Electrolysis of molten and aqueous compounds

This page covers one Form 5 content standard from Redox Equilibrium: electrolysis of molten and aqueous compounds. Electrolysis is redox forced to happen by an electric current, so everything you learned about oxidation, reduction and half-equations applies directly, the difference is that you now control which reactions occur by choosing the electrolyte and electrodes.

What electrolysis is

Electrolysis is the decomposition of an ionic compound, when molten or in aqueous solution, by passing electricity through it. The liquid that conducts is the electrolyte; the rods that carry the current in and out are the electrodes. The electrode connected to the negative terminal is the cathode; the one connected to the positive terminal is the anode.

The two rules that never change:

• Cations (positive ions) move to the cathode, gain electrons and are reduced. Reduction always happens at the cathode.
• Anions (negative ions) move to the anode, lose electrons and are oxidised. Oxidation always happens at the anode.

A memory aid: red cat (reduction at cathode) and an ox (oxidation at anode). Because both happen, electrolysis is a redox process driven by the power supply.

Molten compounds

A molten ionic compound contains only its own cations and anions, there is no water. So the products are simple to predict: the metal (or hydrogen) forms at the cathode and the non-metal forms at the anode. For molten lead(II) bromide, PbBr₂:

• Cathode: Pb²⁺ + 2e- → Pb (grey molten lead forms)
• Anode: 2Br⁻ → Br₂ + 2e- (brown bromine vapour forms)

Aqueous solutions and selective discharge

An aqueous solution also contains H⁺ and OH⁻ ions from water, so there is a choice of which ion is discharged at each electrode. Three factors decide it:

1. Position in the electrochemical series. The lower (less reactive) ion is discharged more readily. At the cathode, a less reactive metal ion (for example Cu²⁺, Ag⁺) is discharged in preference to H⁺; but for a reactive metal ion (for example Na⁺, K⁺, Ca²⁺), hydrogen is discharged instead. At the anode, OH⁻ is discharged (giving oxygen) in preference to sulfate or nitrate.
2. Concentration of ions. A much more concentrated ion may be discharged even against the series. In concentrated sodium chloride, chlorine is discharged at the anode instead of oxygen, because chloride is so concentrated; in dilute sodium chloride, oxygen forms.
3. Type of electrode. With inert electrodes (carbon, platinum) only the electrolyte ions react. With a reactive electrode, the electrode itself can take part, for example a copper anode dissolves (Cu → Cu²⁺ + 2e-) instead of a gas being released, which is the basis of copper purification and electroplating.

Everyday uses of electrolysis

The same rules explain the industrial uses the exam asks about. In electroplating, the object to be plated is the cathode and the plating metal is the anode, in a solution of that metal’s ions, so a spoon can be silver-plated for appearance and corrosion resistance. In purification of copper, an impure copper anode dissolves and pure copper deposits on a pure copper cathode. Electrolysis is also used to extract very reactive metals such as aluminium from their molten ores, because those metals are too reactive to be displaced by carbon. Knowing one or two real applications, and being able to give the electrode reactions, is exactly what a Paper 2 application question rewards.

Worked example

Question. Dilute copper(II) sulfate solution is electrolysed using carbon (inert) electrodes. State what is observed at each electrode and write the half-equation for each, then say whether oxidation or reduction occurs at the cathode.

Step 1, list the ions present. From CuSO₄: Cu²⁺ and SO₄²⁻. From water: H⁺ and OH⁻.

Step 2, decide the cathode product. At the cathode the choice is Cu²⁺ versus H⁺. Copper is less reactive (lower in the series) than hydrogen, so Cu²⁺ is discharged. Observation: a brown/pink layer of copper coats the cathode.

• Cathode half-equation: Cu²⁺ + 2e- → Cu (a gain of electrons, so reduction).

Step 3, decide the anode product. At the anode the choice is OH⁻ versus SO₄²⁻. Hydroxide is lower and discharged more readily, and sulfate is very stable, so OH⁻ is discharged. Observation: bubbles of a colourless gas (oxygen) that relights a glowing splint.

• Anode half-equation: 4OH⁻ → 2H₂O + O₂ + 4e- (a loss of electrons, so oxidation).

Step 4, answer the direct question. At the cathode, electrons are gained, so reduction occurs there (copper is deposited).

Answer. Cathode: copper deposited, Cu²⁺ + 2e- → Cu (reduction). Anode: oxygen gas released, 4OH⁻ → 2H₂O + O₂ + 4e- (oxidation).

Practice question

Molten lead(II) bromide is electrolysed with carbon electrodes. Name the product at each electrode and write the two half-equations.

Answer. There is no water, so only Pb²⁺ and Br⁻ are present. Cathode: lead metal forms, Pb²⁺ + 2e- → Pb (reduction). Anode: bromine forms, 2Br⁻ → Br₂ + 2e- (oxidation).

Exam tip

State the ions present first, including H⁺ and OH⁻ from water in every aqueous case, then apply the three factors in order. Remember the two rules that never move: reduction at the cathode, oxidation at the anode. When you write a half-equation, balance the electrons and match the state of the product to what is observed (gas, metal deposit, colour). The favourite SPM Chemistry traps are (a) forgetting water’s ions, (b) the concentrated-versus-dilute chloride difference, and (c) the reactive copper anode dissolving. Nail those three and this standard is a reliable source of marks.

Where this fits in the chapter

Electrolysis is redox driven by electricity; the next standard, voltaic and electrochemical cells, is the reverse, chemical redox producing electricity. Comparing the two is a very common exam theme.

• Up to the chapter hub: Redox Equilibrium.
• Sideways: the chapter’s worked examples and revision notes.

In our online 1-to-1 SPM Chemistry lessons, English medium, from RM50 per hour, we work through electrolysis one electrolyte at a time until predicting products and writing balanced half-equations becomes second nature, because this standard rewards a careful, repeatable method more than memory.