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Electrolytic cells and electrolysis (SACE Stage 2 Chemistry)

Syllabus dot point

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

10 min read

Quick answer

An electrolytic cell uses an external power supply to drive a non-spontaneous redox reaction. Reduction still occurs at the cathode and oxidation at the anode, but the polarity is reversed from a galvanic cell: the cathode is negative and the anode positive. The amount of product is found from the charge passed: Q=ItQ = It, then moles of electrons =Q/F= Q/F, then the half-equation gives moles of product. This is the basis of metal extraction and electroplating.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. How an electrolytic cell works
  4. Predicting electrode products
  5. The Faraday relationships
  6. Applications
  7. Why it matters for managing resources
  8. Exam-style questions

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 A0.800\ \text{A} current for 25.0 minutes25.0\ \text{minutes}. The cathode reaction is Ag++e−→Ag\text{Ag}^+ + e^- \rightarrow \text{Ag}. Find the mass of silver deposited. (F=96500 C mol−1F = 96500\ \text{C mol}^{-1}; M(Ag)=107.9 g mol−1M(\text{Ag}) = 107.9\ \text{g mol}^{-1}.)

Calculate the charge

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

Convert to moles of electrons

n(e−)=QF=120096500=1.244×10−2 moln(e^-) = \frac{Q}{F} = \frac{1200}{96500} = 1.244 \times 10^{-2}\ \text{mol}

Use the half-equation ratio

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

Convert to mass

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

So 1.34 g1.34\ \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

Electrolytic cell

An electrolytic cell uses electrical energy from an external power source to force a non-spontaneous redox reaction to occur. The power supply pushes electrons to the cathode (negative in electrolysis), where reduction happens, and pulls electrons from the anode (positive), where oxidation happens. It is essentially a galvanic cell run in reverse, converting electrical energy into chemical energy.

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 NaCl\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, Al3+\text{Al}^{3+}) stay in solution and water is reduced to H2\text{H}_2 instead, while less reactive ions (e.g. Cu2+\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 O2\text{O}_2.

The Faraday relationships

Charge and the Faraday constant

Q=Itn(e−)=QFQ = It \qquad n(e^-) = \frac{Q}{F}

QQ is charge in coulombs (C\text{C}), II current in amperes (A\text{A}), tt time in seconds, and FF the Faraday constant (96500 C mol−196500\ \text{C mol}^{-1}, the charge of one mole of electrons). The half-equation then gives the mole ratio between electrons and product.

The full chain is always: Q=ItQ = It, then n(e−)=Q/Fn(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 Al2O3\text{Al}_2\text{O}_3, sodium from molten NaCl\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.

Forgetting the electrons-per-ion ratio or wrong time units

The most common SACE error is computing moles of electrons but forgetting to divide by the charge on the ion (e.g. treating Cu2+\text{Cu}^{2+} or Al3+\text{Al}^{3+} as needing one electron). Always use the half-equation ratio: 2e−2e^- per Cu\text{Cu}, 3e−3e^- per Al\text{Al}. A second error is leaving time in minutes or hours; convert it to seconds before Q=ItQ = It. A third is reversing the polarity logic: in electrolysis the cathode is negative, not positive.

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 questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks
Molten aluminium oxide is electrolysed to extract aluminium. A current of 5.00×104 A5.00 \times 10^{4}\ \text{A} flows for 1.00 hour1.00\ \text{hour}. (a) Write the cathode half-equation. (b) Calculate the mass of aluminium produced. (F=96500 C mol−1F = 96500\ \text{C mol}^{-1}; M(Al)=27.0 g mol−1M(\text{Al}) = 27.0\ \text{g mol}^{-1}.)
Show worked answer →

(a) Cathode (reduction): Al3++3e−→Al\text{Al}^{3+} + 3e^- \rightarrow \text{Al}. (1 mark)

(b) Charge Q=It=5.00×104×3600=1.80×108 CQ = It = 5.00 \times 10^{4} \times 3600 = 1.80 \times 10^{8}\ \text{C}. (1 mark)

Moles of electrons =QF=1.80×10896500=1.865×103 mol= \dfrac{Q}{F} = \dfrac{1.80 \times 10^{8}}{96500} = 1.865 \times 10^{3}\ \text{mol}. (1 mark)

From Al3++3e−\text{Al}^{3+} + 3e^-, n(Al)=1.865×1033=621.6 moln(\text{Al}) = \dfrac{1.865 \times 10^{3}}{3} = 621.6\ \text{mol}, so m=nM=621.6×27.0=1.68×104 g=16.8 kgm = nM = 621.6 \times 27.0 = 1.68 \times 10^{4}\ \text{g} = 16.8\ \text{kg}. (2 marks)

SACE 2020-style4 marks
In the electrolysis of aqueous copper(II) sulfate with inert electrodes, a current of 1.50 A1.50\ \text{A} is passed for 30.0 minutes30.0\ \text{minutes}. (a) Write the cathode half-equation. (b) Calculate the mass of copper deposited. (F=96500 C mol−1F = 96500\ \text{C mol}^{-1}; M(Cu)=63.5 g mol−1M(\text{Cu}) = 63.5\ \text{g mol}^{-1}.)
Show worked answer →

(a) Cathode: Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}. (1 mark)

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

n(e−)=270096500=2.798×10−2 moln(e^-) = \dfrac{2700}{96500} = 2.798 \times 10^{-2}\ \text{mol}; n(Cu)=2.798×10−22=1.399×10−2 moln(\text{Cu}) = \dfrac{2.798 \times 10^{-2}}{2} = 1.399 \times 10^{-2}\ \text{mol}. (1 mark)

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

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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