The shape of the lesson
By the end of the lesson, learners can
- Which way do the electrons go?: Pair two half-cells in the simulator and watch which electrode gives electrons to which.
- Feasible, but will it happen?: For a proposed reaction, put the species being reduced as the positive electrode.
- Ecell = E(positive) − E(negative) = Eright − Eleft.
- Electrons flow in the wire from negative to positive.
- Feasible if Ecell > 0, but rate and conditions can intervene.
- ΔG = −nFE, then ÷ 1000 for kJ.
One picture every idea returns to
When two half-cells are joined, the one with the more positive E⊖ is the positive electrode: reduction happens there. The other is oxidised, and electrons flow through the wire from the negative to the positive electrode. The standard cell potential is E⊖cell = E⊖(positive) − E⊖(negative), e.g. Zn/Cu: +0.34 − (−0.76) = +1.10 V. A reaction is feasible when E⊖cell for it, as written, is positive. E values say nothing about rate (a feasible reaction may have a high activation energy) and apply only under standard conditions. The cell potential links to free energy: ΔG⊖ = −nFE⊖cell, so a positive Ecell gives a negative ΔG.
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, Positive E cell means it can go.. Join two half-cells and the more positive one pulls electrons from the other. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Which way do the electrons go? | Pair two half-cells in the simulator and watch which electrode gives electrons to which. Project tab 4 of the lab (Electrode potentials and electrolysis lab, build a cell) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Feasible, but will it happen? | For a proposed reaction, put the species being reduced as the positive electrode. Project tab 5 of the lab (Electrode potentials and electrolysis lab, feasible or not?) and let a student drive it while the class predicts. | Jigsaw |
| 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 Feasible, but will it happen?, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 |
|---|---|
| E ⊖ cell = E ⊖ (positive) − E ⊖ (negative): never add them, and never multiply E ⊖ by the coefficients used to balance electrons. | Electrons flow in the external wire, from negative to positive; ions, not electrons, carry the current through the salt bridge. In ΔG = −nFE, n is moles of electrons, and the answer in J must be converted to kJ. A feasible reaction may still be too slow to see. |
Standard cell potentials and feasibility: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
A cell is made from a Zn2+/Zn half-cell and an Ag+/Ag half-cell under standard conditions. Calculate E⊖cell, write the overall equation and state the direction of electron flow.
Calculate ΔG⊖ for the reaction Zn + Cu2+ → Zn2+ + Cu, given E⊖cell = +1.10 V.
Use E⊖ values to predict whether acidified dichromate(VI) will oxidise (a) iron(II) ions and (b) chloride ions under standard conditions.
Give two reasons why a reaction predicted to be feasible from E⊖ values may not be observed.
Will bromine oxidise iron(II) ions? Use E⊖(Br2/Br−) = +1.07 V and E⊖(Fe3+/Fe2+) = +0.77 V, and calculate ΔG⊖ for the reaction as written with 2 mol of Fe2+.
Explain the role of the salt bridge in an electrochemical cell, and what carries the charge through it.
Answer key and marking notes
- E⊖cell = +0.80 − (−0.76) = +1.56 V.
- Zn + 2Ag+ → Zn2+ + 2Ag.
- Electrons flow through the external wire from the zinc to the silver electrode.
- n = 2 mol of electrons transferred per mole of reaction.
- ΔG⊖ = −nFE = −2 × 96 500 × 1.10 = −212 300 J mol−1.
- ΔG⊖ = −212 kJ mol−1; negative, so feasible.
- (a) E⊖cell = 1.33 − 0.77 = +0.56 V, positive, so Fe2+ is oxidised to Fe3+.
- (b) E⊖cell = 1.33 − 1.36 = −0.03 V, negative, so Cl− is not oxidised under standard conditions.
- So dichromate titrations of Fe2+ can be done in HCl, unlike manganate(VII).
- The activation energy may be high, so the rate is too slow to observe (E⊖ gives no information about kinetics).
- Conditions may not be standard: different concentrations or temperature change the electrode potentials, especially if E⊖cell is small.
- Br2 + 2Fe2+ → 2Br− + 2Fe3+.
- E⊖cell = 1.07 − 0.77 = +0.30 V, positive, so feasible.
- ΔG⊖ = −2 × 96 500 × 0.30 = −57.9 kJ mol−1.
- It completes the circuit and keeps each solution electrically neutral.
- Ions (e.g. K+ and NO3−) move through it; electrons only travel through the external wire.
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 E⊖cell for a cell made from Mg2+/Mg (E⊖ = −2.38 V) and Cu2+/Cu (E⊖ = +0.34 V)?
- A+2.72 V
- B+2.04 V
- C−2.04 V
- D+1.36 V
In the cell shown, electrode P is zinc and electrode Q is copper. Which statement is correct when the electrodes are joined by a wire?
- Aelectrons flow through the wire from P to Q, and Q gains mass
- Belectrons flow through the wire from Q to P, and P gains mass
- Celectrons flow through the salt bridge from P to Q
- Delectrons flow from P to Q, and P gains mass
Will copper metal reduce iron(III) ions to iron(II) ions under standard conditions? (E⊖: Fe3+/Fe2+ +0.77 V, Cu2+/Cu +0.34 V)
- Ayes: E⊖cell = +0.43 V, which is positive
- Bno: E⊖cell = −0.43 V, which is negative
- Cyes: E⊖cell = +1.11 V, which is positive
- Dno: copper cannot reduce a positive ion
What is ΔG⊖ for Cu + 2Ag+ → Cu2+ + 2Ag, given E⊖cell = +0.46 V? (F = 96 500 C mol−1)
- A−88.8 kJ mol−1
- B−44.4 kJ mol−1
- C+88.8 kJ mol−1
- D−88 780 kJ mol−1
Which reaction is NOT feasible under standard conditions?
- AI2 + 2Fe2+ → 2I− + 2Fe3+
- BCl2 + 2I− → 2Cl− + I2
- CZn + Cu2+ → Zn2+ + Cu
- D2Fe3+ + 2I− → 2Fe2+ + I2
Calculate ΔG⊖ for MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+. (E⊖: MnO4−/Mn2+ +1.52 V, Fe3+/Fe2+ +0.77 V)
- A−362 kJ mol−1
- B−72.4 kJ mol−1
- C+362 kJ mol−1
- D−1.10 × 103 kJ mol−1
For dichromate(VI) oxidising chloride ions, E⊖cell = 1.33 − 1.36 = −0.03 V. Yet hot concentrated hydrochloric acid with potassium dichromate(VI) gives chlorine. What is the best explanation?
- Aconditions are not standard: high [H+] and [Cl−] shift both electrode potentials, so E⊖cell becomes positive
- Ba negative E⊖cell means the reaction is fast
- Cheating always makes non-feasible reactions feasible because the activation energy is overcome
- Dthe Data Booklet values are wrong for chloride
A cell is represented as Pt | Fe2+(aq), Fe3+(aq) || MnO4−(aq), Mn2+(aq), H+(aq) | Pt. What are E⊖cell and the overall reaction?
- A+0.75 V; MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
- B−0.75 V; Mn2+ + 4H2O + 5Fe3+ → MnO4− + 8H+ + 5Fe2+
- C+2.29 V; MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
- D+0.75 V; MnO4− + 8H+ + Fe2+ → Mn2+ + 4H2O + Fe3+
Quiz key
Copper is more positive, so it is the positive electrode: E⊖cell = +0.34 − (−2.38) = +2.72 V. +2.04 V and −2.04 V come from adding the two values (or subtracting the wrong way round with a sign slip). +1.36 V halves the correct answer, as if dividing by the two electrons transferred; E⊖cell is never divided by n.
Zinc (−0.76 V) is the negative electrode: Zn → Zn2+ + 2e−, so P loses mass. Electrons travel through the external wire to copper (+0.34 V), where Cu2+ + 2e− → Cu deposits copper, so Q gains mass. The reverse direction would need copper to be the more negative couple. Electrons never pass through the salt bridge, which carries ions. P cannot gain mass because zinc is being oxidised to Zn2+(aq).
For Cu + 2Fe3+ → Cu2+ + 2Fe2+, Fe3+ is reduced (positive electrode) and Cu is oxidised: E⊖cell = 0.77 − 0.34 = +0.43 V, so the reaction is feasible. This is how iron(III) chloride etches copper circuit boards. −0.43 V subtracts the wrong way. +1.11 V adds the two values. Copper certainly can reduce ions whose couples are more positive than its own.
Two electrons are transferred (Cu → Cu2+ + 2e−), so ΔG⊖ = −2 × 96 500 × 0.46 = −88 780 J mol−1 = −88.8 kJ mol−1. −44.4 uses n = 1. +88.8 drops the minus sign: a positive E⊖cell must give a negative ΔG⊖. −88 780 is the value in J, wrongly labelled as kJ.
Iodine oxidising Fe2+: E⊖cell = 0.54 − 0.77 = −0.23 V, negative, so not feasible; in fact the reverse reaction (Fe3+ oxidising I−, +0.23 V) happens. Chlorine oxidising iodide: 1.36 − 0.54 = +0.82 V. Zinc reducing copper(II): +1.10 V. All three of these are positive and feasible.
E⊖cell = 1.52 − 0.77 = +0.75 V. Five electrons are transferred (MnO4− + 8H+ + 5e− → Mn2+ + 4H2O), so ΔG⊖ = −5 × 96 500 × 0.75 = −361 875 J = −362 kJ mol−1. −72.4 uses n = 1. +362 loses the minus sign. −1.10 × 103 adds the two potentials (2.29 V) instead of subtracting them.
With a small E⊖cell (only 0.03 V), non-standard conditions can easily change the sign. High [H+] makes E⊖(Cr2O72−/Cr3+) more positive, and high [Cl−] makes E(Cl2/Cl−) less positive, so E⊖cell becomes positive. The sign of E⊖cell says nothing about rate. Supplying activation energy speeds a feasible reaction but cannot make a thermodynamically unfavourable one go. The Data Booklet values are correct for standard conditions.
E⊖cell = Eright − Eleft = 1.52 − 0.77 = +0.75 V. Manganate(VII) is reduced on the right (5 electrons) and Fe2+ is oxidised on the left (1 electron each), so 5Fe2+ are needed per MnO4−. The reverse reaction would have −0.75 V and is not what the diagram represents as a working cell. +2.29 V adds the potentials. The last equation is not charge-balanced: one Fe2+ cannot supply five electrons.
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.