SCSA WACE 2021-style practice: Concentrated aqueous copper(II) sulfate is electrolysed using inert platinum electrodes. Relevant reduction potentials: Cu^2+ + 2e^- → Cu, E^ = +0.34\ V; 2H_2O + 2e^- → H_2 + 2OH^-, E^ = -0.83\ V; O_2 + 4H^+ + 4e^- → 2H_2O, E^ = +1.23\ V; S_2O_8^2- + 2e^- → 2SO_4^2-, E^ = +2.01\ V. (a) Predict and justify the product at each electrode. (b) Write the half-equations. (c) Explain how the pH of the solution changes.

A 6 mark question rewards the justified prediction, both half-equations, and the pH reasoning. **(a) Cathode.** The two candidates are Cu^2+ (+0.34\ V) and water (-0.83\ V). The species with the more positive reduction potential is reduced more readily, so Cu^2+ is reduced and copper metal deposits. **Anode.** Oxidation is the reverse of a reduction, so the species easiest to oxidise has the least positive reduction potential. Comparing water (+1.23\ V) and sulfate (via S_2O_8^2-, +2.01\ V), water is oxidised in preference to sulfate, releasing oxygen. **(b)** Cathode: Cu^2+(aq) + 2e^- → Cu(s). Anode: 2H_2O(l) → O_2(g) + 4H^+(aq) + 4e^-. **(c)** The anode produces H^+ while no OH^- is generated at the cathode (copper deposits instead), so [H^+] rises and the solution becomes more acidic (pH falls). Markers reward Cu^2+ reduced and water oxidised with potential-based justification, both half-equations, and the falling pH.

§-Syllabus dot point

WAChemistrySyllabus dot point

How can an external power source drive a non-spontaneous redox reaction in an electrolytic cell?

Describe the operation of electrolytic cells, predict the products of electrolysis, and explain industrial applications

A focused answer to the WACE Year 12 Chemistry dot point on electrolytic cells, how an external voltage drives non-spontaneous reactions, predicting products of molten and aqueous electrolysis, and industrial uses, with a worked example and common exam mistakes.

Generated by Claude Opus 4.88 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

An electrolytic cell does the opposite of a galvanic cell: it uses electrical energy from an external source to drive a reaction that would not happen on its own (a reaction with a negative $E^\circ_{cell}$ ).

Electrode signs are reversed

The power supply pushes electrons to one electrode and pulls them from the other.

Cathode (reduction): connected to the negative terminal of the supply, so it is the negative electrode.
Anode (oxidation): connected to the positive terminal, so it is the positive electrode.

Predicting products of electrolysis

For a molten ionic compound the only species present are the cation and anion, so the cation is reduced at the cathode and the anion is oxidised at the anode. Electrolysis of molten sodium chloride gives sodium metal and chlorine gas.

For an aqueous solution water is also present and can compete:

\text{Cathode possibilities: } \text{M}^{n+} + n\text{e}^- \rightarrow \text{M} \quad \text{or} \quad 2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-

\text{Anode possibilities: anion oxidation} \quad \text{or} \quad 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^-

The species more easily reduced (more positive potential) wins at the cathode; the species more easily oxidised wins at the anode. So electrolysis of dilute sodium chloride solution tends to give hydrogen and oxygen (from water) rather than sodium and chlorine, though high chloride concentration can favour chlorine at the anode (an overpotential effect noted in some courses).

Industrial applications

Electroplating: depositing a thin metal layer (such as silver or chromium) onto an object made the cathode.
Electrolytic refining: purifying copper by making impure copper the anode and pure copper the cathode.
Extraction of reactive metals: aluminium is extracted by electrolysis of molten aluminium oxide because it is too reactive to reduce chemically.

Active versus inert electrodes

A subtle point that distinguishes the top responses is whether the electrode itself reacts. With inert electrodes (platinum, graphite) the electrode only transfers electrons, so the anode product comes from the electrolyte (an anion or water). With an active electrode the metal of the anode is itself oxidised. In copper refining, the impure copper anode dissolves, $\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2\text{e}^-$ , and the same ions plate out as pure copper at the cathode, $\text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)$ . The more active metals in the impurities (such as zinc and iron) dissolve but stay in solution, while less reactive impurities (silver, gold) drop to the bottom as anode sludge, a valuable by-product. Always check whether a question specifies inert electrodes before deciding what is oxidised at the anode.

Why this matters

Electrolysis underpins major industries: aluminium and chlorine production, copper refining, and electroplating. The dot point connects directly to standard electrode potentials (to predict products) and to Faraday's laws (to calculate amounts), and it is the natural contrast to galvanic cells.

Exam-style practice questions

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

WACE 20216 marksConcentrated aqueous copper(II) sulfate is electrolysed using inert platinum electrodes. Relevant reduction potentials:

\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}

E^\circ = +0.34\ \text{V}

;

2\text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{H}_2 + 2\text{OH}^-

E^\circ = -0.83\ \text{V}

;

\text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O}

E^\circ = +1.23\ \text{V}

;

\text{S}_2\text{O}_8^{2-} + 2\text{e}^- \rightarrow 2\text{SO}_4^{2-}

E^\circ = +2.01\ \text{V}

. (a) Predict and justify the product at each electrode. (b) Write the half-equations. (c) Explain how the

\text{pH}

of the solution changes.

Show worked answer →

A 6 mark question rewards the justified prediction, both half-equations, and the pH reasoning.

(a) Cathode. The two candidates are $\text{Cu}^{2+}$ ( $+0.34\ \text{V}$ ) and water ( $-0.83\ \text{V}$ ). The species with the more positive reduction potential is reduced more readily, so $\text{Cu}^{2+}$ is reduced and copper metal deposits.

Anode. Oxidation is the reverse of a reduction, so the species easiest to oxidise has the least positive reduction potential. Comparing water ( $+1.23\ \text{V}$ ) and sulfate (via $\text{S}_2\text{O}_8^{2-}$ , $+2.01\ \text{V}$ ), water is oxidised in preference to sulfate, releasing oxygen.

(b) Cathode: $\text{Cu}^{2+}(aq) + 2\text{e}^- \rightarrow \text{Cu}(s)$ . Anode: $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^-$ .

(c) The anode produces $\text{H}^+$ while no $\text{OH}^-$ is generated at the cathode (copper deposits instead), so $[\text{H}^+]$ rises and the solution becomes more acidic (pH falls).

Markers reward $\text{Cu}^{2+}$ reduced and water oxidised with potential-based justification, both half-equations, and the falling pH.

WACE 20234 marksCompare an electrolytic cell with a galvanic cell in terms of (i) the sign of the energy change and spontaneity, and (ii) the polarity of the anode. Explain why the chemical roles of the electrodes are the same in both.

Show worked answer →

A 4 mark compare answer needs both contrasts and the unifying point.

(i) A galvanic cell runs a spontaneous redox reaction ( $E^\circ_{\text{cell}} > 0$ ) and converts chemical energy into electrical energy. An electrolytic cell uses an external power supply to drive a non-spontaneous reaction ( $E^\circ_{\text{cell}} < 0$ ), converting electrical energy into chemical energy.

(ii) In a galvanic cell the anode is negative (electrons are pushed out by the spontaneous oxidation). In an electrolytic cell the anode is connected to the positive terminal of the supply, so it is positive.

Same roles. By definition oxidation always occurs at the anode and reduction at the cathode in both cells; only the polarity differs, because in one case the reaction drives the electrons and in the other the supply does.

Markers reward the spontaneity/energy contrast, the anode-polarity contrast, and the oxidation-at-anode unifying statement.