming H⁺ as oxidant.]

Marking notes. 1m — Mg oxidised, ON change 0 → +2; 1m — Mg named as reductant; 1m — H⁺ (or HCl) reduced, ON change +1 → 0; 1m — H⁺ named as oxidant. Cl is a spectator (ON = −1 throughout).

Sample response. In SO₄²⁻, O = −2 (Rule 3). Four oxygen atoms contribute 4 × (−2) = −8. The overall charge of the ion is −2. Setting up the equation: S + (−8) = −2, so S = −2 + 8 = +6. The oxidation state of S in SO₄²⁻ is +6.

Marking notes. 1 mark for correctly setting up the algebraic equation (S + 4(−2) = −2). 1 mark for correct answer S = +6.

Sample response. Fe²⁺(aq) → Fe³⁺(aq) + e⁻. Atom check: 1Fe on each side. ✓ Charge check: left = +2; right = (+3) + (−1) = +2. ✓

Marking notes. 1 mark for the correct half-equation with electrons on the right (oxidation). 1 mark for both a correct atom check AND a correct charge check (both checks required for this mark).

Sample response (a). All three concentrations show an increasing percentage of stain molecules destroyed over time. The 3.0% solution destroys stain molecules fastest — reaching 40% destruction by approximately 4 minutes — while the 1.0% solution reaches 40% at approximately 5–6 minutes, and the 0.5% solution has not reached 40% by 10 minutes (only approximately 35%). Award: 1 mark for correctly describing the general increasing trend for all three; 1 mark for comparing the time to 40% destruction with appropriate data reference for at least two concentrations.

Sample response (b). The 3.0% OCl⁻ solution has a higher concentration of oxidant molecules than the 0.5% solution. A higher oxidant concentration means more OCl⁻ ions are available per unit volume per unit time to accept electrons from chromophore (stain) molecules. The rate of electron transfer from chromophore to OCl⁻ is therefore greater, destroying more stain molecules in the same time. The 0.5% solution has fewer OCl⁻ molecules available, so electron-transfer events occur less frequently and stain destruction proceeds more slowly. Award: 1 mark for identifying higher [OCl⁻] in the 3.0% solution; 1 mark for explaining the link between oxidant concentration and rate/frequency of electron-transfer events; 1 mark for explicitly connecting this to the observed difference in stain destruction rate (not just stating “more concentrated = faster”).

Sample response (a). In MnO₄⁻: charge = −1, O = −2, so 4(−2) + Mn = −1 → Mn = +7. In Mn²⁺: oxidation state = +2 (monatomic ion). Mn changes from +7 → +2 (a decrease). A decrease in oxidation state indicates reduction (RIG). MnO₄⁻ is therefore the oxidant — it gains electrons and is reduced. [2 marks: 1 for correct ON of Mn with working; 1 for identifying MnO₄⁻ as oxidant with RIG reasoning.]

Sample response (b). Multiply oxidation half by 5: 5Fe²⁺(aq) → 5Fe³⁺(aq) + 5e⁻. This matches the 5e⁻ in the reduction half. Add and cancel 5e⁻: 5Fe²⁺(aq) + MnO₄⁻(aq) + 8H⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l). [2 marks: 1 for correctly multiplying Fe half by 5 and cancelling electrons; 1 for the balanced overall equation.]

Sample response (c). Left: 5(+2) + (−1) + 8(+1) = 10 − 1 + 8 = +17. Right: 5(+3) + (+2) + 0 = 15 + 2 = +17. Charge balanced. ✓ [2 marks: 1 for calculating left-side total charge correctly; 1 for calculating right-side total charge and confirming balance.]

Sample response. The claim is incorrect. The modern definition of an oxidising agent does not require molecular oxygen (O₂) to be present; it requires only that the species accepts electrons from another species, causing that other species to be oxidised. This is the electron-transfer definition — summarised by OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons) — which supersedes the older “reaction with oxygen” definition.

In household bleach, the active species is OCl⁻ (hypochlorite ion), formed when NaOCl dissolves in water. The oxidation state of chlorine in OCl⁻ is calculated as: charge = −1, O = −2, so Cl + (−2) = −1 → Cl = +1. The oxidation state of chlorine in the product Cl⁻ is −1. The change is +1 → −1, a decrease of 2. A decrease in oxidation number indicates reduction (RIG). The reduction half-equation in alkaline solution is: OCl⁻(aq) + H₂O(l) + 2e⁻ → Cl⁻(aq) + 2OH⁻(aq). This half-equation shows OCl⁻ accepting 2 electrons — it is being reduced. A species that is reduced in a redox reaction is, by definition, the oxidant (oxidising agent). OCl⁻ causes whatever it reacts with (e.g. stain molecules or bacterial cell walls) to lose electrons, i.e. to be oxidised. The absence of O₂ from this mechanism is irrelevant: the electron-transfer criterion is fully met. The claim therefore conflates the old oxygen-combination definition with the modern electron-transfer definition, and is rejected.

Marking criteria.

• 1 mark — States an explicit overall evaluative judgement rejecting the claim.
• 1 mark — Explains that the modern definition of an oxidising agent is based on electron gain (not involvement of O₂), referencing OIL RIG or electron-transfer explicitly.
• 1 mark — Correctly calculates the oxidation state of Cl in OCl⁻ (+1) with algebraic working shown.
• 1 mark — States the oxidation state of Cl in Cl⁻ (−1) and identifies the decrease in oxidation state as evidence of reduction.
• 1 mark — Writes the balanced reduction half-equation for OCl⁻ in alkaline solution (or correctly cites/verifies the given equation with atom and charge checks).
• 1 mark — Explains, using the half-equation, that OCl⁻ accepts electrons and therefore causes the other species to be oxidised — satisfying the definition of an oxidising agent without O₂ being involved.
• 1 mark — Clearly distinguishes the old oxygen-combination definition from the modern electron-transfer definition and explicitly explains why the modern definition makes the claim incorrect.