The shape of the lesson
By the end of the lesson, learners can
- The rules: Pick species in the simulator and assign each atom's oxidation number before revealing it.
- Working them out: Write the equation for the sum and solve it.
- Oxidised or reduced?: Use the tracker to follow each element across an equation.
- Fixed values first: F −1, O −2, H +1, Group 1 +1, Group 2 +2.
- Up = oxidised (lost e−); down = reduced (gained e−).
- Oxidising agent is reduced; reducing agent is oxidised.
- Disproportionation: one element goes both up and down.
One picture every idea returns to
An oxidation number is the charge an atom would have if all its bonds were ionic. Elements are 0; a simple ion has its charge; the numbers in a compound add up to 0, and in an ion to the ion's charge. Oxygen is usually −2, hydrogen +1 and fluorine always −1. Oxidation is loss of electrons, an increase in oxidation number; reduction is gain of electrons, a decrease. An oxidising agent is itself reduced; a reducing agent is itself oxidised.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Keep score of the electrons.. Redox is about electrons moving, but in a covalent molecule nobody owns them outright. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The rules | Pick species in the simulator and assign each atom's oxidation number before revealing it. Project tab 1 of the lab (Electrochemistry lab, oxidation numbers) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Working them out | Write the equation for the sum and solve it. | Pairs |
| 21 to 30 min | Teach and check: Oxidised or reduced? | Use the tracker to follow each element across an equation. Project tab 4 of the lab (Electrochemistry lab, oxidised or reduced?) and let a student drive it while the class predicts. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Oxidised or reduced?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The oxidising agent is the species that is reduced, because it takes electrons from something else. | Students often name the species that was oxidised instead. Also remember that oxidation numbers are written with the sign first (+3), while ionic charges are written with the sign last (3+). |
Oxidation numbers: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Give the oxidation number of the underlined element: (a) NO3−, (b) VO2+, (c) H2O2, (d) PCl5.
Write the formulae of (a) copper(I) oxide and (b) potassium manganate(VII).
For Cu + 2Ag+ → Cu2+ + 2Ag, state which species is oxidised, which is reduced, and identify the oxidising agent.
Chlorine reacts with hot aqueous sodium hydroxide: 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O. Use oxidation numbers to explain why this is disproportionation.
In SO2 + 2H2S → 3S + 2H2O, give the change in oxidation number of sulfur from each reactant.
Explain, in terms of electrons, why an oxidising agent is itself reduced.
Answer key and marking notes
- (a) N + 3(−2) = −1, N = +5.
- (b) V + 2(−2) = +1, V = +5.
- (c) peroxide: O = −1.
- (d) P + 5(−1) = 0, P = +5.
- (a) Cu is +1, so Cu2O.
- (b) Mn is +7, so the ion is MnO4− and the salt is KMnO4.
- Cu is oxidised: 0 → +2.
- Ag+ is reduced: +1 → 0.
- Ag+ is the oxidising agent.
- Cl in Cl2 is 0.
- In NaCl, Cl is −1 (reduced); in NaClO3, Cl is +5 (oxidised).
- The same element is simultaneously oxidised and reduced, which is disproportionation.
- In SO2, S is +4 and goes to 0: reduced, a change of −4.
- In H2S, S is −2 and goes to 0: oxidised, a change of +2 (so two H2S are needed per SO2).
- An oxidising agent oxidises another species by removing electrons from it.
- It gains those electrons, so it is reduced and its oxidation number decreases.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What is the oxidation number of manganese in the manganate(VII) ion, MnO4−?
- A+7
- B+8
- C+6
- D−7
Which rule for assigning oxidation numbers is always true?
- Afluorine has an oxidation number of −1 in all of its compounds
- Boxygen has an oxidation number of −2 in all of its compounds
- Chydrogen has an oxidation number of +1 in all of its compounds
- Dthe oxidation numbers in an ion always add up to zero
What is the oxidation number of nitrogen in the ammonium ion, NH4+?
- A−3
- B+3
- C−4
- D+5
Which is the correct systematic name for NaClO3?
- Asodium chlorate(V)
- Bsodium chlorate(VI)
- Csodium chlorate(VII)
- Dsodium chlorate(I)
In the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), which species is the oxidising agent?
- ACu2+, because it gains electrons and is itself reduced from +2 to 0
- BZn, because it is oxidised from 0 to +2
- CZn2+, because it has the higher oxidation number among the products
- DCu, because it is formed by reduction
What is the oxidation number of sulfur in the thiosulfate ion, S2O32−?
- A+2
- B+3
- C+4
- D−2
Chlorine reacts with cold aqueous sodium hydroxide: Cl2 + 2NaOH → NaCl + NaClO + H2O. Which statement describes the redox change?
- Achlorine disproportionates: it is reduced from 0 to −1 in NaCl and oxidised from 0 to +1 in NaClO
- Bchlorine is only reduced, from 0 to −1, and NaOH is the reducing agent
- Csodium is oxidised from 0 to +1 and chlorine is reduced
- Doxygen is oxidised from −2 to 0 in NaClO and chlorine is reduced
The number line shows the oxidation number of nitrogen in several species. In acidic solution, nitrate ions are reduced to nitrogen monoxide (blue arrow). How many electrons does each nitrogen atom gain?
- A3
- B2
- C5
- D7
Quiz key
Let the oxidation number of Mn be x. Four oxygens contribute 4 × (−2) = −8, and the total must equal the ion charge, −1: x − 8 = −1, so x = +7 (which is why the name is manganate(VII)). +8 comes from setting the sum to zero, forgetting the −1 charge on the ion. +6 is manganese in the manganate(VI) ion, MnO42−, or comes from using the wrong charge. −7 has the right size but the wrong sign: oxygen is more electronegative, so manganese is the positive partner.
Fluorine is the most electronegative element, so it always takes the electron and is −1 in every compound. Oxygen is usually −2, but it is −1 in peroxides such as H2O2 and +2 in OF2, where it is bonded to the more electronegative fluorine. Hydrogen is usually +1, but it is −1 in metal hydrides such as NaH. In an ion the oxidation numbers add up to the charge on the ion; they add up to zero only in a neutral compound.
Let N be x. Four hydrogens give +4, and the sum must equal the ion charge, +1: x + 4 = +1, so x = −3 (the same as in NH3). +3 has the wrong sign: nitrogen is more electronegative than hydrogen, so it is the negative partner. −4 sets the sum to zero, forgetting the +1 charge. +5 is nitrogen in the nitrate ion, NO3−.
Na is +1 and each O is −2, so for the neutral compound 1 + x + 3(−2) = 0, giving Cl = +5: sodium chlorate(V). The Roman numeral names the oxidation number of the element that can vary, just as in iron(III) chloride, FeCl3, where Fe is +3. (VI) comes from ignoring sodium (or the −1 charge on the chlorate ion) and setting x − 6 = 0. (VII) is chlorine in ClO4−, chlorate(VII). (I) is chlorine in ClO−, chlorate(I), found in bleach.
Zinc loses two electrons (0 to +2, oxidation) and copper(II) ions gain them (+2 to 0, reduction). The oxidising agent is the species that accepts the electrons and so causes the oxidation: Cu2+, which is itself reduced. Zn is the species oxidised, which makes it the reducing agent, not the oxidising agent; this is the classic mix-up. Zn2+ and Cu are products; the agents are named from the reactants. Cu is the reduced product, not the agent that did the oxidising.
With two S atoms: 2x + 3(−2) = −2, so 2x = +4 and x = +2 (an average over the two sulfur atoms). Compare SO42−, where x − 8 = −2 gives +6. +3 sets the sum to zero, forgetting the 2− charge (2x = 6). +4 is the total for both sulfur atoms, forgetting to divide by two. −2 has the right size but the wrong sign; sulfur is the less electronegative partner of oxygen.
Chlorine is 0 in Cl2, −1 in NaCl and +1 in NaClO (1 + x − 2 = 0). The same element is both reduced and oxidised in one reaction, which is disproportionation. Saying chlorine is only reduced misses the +1 in NaClO, and Na, O and H do not change, so NaOH is not a reducing agent. Sodium is +1 throughout, as it is already an ion in NaOH. Oxygen is −2 in NaOH, NaClO and H2O, so it is not oxidised.
Nitrogen is +5 in NO3− (x − 6 = −1) and +2 in NO (x − 2 = 0). A decrease of 3 in oxidation number means each N atom gains 3 electrons, so the half-equation is NO3− + 4H+ + 3e− → NO + 2H2O. 2 is the oxidation number of N in NO, the end point, not the change. 5 is the starting oxidation number in NO3−. 7 adds the two values together instead of taking the difference.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.