SACE Board SACE 2022-style practice: Molten aluminium oxide is electrolysed to extract aluminium. A current of 5.00 × 10^4\ A flows for 1.00\ hour. (a) Write the cathode half-equation. (b) Calculate the mass of aluminium produced. (F = 96500\ C mol^-1; M(Al) = 27.0\ g mol^-1.)

(a) Cathode (reduction): Al^3+ + 3e^- → Al. (1 mark) (b) Charge Q = It = 5.00 × 10^4 × 3600 = 1.80 × 10^8\ C. (1 mark) Moles of electrons = QF = 1.80 × 10^896500 = 1.865 × 10^3\ mol. (1 mark) From Al^3+ + 3e^-, n(Al) = 1.865 × 10^33 = 621.6\ mol, so m = nM = 621.6 × 27.0 = 1.68 × 10^4\ g = 16.8\ kg. (2 marks)

SACE Board SACE 2020-style practice: In the electrolysis of aqueous copper(II) sulfate with inert electrodes, a current of 1.50\ A is passed for 30.0\ minutes. (a) Write the cathode half-equation. (b) Calculate the mass of copper deposited. (F = 96500\ C mol^-1; M(Cu) = 63.5\ g mol^-1.)

(a) Cathode: Cu^2+ + 2e^- → Cu. (1 mark) (b) Q = It = 1.50 × (30.0 × 60) = 1.50 × 1800 = 2700\ C. (1 mark) n(e^-) = 270096500 = 2.798 × 10^-2\ mol; n(Cu) = 2.798 × 10^-22 = 1.399 × 10^-2\ mol. (1 mark) m(Cu) = nM = 1.399 × 10^-2 × 63.5 = 0.889\ g. (1 mark)

§-Syllabus dot point

SAChemistrySyllabus dot point

How does electrical energy drive non-spontaneous redox reactions in electrolysis?

Describe electrolytic cells, predict electrode products, and apply Faraday's relationships to calculate amounts in electrolysis.

How electrolytic cells use electrical energy to drive non-spontaneous redox reactions, predicting electrode products, and applying the charge and Faraday relationships, with fully worked SACE-style electrolysis calculations.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Lead worked calculation
How an electrolytic cell works
Predicting electrode products
The Faraday relationships
Applications
Why it matters for managing resources

What this dot point is asking

SACE expects you to describe an electrolytic cell, predict the products at each electrode, and carry out quantitative Faraday calculations.

Lead worked calculation

Mass of silver electroplated

A spoon is electroplated with silver using a $0.800\ \text{A}$ current for $25.0\ \text{minutes}$ . The cathode reaction is $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ . Find the mass of silver deposited. ( $F = 96500\ \text{C mol}^{-1}$ ; $M(\text{Ag}) = 107.9\ \text{g mol}^{-1}$ .)

Calculate the charge

Q = It = 0.800 \times (25.0 \times 60) = 0.800 \times 1500 = 1200\ \text{C}

Convert to moles of electrons

n(e^-) = \frac{Q}{F} = \frac{1200}{96500} = 1.244 \times 10^{-2}\ \text{mol}

Use the half-equation ratio

$\text{Ag}^+ + e^- \rightarrow \text{Ag}$ needs one electron per silver atom, so $n(\text{Ag}) = 1.244 \times 10^{-2}\ \text{mol}$ .

Convert to mass

m(\text{Ag}) = nM = 1.244 \times 10^{-2} \times 107.9 = 1.34\ \text{g}

So $1.34\ \text{g}$ of silver is plated. The electron-to-product ratio from the half-equation is the step students most often miss.

How an electrolytic cell works

The electrode roles (anode oxidation, cathode reduction) are the same as a galvanic cell, but the polarity is reversed: in electrolysis the cathode is negative and the anode positive.

Predicting electrode products

In a molten electrolyte the only ions present are from the compound, so the metal cation is reduced at the cathode and the non-metal anion oxidised at the anode (e.g. molten $\text{NaCl}$ gives sodium and chlorine). In aqueous solution, water competes:

At the cathode, either the metal ion or water is reduced; very reactive metal ions (group 1, 2, $\text{Al}^{3+}$ ) stay in solution and water is reduced to $\text{H}_2$ instead, while less reactive ions (e.g. $\text{Cu}^{2+}$ ) are deposited as the metal.
At the anode, either the anion or water is oxidised; concentrated halides give the halogen, while in dilute or oxoanion solutions water is oxidised to $\text{O}_2$ .

The Faraday relationships

The full chain is always: $Q = It$ , then $n(e^-) = Q/F$ , then apply the electrons-per-formula-unit ratio from the half-equation, then convert to mass or volume.

Applications

Electrolysis extracts very reactive metals that cannot be obtained by chemical reduction (e.g. aluminium from molten $\text{Al}_2\text{O}_3$ , sodium from molten $\text{NaCl}$ ), purifies copper, produces chlorine and sodium hydroxide, and electroplates objects with thin metal coatings. Each is a managed use of electrical energy to obtain a chemical resource.

Why it matters for managing resources

Electrolysis is how we obtain reactive metals, refine copper to high purity, and protect or decorate surfaces by electroplating. The Faraday calculation links electrical input directly to product output, letting engineers size the energy and cost of these resource-management processes.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20225 marksMolten aluminium oxide is electrolysed to extract aluminium. A current of

5.00 \times 10^{4}\ \text{A}

flows for

1.00\ \text{hour}

. (a) Write the cathode half-equation. (b) Calculate the mass of aluminium produced. (

F = 96500\ \text{C mol}^{-1}

;

M(\text{Al}) = 27.0\ \text{g mol}^{-1}

Show worked answer →

(a) Cathode (reduction): $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ . (1 mark)

(b) Charge $Q = It = 5.00 \times 10^{4} \times 3600 = 1.80 \times 10^{8}\ \text{C}$ . (1 mark)

Moles of electrons $= \dfrac{Q}{F} = \dfrac{1.80 \times 10^{8}}{96500} = 1.865 \times 10^{3}\ \text{mol}$ . (1 mark)

From $\text{Al}^{3+} + 3e^-$ , $n(\text{Al}) = \dfrac{1.865 \times 10^{3}}{3} = 621.6\ \text{mol}$ , so $m = nM = 621.6 \times 27.0 = 1.68 \times 10^{4}\ \text{g} = 16.8\ \text{kg}$ . (2 marks)

SACE 20204 marksIn the electrolysis of aqueous copper(II) sulfate with inert electrodes, a current of

1.50\ \text{A}

is passed for

30.0\ \text{minutes}

. (a) Write the cathode half-equation. (b) Calculate the mass of copper deposited. (

F = 96500\ \text{C mol}^{-1}

;

M(\text{Cu}) = 63.5\ \text{g mol}^{-1}

Show worked answer →

(a) Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ . (1 mark)

(b) $Q = It = 1.50 \times (30.0 \times 60) = 1.50 \times 1800 = 2700\ \text{C}$ . (1 mark)

$n(e^-) = \dfrac{2700}{96500} = 2.798 \times 10^{-2}\ \text{mol}$ ; $n(\text{Cu}) = \dfrac{2.798 \times 10^{-2}}{2} = 1.399 \times 10^{-2}\ \text{mol}$ . (1 mark)

$m(\text{Cu}) = nM = 1.399 \times 10^{-2} \times 63.5 = 0.889\ \text{g}$ . (1 mark)