S5 Chemistry | Metal extraction prep guide (for tutors) | reactivity series, blast furnace, electrolysis

2026-11-01

Tutor lesson-prep guides · Secondary Chemistry

Students can recite the reactivity series but freeze when asked why aluminium needs electrolysis. One sentence about grabbing electrons holds the whole chapter together.

Objectives — Students can: (1) use the reactivity series to justify three extraction routes; (2) write Fe2O3 + 3CO to 2Fe + 3CO2 and compute mass and percentage yield; (3) explain why aluminium oxide must be molten rather than aqueous and compute a theoretical mass from Q = It.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Mole and molar mass	n = m / M	M(Fe2O3) = 159.7
Redox	the metal is reduced, the reducing agent is oxidised	CO is the reducing agent
Electrolysis	metal at the cathode, oxygen at the anode	Al3+ + 3e- gives Al

02 Lesson flow (50 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Reactivity series	K Na Ca Mg Al Zn Fe Pb Cu Ag Au	earlier means harder to reduce
8-18 min	Three routes	electrolysis, carbon, native	position decides the route
18-30 min	Blast furnace calculation	250 g ore at 80%	theory 139.9 g Fe
30-40 min	Percentage yield	actual 123.1 g	yield 88%
40-47 min	Aluminium electrolysis	I = 180000 A for 3600 s	60.4 kg Al
47-50 min	Exit check	three questions collected	independent work

03 Teaching script (read it out)

Scene 1: Reactivity series: a queue for electrons

From strongest to weakest: K, Na, Ca, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au. The earlier the metal, the more readily it loses electrons, so its compounds are more stable and harder to reduce. That one sentence is the foundation: the most reactive need the strongest method, electrolysis; the middle ones use carbon; the least reactive occur native or decompose on heating.
Students should write the method beside the series in three blocks: K to Al electrolysis, Zn to Cu carbon reduction, Ag and Au found native. Mention that copper can be reduced by carbon but is often purified by electrolysis for conductivity.

Student task: order the twelve metals, then draw one line between Al and Zn and label each side.

Scene 2: Blast furnace: equation first, then moles

Key reactions: C + O2 gives CO2 for heat; CO2 + C gives 2CO for the reducing gas; Fe2O3 + 3CO gives 2Fe + 3CO2. Limestone CaCO3 decomposes to CaO and CO2, and CaO removes silica as slag.
Calculation: 250 g of ore at 80% gives 200 g of Fe2O3; n = 200 / 159.7 = 1.252 mol; each mole gives two moles of iron, so the theoretical mass is 2.505 times 55.85 = 139.9 g. The actual mass is 123.1 g, which is a yield of 88%. The CO required is 105 g, about 3.76 mol.
The commonest error is forgetting the factor of two, because Fe2O3 holds two iron atoms. The second is confusing yield with purity: yield is actual over theoretical.

Student task: work from the mass of Fe2O3 to the mass of iron, writing units at every step.

Scene 3: Aluminium: why not an aqueous electrolyte

Aluminium sits high in the series, so carbon cannot reduce it and electrolysis is the only route. Electrolysis of aqueous aluminium oxide would give hydrogen at the cathode, because water discharges more readily than Al3+, so the oxide must be molten. Cryolite is added to lower the melting point from about 2050 C to about 950 C and save energy.
Cathode: Al3+ + 3e- gives Al. Anode: 2O2- gives O2 + 4e-, and the carbon anodes burn away and need replacing. Calculation: Q = It = 180000 times 3600 = 6.48e+08 C; moles of electrons = Q / F = 6715 mol; three electrons per aluminium gives 2238 mol, that is 60.4 kg. This is a minimum, since current efficiency is about 90%. Compare with iron, where a furnace produces tonnes per hour: that gap explains the price of aluminium.

Student task: from Q = It, reach the mass of aluminium in three lines and explain the aqueous failure.

04 Question types and worked steps

Type	Example	Steps	Answer
Choose the route	MgO, Fe2O3, Ag2O	locate on the reactivity series	electrolysis, carbon, heating
Blast furnace mass	250 g ore, 80% Fe2O3	pure mass, moles, times two, times M	139.9 g
Yield	theory 139.9 g, actual 123.1 g	actual over theory	88%
Electrolysis mass	I = 180000 A, t = 3600 s	Q = It, n(e) = Q/F, divide by 3	60.4 kg

Sanity check (teach this): The theoretical 139.9 g must be less than the 250 g of ore, since iron is only part of it; an answer of 320 g is impossible. A yield of 88% must lie between 0 and 100 percent. 60.4 kg of aluminium from 180000 A for an hour is the right order of magnitude, roughly 20 kg per thousand amps per hour; hundreds of tonnes means the Faraday constant was dropped.

05 Independent practice (3 questions)

Question	Answer	What to watch
500 g of ore at 60% Fe2O3: theoretical iron?	about 209.8 g	multiply 0.6 then 2M(Fe)/M(Fe2O3)
Theory 209.8 g, actual 178.3 g: yield?	85.0%	actual over theory
What forms if Al2O3 is electrolysed in water?	hydrogen, not aluminium	water discharges first

06 Common mistakes and what to say

Mistake	Student writes	Say this
One iron per Fe2O3	using a 1:1 mole ratio	Fe2O3 holds two iron atoms, so double it.
Yield treated as impurity	saying 88% yield means 88% impurity	Yield is what you actually recovered over the maximum.
Aluminium from aqueous solution	writing Al from Al2O3(aq)	Water discharges first, so the oxide must be molten with cryolite.
Reducing agent role reversed	writing that CO is reduced to CO2	CO is oxidised; the iron is the species reduced.

07 Homework (paper, section and question numbers)

Secondary Chemistry chapter index (moles, acids, redox, electrolysis) — do moles before extraction
Secondary prep guides overview — all subjects

08 Exit check (last 5 minutes)

Question	Answer	Decision
Why electrolysis for aluminium but not iron?	aluminium is higher in the series, carbon cannot reduce it	wrong means redo act one
Mole ratio between Fe2O3 and Fe?	1 to 2	wrong means redo act two
Why add cryolite?	to lower the melting point and save energy	two wrong means back to act three

09 Tutor FAQ

Which year is this for?

S5, the metals and redox sections of the HKDSE chemistry syllabus.

How many equations?

Four: carbon burning, CO2 with C, Fe2O3 with CO, and the aluminium electrolysis.

Is there paired practice?

The homework links to the chemistry index, which also lists redox and electrolysis chapters.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.