How to explain electrolysis of aqueous solutions

Electrolysis of aqueous solutions is where many SPM students lose marks, because there is an extra twist: water itself provides ions. Alongside the ions from the dissolved compound, water supplies a small amount of hydrogen ions (H⁺) and hydroxide ions (OH⁻). So at each electrode there is a choice of which ion to discharge, and you have to decide which one wins. This guide, from the redox and electrochemistry chapter, shows you how to make that decision every time.

The extra ions from water

In any aqueous solution there are four ions to keep track of:

• From the dissolved compound: a cation (a metal ion, or H⁺ from an acid) and an anion.
• From the water: H⁺ and OH⁻.

At the cathode the contest is between the compound’s cation and H⁺. At the anode the contest is between the compound’s anion and OH⁻. Only one ion is discharged at each electrode, this is called selective discharge.

The three factors that decide who is discharged

Three factors settle which ion is discharged. Learn them in order, because the first usually decides it.

1. Position in the electrochemical series. The lower an ion is in the electrochemical series, the more readily it is discharged. At the cathode, a less reactive metal ion (like Cu²⁺ or Ag⁺) is discharged in preference to H⁺, so the metal is deposited. But for a reactive metal ion (like K⁺, Na⁺, Ca²⁺, Mg²⁺ or Al³⁺), H⁺ is discharged instead, giving hydrogen gas. At the anode, OH⁻ is usually discharged in preference to common anions like sulphate and nitrate, giving oxygen.

2. Concentration of ions. A very high concentration of an ion can override the first factor. The classic case is a concentrated halide solution: even though OH⁻ would normally be discharged, a concentrated chloride solution releases chlorine gas at the anode instead of oxygen.

3. Type of electrode. With inert electrodes (carbon or platinum) the electrode takes no part. But a reactive electrode, such as a copper anode, dissolves into the solution instead of oxygen being released, the basis of electroplating and copper purification.

Worked example 1: dilute sulphuric acid

Electrolysing dilute sulphuric acid is effectively electrolysing water. The ions present are H⁺, SO₄²⁻ and OH⁻.

• At the cathode: H⁺ is discharged. 2H⁺ + 2e⁻ → H₂, hydrogen gas.
• At the anode: OH⁻ is discharged in preference to sulphate. 4OH⁻ → O₂ + 2H₂O + 4e⁻, oxygen gas.

Hydrogen and oxygen are released in a 2:1 ratio by volume, which is a favourite examiner detail.

Worked example 2: concentrated sodium chloride (brine)

Brine contains Na⁺, Cl⁻, H⁺ and OH⁻, and it shows two of the factors at once. This is the electrolysis of sodium chloride solution experiment.

• At the cathode: sodium is too reactive, so H⁺ is discharged. 2H⁺ + 2e⁻ → H₂, hydrogen gas.
• At the anode: although OH⁻ is lower in the series, the chloride is concentrated, so chlorine is discharged. 2Cl⁻ → Cl₂ + 2e⁻, chlorine gas.

The sodium and hydroxide ions stay behind, so the solution left is sodium hydroxide.

Worked example 3: copper(II) sulphate with carbon electrodes

Copper(II) sulphate solution contains Cu²⁺, SO₄²⁻, H⁺ and OH⁻.

• At the cathode: copper is below hydrogen in the series, so Cu²⁺ is discharged. Cu²⁺ + 2e⁻ → Cu, a pink-brown copper coating.
• At the anode: OH⁻ is discharged in preference to sulphate. 4OH⁻ → O₂ + 2H₂O + 4e⁻, oxygen gas.

As copper is deposited and the solution loses Cu²⁺, its blue colour fades. You can compare the reactions for many solutions on the electrolysis of aqueous solutions page.

How to answer these questions

1. List all four ions, from the compound and from water.
2. At the cathode, choose between the metal ion and H⁺ using the electrochemical series.
3. At the anode, choose between the anion and OH⁻, checking concentration for halides.
4. Write a balanced half-equation for each electrode.
5. Note any change in the solution, such as a fading colour or a leftover alkali.

Common mistakes to avoid

• Forgetting that water adds H⁺ and OH⁻ to every aqueous solution.
• Discharging a reactive metal like sodium at the cathode instead of hydrogen.
• Ignoring concentration and giving oxygen where a concentrated halide gives the halogen.
• Leaving electrons out of the half-equations or unbalancing them.

Selective discharge is the one genuinely tricky idea in this topic, and it clicks fast once someone walks you through a few solutions. Our online one-to-one teachers do exactly that, English lessons from RM50 an hour, with a paid one-hour trial. See the how it works page if you would like electrolysis drilled against past-paper questions.