The shape of the lesson
By the end of the lesson, learners can
- Where the electrons go: Watch electrons pass from zinc to copper(II) ions in the simulator, then split the reaction into its two halves.
- Half-equations in acid: Build the manganate(VII) and dichromate(VI) half-equations in the simulator one step at a time, following Fig. 1.
- Combining halves: Find the lowest common multiple of the electrons, scale each half-equation, add and cancel.
- O with H2O, H with H+, charge with e−.
- Electrons = change in oxidation number.
- Match electrons lost and gained, then add.
- Check atoms and charge; no e− left over.
One picture every idea returns to
A half-equation shows one species gaining or losing electrons. In acidic solution balance it in four steps: the element changing oxidation number, then O with H2O, then H with H+, then charge with electrons. The number of electrons equals the change in oxidation number. To combine two half-equations, multiply them so the electrons lost equal the electrons gained, add them, and cancel anything that appears on both sides. No electrons may remain in the final equation.
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, Balance the electrons.. A redox equation is two half-equations stitched together: one losing electrons, one gaining them. 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: Where the electrons go | Watch electrons pass from zinc to copper(II) ions in the simulator, then split the reaction into its two halves. Project tab 2 of the lab (Electrochemistry lab, electron transfer) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Half-equations in acid | Build the manganate(VII) and dichromate(VI) half-equations in the simulator one step at a time, following Fig. 1. Project tab 3 of the lab (Electrochemistry lab, half-equations) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Combining halves | Find the lowest common multiple of the electrons, scale each half-equation, add and cancel. | Cooperative Concept Mapping |
| 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 Combining halves, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| Electrons go on the left in a reduction and on the right in an oxidation, and they must all cancel in the full equation. | A final equation that still shows e − has not been combined properly. Always finish by checking that the charges balance as well as the atoms. |
Redox equations and half-equations: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Write the half-equation for the reduction of manganate(VII) ions to manganese(II) ions in acidic solution.
Write the half-equation for the oxidation of sulfite ions, SO32−, to sulfate ions in water.
Combine the manganate(VII) half-equation with H2O2 → O2 + 2H+ + 2e− to give the overall equation.
Write the overall equation for the reaction of acidified dichromate(VI) with iron(II) ions, and give the mole ratio of Cr2O72− to Fe2+.
25.0 cm3 of an iron(II) solution needs 22.50 cm3 of 0.0200 mol dm−3 KMnO4 in acid. Calculate the concentration of Fe2+.
Use oxidation number changes to balance: Cu + HNO3 → Cu(NO3)2 + NO2 + H2O.
Answer key and marking notes
- MnO4− + 8H+ + 5e− → Mn2+ + 4H2O.
- Check the charge: −1 + 8 − 5 = +2 on both sides.
- SO32− + H2O → SO42− + 2H+ + 2e−.
- S goes +4 → +6, losing two electrons; charge −2 on both sides.
- LCM of 5 and 2 is 10: double the manganate half, multiply the peroxide half by 5.
- 2MnO4− + 16H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2 + 10H+.
- Cancel H+: 2MnO4− + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2.
- Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O; Fe2+ → Fe3+ + e− (× 6).
- Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O.
- Ratio 1 : 6.
- n(MnO4−) = 0.0200 × 0.02250 = 4.50 × 10−4 mol.
- n(Fe2+) = 5 × 4.50 × 10−4 = 2.25 × 10−3 mol.
- c = 2.25 × 10−3 ÷ 0.0250 = 0.0900 mol dm−3.
- Cu goes 0 → +2 (loses 2); N goes +5 → +4 in NO2 (gains 1), so 2NO2 per Cu.
- Two more HNO3 supply the nitrate ions in Cu(NO3)2: 4HNO3 in total.
- Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O.
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.
The flowchart shows the steps for balancing a half-equation in acidic solution. What is step 3?
- Abalance H atoms by adding H+ ions
- Bbalance H atoms by adding H2 molecules
- Cbalance O atoms by adding O2 molecules
- Dbalance H atoms by adding OH− ions
Which is the correctly balanced half-equation for the reduction of MnO4− to Mn2+ in acidic solution?
- AMnO4− + 8H+ + 5e− → Mn2+ + 4H2O
- BMnO4− + 8H+ → Mn2+ + 4H2O + 5e−
- CMnO4− + 4H+ + 5e− → Mn2+ + 4H2O
- DMnO4− + 8H+ + 7e− → Mn2+ + 4H2O
In the reaction 2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq), which statement is correct?
- AI− is oxidised and Fe3+ is the oxidising agent
- BFe3+ is oxidised and I− is the oxidising agent
- CI− is oxidised and I− is the oxidising agent
- DFe3+ is oxidised, because its charge changes
Which statement about the half-equation Fe2+ → Fe3+ + e− is correct?
- Ait is an oxidation, because iron loses an electron and its oxidation number rises from +2 to +3
- Bit is a reduction, because the charge on the iron ion increases
- Cit is a reduction, because an electron appears in the equation
- Dit is an oxidation, because iron gains an electron
Combine the half-equations MnO4− + 8H+ + 5e− → Mn2+ + 4H2O and Fe2+ → Fe3+ + e−. Which is the overall ionic equation?
- AMnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
- BMnO4− + 8H+ + Fe2+ → Mn2+ + Fe3+ + 4H2O
- CMnO4− + 8H+ + 5Fe2+ + 5e− → Mn2+ + 5Fe3+ + 4H2O + 5e−
- DMnO4− + 8H+ + 5Fe3+ → Mn2+ + 5Fe2+ + 4H2O
Complete the half-equation Cr2O72− + 14H+ + ne− → 2Cr3+ + 7H2O. What is n?
- A6
- B3
- C12
- D14
Copper reacts with dilute nitric acid to form copper(II) nitrate, nitrogen monoxide and water. Using oxidation number changes, what is the mole ratio Cu : NO in the balanced equation?
- A3 : 2
- B2 : 3
- C1 : 1
- D3 : 8
25.0 cm3 of acidified iron(II) sulfate solution, concentration 0.0800 mol dm−3, is titrated with 0.0200 mol dm−3 potassium manganate(VII). What volume of KMnO4 is needed to reach the end point?
- A20.0 cm3
- B100 cm3
- C500 cm3
- D12.5 cm3
Quiz key
Adding H2O in step 2 puts hydrogen on that side, so in acidic solution the hydrogen is balanced with H+ ions on the other side. Only then is the charge balanced with electrons. H2 is not present in the reaction mixture and would change the chemistry. Oxygen has already been balanced with water; O2 is not used because it is not a species in the solution. OH− ions are used for reactions in alkaline solution, not acidic.
Four O atoms need 4H2O, whose 8 H atoms need 8H+. Charge on the left before electrons: −1 + 8 = +7; on the right: +2. Adding 5e− to the left balances it, matching the fall in oxidation number from +7 to +2. Putting the electrons on the right would describe oxidation, but manganese is reduced (and the charge would not balance: +7 vs −3). 4H+ leaves the hydrogen unbalanced (4 H on the left, 8 H on the right). 7e− uses the oxidation number of Mn instead of the change; the charges would be 0 and +2.
Iodine goes from −1 in I− to 0 in I2: an increase, so iodide ions are oxidised (they lose electrons). Iron goes from +3 to +2: a decrease, so Fe3+ is reduced, and the species that is reduced is the oxidising agent. Saying Fe3+ is oxidised gets the direction backwards (its oxidation number falls). I− is the species oxidised, so it is the reducing agent, not the oxidising agent. A change in charge alone does not make something oxidised; the charge on iron falls, which is reduction.
The electron is a product, so iron(II) loses an electron and its oxidation number increases from +2 to +3: that is oxidation. A rising charge on a cation is a sign of oxidation, not reduction. The presence of an electron in a half-equation says nothing on its own; what matters is whether it is lost (product side) or gained (reactant side). Iron does not gain an electron here; gaining one would be reduction.
Multiply the iron half-equation by 5 so that 5e− are released, matching the 5e− taken in by MnO4−. Adding and cancelling the electrons gives MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O (charge +17 on each side). Using one Fe2+ does not match the electrons, and the charge (+9 vs +5) does not balance. Leaving electrons on both sides is not an overall equation; they must cancel. Swapping Fe2+ and Fe3+ shows iron being reduced, but both iron and manganese cannot be reduced in the same reaction.
In Cr2O72−, 2x + 7(−2) = −2 gives Cr = +6; in Cr3+ it is +3. Each of the two Cr atoms gains 3 electrons, so n = 6. Check the charge: −2 + 14 − 6 = +6 on the left and 2 × (+3) = +6 on the right. 3 is the change for only one chromium atom. 12 is the total oxidation number of both Cr atoms (2 × +6), not the change. 14 copies the number of H+ ions; the charges would then be −2 and +6.
Each Cu loses 2 electrons (0 to +2) and each N that becomes NO gains 3 (+5 to +2). The lowest common multiple is 6, so 3 Cu balance 2 NO, giving 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O. 2 : 3 puts the electron numbers the wrong way round (it would transfer 4 electrons from copper but need 9). 1 : 1 does not match electrons at all (2 lost, 3 gained). 3 : 8 is the ratio of Cu to HNO3: most of the nitric acid ends up as nitrate ions, which are not reduced.
n(Fe2+) = 0.0250 × 0.0800 = 2.00 × 10−3 mol. The ratio MnO4− : Fe2+ is 1 : 5 (5e− gained by Mn, 1e− lost by each Fe), so n(MnO4−) = 4.00 × 10−4 mol. V = 4.00 × 10−4 / 0.0200 = 0.0200 dm3 = 20.0 cm3. 100 cm3 assumes a 1 : 1 ratio. 500 cm3 multiplies by 5 instead of dividing, as if one Fe2+ reacted with five MnO4−. 12.5 cm3 uses the 8 H+ in the equation as the ratio (dividing by 8), but H+ is not the species being titrated.
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.