The shape of the lesson
By the end of the lesson, learners can
- Which ion is discharged?: Electrolyse molten and aqueous electrolytes in the simulator.
- Q = It and the Faraday: Change the current and time in the electrolysis simulator and watch how much copper is deposited.
- Q = It with t in seconds.
- n(e−) = Q ÷ F, F = 96 500 C mol−1.
- Half-equation gives electrons per atom or molecule.
- Cathode: more positive E⊖ reduced; anode: less positive E⊖ oxidised, unless concentration wins.
One picture every idea returns to
In electrolysis the cathode (negative) reduces and the anode (positive) oxidises. In aqueous solution the ions compete with water: at the cathode the species with the more positive E⊖ is reduced (Cu2+ and Ag+ before H2O; H2 forms instead of Na or K), and at the anode the species with the less positive E⊖ is oxidised, unless a high concentration of Cl− makes chlorine form instead of oxygen. The charge passed is Q = It, and one mole of electrons carries the Faraday constant, F = Le = 96 500 C mol−1. So 2.00 A for 965 s passes 1930 C = 0.0200 mol e−, depositing 0.0100 mol (0.635 g) of copper. Measuring the charge and the mass deposited gives L = F ÷ e.
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, Count the electrons, weigh the metal.. Electrolysis is a reaction driven by electricity, and every electron can be counted. 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 ion is discharged? | Electrolyse molten and aqueous electrolytes in the simulator. Project tab 1 of the lab (Electrode potentials and electrolysis lab, electrolysis cell) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Q = It and the Faraday | Change the current and time in the electrolysis simulator and watch how much copper is deposited. Project tab 2 of the lab (Electrode potentials and electrolysis lab, Q = It and the Faraday) and let a student drive it while the class predicts. | Rally Coach |
| 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, 18 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 Q = It and the Faraday, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| Time must be in seconds (minutes times 60). | Use the half-equation to convert moles of electrons into moles of product: 2 e − per Cu, 3 e − per Al, 1 e − per Ag, 4 e − per O 2. In aqueous solution, reactive metals such as Na and K are never deposited: hydrogen forms instead. Dilute chloride solutions give mainly oxygen at the anode, not chlorine. |
Electrolysis and the Faraday constant: worksheet
6 questions, 18 marks. Written in the style of the papers; no past paper question is reproduced.
A current of 1.50 A is passed through aqueous copper(II) sulfate for 1.00 hour. Calculate the mass of copper deposited on the cathode.
Concentrated aqueous sodium chloride is electrolysed with a current of 3.00 A for 20.0 minutes. Calculate the volume of chlorine produced at room conditions (24.0 dm3 mol−1).
Aluminium is extracted by electrolysis of Al3+. How long, in seconds, must a current of 10.0 A flow to deposit 2.70 g of aluminium?
Explain, using E⊖ values, the products at each electrode when dilute aqueous sodium chloride is electrolysed with inert electrodes, and how they change when the solution is concentrated.
A student passes 0.500 A for 40.0 minutes through copper(II) sulfate solution using copper electrodes. The cathode gains 0.395 g. Using e = 1.60 × 10−19 C, calculate a value for the Avogadro constant and state two measurements that must be made accurately.
Write the half-equations for the reactions at each electrode when aqueous copper(II) sulfate is electrolysed using inert electrodes.
Answer key and marking notes
- Q = It = 1.50 × 3600 = 5400 C.
- n(e−) = 5400 ÷ 96 500 = 0.05596 mol, so n(Cu) = 0.02798 mol.
- Mass = 0.02798 × 63.5 = 1.78 g.
- Q = 3.00 × 1200 = 3600 C; n(e−) = 3600 ÷ 96 500 = 0.03731 mol.
- 2Cl− → Cl2 + 2e−, so n(Cl2) = 0.01865 mol.
- Volume = 0.01865 × 24.0 = 0.448 dm3.
- n(Al) = 2.70 ÷ 27.0 = 0.100 mol; Al3+ + 3e− → Al, so n(e−) = 0.300 mol.
- Q = 0.300 × 96 500 = 28 950 C.
- t = Q ÷ I = 28 950 ÷ 10.0 = 2.90 × 103 s (2895 s).
- Cathode: E⊖(Na+/Na) = −2.71 V is far more negative than for water, so water is reduced: 2H2O + 2e− → H2 + 2OH−.
- Anode (dilute): E⊖(O2/H2O) = +1.23 V is less positive than E⊖(Cl2/Cl−) = +1.36 V, so water is oxidised to oxygen.
- Concentrated: the high [Cl−] makes Cl− oxidation easier, so chlorine is the main anode product; hydrogen still forms at the cathode.
- Q = 0.500 × 2400 = 1200 C; number of electrons = 1200 ÷ 1.60 × 10−19 = 7.50 × 1021.
- n(Cu) = 0.395 ÷ 63.5 = 0.006220 mol, so n(e−) = 0.01244 mol.
- L = 7.50 × 1021 ÷ 0.01244 = 6.03 × 1023 mol−1.
- Measure the mass of the cathode before and after (washed and dried), and keep the current constant while timing accurately.
- Cathode: Cu2+ + 2e− → Cu.
- Anode: 2H2O → O2 + 4H+ + 4e− (or 4OH− → O2 + 2H2O + 4e−).
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.
A current of 0.250 A is passed through aqueous silver nitrate for 386 s. What mass of silver is deposited on the cathode? (Ar Ag = 107.9; F = 96 500 C mol−1)
- A0.108 g
- B0.0540 g
- C0.216 g
- D6.47 g
Aqueous copper(II) sulfate is electrolysed using inert platinum electrodes. What forms at each electrode?
- Acopper at the cathode, oxygen at the anode
- Bhydrogen at the cathode, oxygen at the anode
- Ccopper at the cathode, sulfur dioxide at the anode
- Doxygen at the cathode, copper at the anode
Which expression gives the Faraday constant, F?
- AF = L × e, the Avogadro constant multiplied by the charge on one electron
- BF = L ÷ e
- CF = Q × t
- DF = e ÷ L
Concentrated aqueous sodium chloride is electrolysed with inert electrodes. Why is chlorine, not oxygen, the main product at the anode, even though E⊖(Cl2/Cl−) = +1.36 V is more positive than E⊖(O2/H2O) = +1.23 V?
- Athe high concentration of chloride ions makes them more easily oxidised than water
- Bchloride ions are always discharged before water
- Coxygen reacts with sodium at the anode
- Dthe more positive E⊖ means chloride is the stronger reducing agent
The two cells in the diagram are connected in series. In cell A, 1.08 g of silver is deposited. What mass of copper is deposited in cell B? (Ar Ag = 107.9, Cu = 63.5)
- A0.318 g
- B0.636 g
- C0.159 g
- D1.08 g
To find the Avogadro constant, a current of 0.400 A is passed for 30.0 minutes through copper(II) sulfate solution with copper electrodes. The cathode gains 0.237 g. Using e = 1.60 × 10−19 C, what value of L is obtained?
- A6.03 × 1023 mol−1
- B3.01 × 1023 mol−1
- C1.21 × 1024 mol−1
- D1.00 × 1022 mol−1
Dilute sulfuric acid is electrolysed with a current of 0.500 A. How long does it take to collect 0.120 dm3 of oxygen at room conditions (24.0 dm3 mol−1)?
- A3860 s
- B1930 s
- C965 s
- D7720 s
A current of 0.500 A for 1930 s deposits 0.173 g of chromium (Ar 52.0) from a solution of a chromium salt. What is the charge on the chromium ion?
- A3+
- B2+
- C6+
- D1+
Quiz key
Q = It = 0.250 × 386 = 96.5 C, so n(e−) = 96.5 ÷ 96 500 = 1.00 × 10−3 mol. Each Ag+ needs one electron, so n(Ag) = 1.00 × 10−3 mol and the mass is 0.108 g. 0.0540 g wrongly uses two electrons per silver atom, 0.216 g wrongly doubles the moles, and 6.47 g treats 386 as minutes and multiplies by 60.
At the cathode Cu2+ (+0.34 V) is more easily reduced than water, so copper is deposited: Cu2+ + 2e− → Cu. At the anode, sulfate is not oxidised, so water is: 2H2O → O2 + 4H+ + 4e−. Hydrogen would only form at the cathode if the metal ion were from a reactive metal such as Na+. Sulfur dioxide is a reduction product of sulfate, so it could never form at the anode. Reduction always happens at the cathode, so oxygen (an oxidation product) cannot form there.
One mole contains L electrons, each carrying e coulombs, so F = Le = 6.02 × 1023 × 1.60 × 10−19 = 9.63 × 104 C mol−1 (the Data Booklet gives 96 500). Dividing L by e, or e by L, gives meaningless numbers of the wrong size. Q × t is not a physical quantity here: charge is current times time, Q = It.
The two values differ by only 0.13 V, so concentration decides. A high [Cl−] makes Cl− easier to oxidise, so Cl2 forms; in dilute NaCl(aq) mainly oxygen forms, so chloride is not always discharged first. No sodium is present at the anode; Na+ is not reduced at all in aqueous solution. A more positive E⊖ means Cl2 is the stronger oxidising agent, so Cl− is a weaker reducing agent than water under standard conditions: the opposite of the claim.
n(Ag) = 1.08 ÷ 107.9 = 0.01001 mol, so 0.01001 mol of electrons passed. Copper needs two electrons per atom, so n(Cu) = 0.005005 mol and the mass = 0.005005 × 63.5 = 0.318 g. 0.636 g assumes one electron per Cu, 0.159 g halves twice, and 1.08 g assumes equal masses rather than equal charge.
Q = 0.400 × 1800 = 720 C, so the number of electrons = 720 ÷ 1.60 × 10−19 = 4.50 × 1021. n(Cu) = 0.237 ÷ 63.5 = 0.003732 mol, so n(e−) = 0.007465 mol. L = 4.50 × 1021 ÷ 0.007465 = 6.03 × 1023 mol−1. 3.01 × 1023 divides by 2 × 0.007465 (four electrons per Cu); 1.21 × 1024 divides by moles of Cu instead of electrons; 1.00 × 1022 forgets to convert 30.0 minutes to seconds and is far too small.
n(O2) = 0.120 ÷ 24.0 = 0.00500 mol. 2H2O → O2 + 4H+ + 4e−, so n(e−) = 0.0200 mol and Q = 0.0200 × 96 500 = 1930 C. t = Q ÷ I = 1930 ÷ 0.500 = 3860 s. 1930 s uses 2 electrons per O2 (the hydrogen ratio), 965 s uses 1, and 7720 s uses 8. 1930 C is the charge, not the time.
Q = 0.500 × 1930 = 965 C, so n(e−) = 965 ÷ 96 500 = 0.0100 mol. n(Cr) = 0.173 ÷ 52.0 = 0.00333 mol. Ratio = 0.0100 ÷ 0.00333 = 3.0, so three electrons are needed per atom: Cr3+ + 3e− → Cr. A 2+ ion would give 0.00500 mol Cr (0.260 g), 1+ would give 0.520 g. 6+ is the oxidation state of Cr in dichromate, an anion that is not reduced to the metal at the cathode in this way, and would give only 0.0867 g.
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.