Chemistry • Year 11 • Module 3 • Lesson 8
Redox Reactions & Oxidation States
Build HSC Band 5–6 extended-response technique for redox: synthesising data, half-equations, and OIL RIG reasoning into justified chemical arguments.
1. Data + scenario — bleach concentration and oxidation in a hospital setting (Band 5–6)
8 marks Band 5–6
Scenario. A hospital infection-control team is evaluating two bleach formulations for surface disinfection. Formulation P contains 1% NaOCl (sodium hypochlorite); Formulation Q contains 5% NaOCl. Both are alkaline solutions (pH ~12). The active redox species in both is the hypochlorite ion OCl−, which acts as an oxidant by accepting electrons from organic molecules in bacterial cell walls. The reduction half-equation for OCl− in alkaline solution is: OCl−(aq) + H2O(l) + 2e− → Cl−(aq) + 2OH−(aq).
The table below shows measured disinfection performance after a 1-minute contact time on a stainless steel surface contaminated with Staphylococcus aureus.
| Formulation | NaOCl (%) | Log reduction in bacteria | Visible surface corrosion? | OCl− remaining after 1 min (%) |
|---|---|---|---|---|
| P | 1% | 3.1 | None observed | 72% |
| Q | 5% | 5.8 | Slight pitting after repeat use | 34% |
Illustrative data. After McDonnell & Russell (1999), Clinical Microbiology Reviews.
Q1. Analyse and evaluate the two bleach formulations for hospital use, using redox chemistry as your framework. In your response you must:
- State the oxidation states of chlorine in OCl− and Cl−, and use this to confirm that OCl− acts as an oxidant in the reduction half-equation provided.
- Explain, using the concept of oxidant concentration, why Formulation Q achieves a greater log reduction in bacteria (5.8 vs 3.1) than Formulation P.
- Interpret the “OCl− remaining after 1 min” data to explain why Formulation Q has a lower percentage remaining, and link this to the rate of electron transfer.
- Evaluate which formulation is more appropriate for routine hospital surface disinfection, weighing bactericidal effectiveness against surface corrosion risk — reach a justified recommendation.
2. Multi-step redox analysis — Fe2+ reacting with permanganate (Band 5–6)
7 marks Band 5–6
Context. Iron ore (mainly Fe2O3) can be analysed using a redox reaction with acidic potassium permanganate (KMnO4). The iron is first dissolved in acid and reduced to Fe2+, which then reacts with MnO4−. The two half-equations are:
Oxidation: Fe2+(aq) → Fe3+(aq) + e− (i)
Reduction: MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) (ii)
(a) Combine half-equations (i) and (ii) to give the balanced overall ionic equation. Show how you balance the electron count, then verify the combined equation with a charge check. 3 marks
(b) Assign the oxidation state of Mn in MnO4− and in Mn2+. Show your algebraic working for MnO4−. Use OIL RIG to identify whether MnO4− is the oxidant or the reductant, and justify your answer by referring to the change in oxidation state. 3 marks
(c) Explain why H+ ions appear in the reduction half-equation (ii) but are not classified as either the oxidant or the reductant in this reaction. Use the concept of oxidation state to support your answer. 1 mark
Q1 — Sample Band 6 response (8 marks), annotated
In OCl−, the charge = −1, O = −2, so Cl + (−2) = −1, giving Cl = +1. In Cl−, Cl = −1. The change is +1 → −1 (a decrease of 2), which by OIL RIG (Reduction Is Gain) confirms that chlorine is reduced — gaining 2 electrons per OCl− ion. A species that is reduced is by definition an oxidant (it oxidises the bacterial molecules by accepting their electrons). The half-equation confirms this: OCl− + H2O + 2e− → Cl− + 2OH−. [2 marks — oxidation states confirmed; oxidant identification with half-equation evidence]
Formulation Q (5% NaOCl) achieves a 5.8 log reduction compared to 3.1 for Formulation P (1% NaOCl). A higher concentration of OCl− means more oxidant molecules are available per unit surface area per unit time. More electron-transfer events occur between OCl− and organic molecules in bacterial cell walls, more rapidly destroying the structural integrity of the bacteria. The relationship is consistent with collision theory applied to redox: higher [OCl−] → greater frequency of productive collisions with bacterial substrates → greater bactericidal effect. [2 marks — links higher oxidant concentration to rate of electron transfer and bactericidal outcome]
Formulation Q has only 34% OCl− remaining after 1 minute, compared to 72% for Formulation P. This directly reflects the higher rate of electron-transfer reactions in Q: more OCl− is reduced to Cl− because more electrons are accepted from bacterial substrates. The faster consumption of OCl− in Q is both the cause of its greater disinfection and its faster depletion — the two are linked by the same redox mechanism. [2 marks — interprets remaining OCl² as consumed oxidant; links to electron-transfer rate]
For routine hospital surface disinfection, Formulation P (1% NaOCl) is the more appropriate choice. A 3.1 log reduction is sufficient for most routine disinfection contexts (3 log = 99.9% bacteria destroyed), the surface is not visibly corroded, and 72% of OCl− remains active, providing residual protection. Formulation Q is more appropriate for high-risk or outbreak situations where maximum pathogen kill is paramount and surface corrosion is an accepted trade-off, but its repeat-use corrosion of stainless steel makes it unsuitable for routine cleaning of hospital equipment. [2 marks — context-aware recommendation with trade-off reasoning]
Marking criteria.
- 1 mark — Correctly states oxidation states of Cl in OCl− (+1) and Cl− (−1) with algebraic working.
- 1 mark — Uses the decrease in oxidation number and OIL RIG to confirm OCl− is the oxidant (is reduced, gaining electrons).
- 1 mark — Explains higher [OCl−] in Q leads to more electron-transfer events per unit time.
- 1 mark — Links increased electron-transfer rate to greater bactericidal performance (5.8 vs 3.1 log reduction).
- 1 mark — Interprets the lower % OCl− remaining in Q as evidence that more oxidant has been consumed in redox reactions.
- 1 mark — Links faster OCl− consumption in Q to the same redox mechanism responsible for its greater efficacy.
- 1 mark — Evaluates both formulations on at least two criteria (efficacy AND corrosion risk) and reaches a context-conditional recommendation.
- 1 mark — Recommendation is explicitly justified using both the redox data (log reduction, % OCl− remaining) and contextual trade-off (routine vs outbreak) in precise chemical terminology.
Q2(a) — Overall ionic equation and charge check (3 marks)
Multiply oxidation half-equation (i) by 5 to match 5e− in half-equation (ii):
5Fe2+(aq) → 5Fe3+(aq) + 5e− [1 mark — correct multiplication and electron matching]
Add to reduction half (ii) and cancel 5e−:
5Fe2+(aq) + MnO4−(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) [1 mark — correct overall equation]
Charge check: left = 5(+2) + (−1) + 8(+1) = 10 − 1 + 8 = +17. Right = 5(+3) + (+2) + 0 = 17. ✓ [1 mark — both sides confirmed equal]
Q2(b) — Oxidation states and oxidant identification (3 marks)
MnO4−: charge = −1; O = −2, so 4(−2) + Mn = −1; Mn = −1 + 8 = +7. [1 mark — correct algebraic working and answer]
Mn2+: monatomic ion, oxidation state = charge = +2. [no separate mark; needed for next part]
Mn changes from +7 → +2 — a decrease. By OIL RIG (Reduction Is Gain), a decrease in oxidation number indicates that Mn (in MnO4−) has gained electrons → MnO4− is reduced. [1 mark — correct identification of decrease = reduction]
A species that is reduced in a redox reaction is by definition the oxidant (it accepts electrons and causes the other species to be oxidised). [1 mark — correct naming as oxidant with justification]
Q2(c) — Role of H+ (1 mark)
H+ appears in the reduction half-equation to balance the hydrogen atoms needed to form H2O. However, the oxidation state of hydrogen does not change during the reaction: H is +1 in H+ (reactant) and +1 in H2O (product). Because there is no change in oxidation state, no electron transfer involving H occurs, so H+ is neither oxidised nor reduced — it is neither the oxidant nor the reductant. It acts only as a balancing species for atoms and charge. [1 mark]