How to predict the products of electrolysis

“What is liberated at the cathode? What is produced at the anode?” These are staple SPM questions from the redox equilibrium chapter, and once you know the reasoning they become almost automatic. The trick is that an aqueous solution contains more ions than the salt you dissolved, water itself supplies H⁺ and OH⁻, so at each electrode there is often a choice, and you must predict which ion wins. This guide gives you the decision process step by step. Writing the half-equations afterwards is a separate skill, covered in our companion guide on half-equations for electrolysis.

Start by listing every ion present

Before predicting anything, list all the ions in the electrolyte. This is where careless mistakes begin.

• In a molten compound there are only two kinds of ion, from the compound itself. Molten lead(II) bromide contains only Pb²⁺ and Br⁻. There is no water, so no H⁺ or OH⁻, and prediction is easy: the metal forms at the cathode, the non-metal at the anode.
• In an aqueous solution there are four kinds of ion: the two from the dissolved salt, plus H⁺ and OH⁻ from the slight ionisation of water. Now each electrode has two candidate ions, and you need the discharge rules.

The three factors that decide discharge

When two ions are attracted to the same electrode, which is discharged depends on three factors, in this order of usual importance:

1. Position in the electrochemical series. The ion of the less reactive element is discharged more easily. At the cathode, if the metal is below hydrogen in reactivity (such as copper or silver), the metal ion is discharged; if the metal is more reactive than hydrogen (such as sodium, potassium, magnesium), then H⁺ is discharged and hydrogen gas is released. At the anode, the ease of discharge of anions generally runs OH⁻ before the sulfate and nitrate ions (which are essentially not discharged), while halide ions can be discharged in preference to OH⁻ under the right conditions.
2. Concentration of ions. A high concentration of an ion can override the series. Concentrated chloride solution gives chlorine at the anode because the sheer abundance of Cl⁻ wins, even though OH⁻ sits lower in the series. Dilute the same solution and oxygen from OH⁻ is favoured instead.
3. Type of electrode. Inert electrodes (carbon or platinum) take no part in the reaction. An active electrode, a copper anode, for instance, dissolves into the solution itself instead of an ion being discharged from solution.

Worked example one: a molten compound

Electrolyse molten lead(II) bromide with carbon electrodes.

Ions present: Pb²⁺ and Br⁻ only. There is no choice.

• Cathode: Pb²⁺ is reduced, so lead metal forms (a molten silvery bead).
• Anode: Br⁻ is oxidised, so brown bromine vapour is produced.

You can follow the full experiment for electrolysis of molten lead bromide to see the observations.

Worked example two: a dilute aqueous solution

Electrolyse dilute sodium chloride solution with carbon electrodes.

Ions present: Na⁺, Cl⁻ (from the salt), and H⁺, OH⁻ (from water).

• Cathode (Na⁺ vs H⁺): sodium is far more reactive than hydrogen, so H⁺ is discharged. Hydrogen gas is released.
• Anode (Cl⁻ vs OH⁻): the solution is dilute, so OH⁻ is discharged in preference. Oxygen gas is produced.

This is why dilute salt solution effectively electrolyses the water. Explore the pattern further in the electrolysis of aqueous solutions, and see it done in the lab in the electrolysis of sodium chloride solution.

Worked example three: concentration changes the answer

Electrolyse concentrated sodium chloride solution with carbon electrodes.

Same four ions, but now the chloride concentration is high.

• Cathode: still H⁺, so hydrogen again.
• Anode: the high concentration of Cl⁻ overrides the series, so chlorine gas is produced instead of oxygen.

That single change, dilute to concentrated, flips the anode product from oxygen to chlorine. Examiners love this comparison, so make sure you can explain why: it is factor 2, concentration, defeating factor 1.

Worked example four: the electrode itself reacts

Electrolyse copper(II) sulfate solution, first with carbon electrodes, then with copper electrodes.

Ions present: Cu²⁺, SO₄²⁻, H⁺, OH⁻.

• With carbon electrodes: at the cathode Cu²⁺ is below hydrogen, so copper is deposited; at the anode OH⁻ is discharged (sulfate is not), so oxygen forms. The blue colour fades as Cu²⁺ is used up.
• With copper electrodes: at the cathode copper is still deposited, but at the anode the copper electrode itself dissolves (Cu → Cu²⁺ + 2e⁻) instead of discharging any solution ion. Cu²⁺ is replaced as fast as it is removed, so the blue colour stays constant. This is the principle behind electroplating and purifying copper.

A quick decision routine

For any electrolysis, run this in order: list every ion; group them by electrode; at each electrode apply the series first, then check whether a high concentration overrides it, then check whether the electrode is active. Say what forms, and add the observation, a gas, a metal deposit, a colour change. That final observation line often carries its own mark.

Predicting products becomes reliable once you always begin with the full ion list. If you would like a teacher to walk you through borderline cases, dilute versus concentrated, inert versus active, our online one-to-one lessons with our experienced SPM Chemistry teachers are built for that, from RM50 an hour with a paid one-hour trial lesson.