What are unbalanced half-equations?

Balance atoms and electrons; the oxygen half-equation needs four hydroxide ions and four electrons.

SingaporeChemistrySyllabus dot point

How do we predict the products when an aqueous solution is electrolysed, given that water also provides ions?

Predict the products of electrolysing aqueous solutions using the selective discharge rules based on ion reactivity and concentration, and write the electrode half-equations

A focused answer to the O-Level Chemistry outcome on electrolysing aqueous solutions. The selective discharge rules at each electrode based on the reactivity series and concentration, with worked predictions and electrode half-equations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to predict the products of electrolysing an aqueous (water-based) solution, using selective discharge rules. The complication compared with a molten compound is that water also provides ions ( $\text{H}^+$ and $\text{OH}^-$ ), so at each electrode there is a choice of which ion is discharged. You also need to write the electrode half-equations.

The answer

Why aqueous electrolysis is different

In a solution, two sources of ions are present: the dissolved compound and the water itself (a tiny amount of water splits into $\text{H}^+$ and $\text{OH}^-$ ions). So at the cathode there may be a metal ion and $\text{H}^+$ competing, and at the anode a non-metal ion and $\text{OH}^-$ competing. The selective discharge rules decide which one is actually discharged.

Discharge at the cathode (reactivity rule)

At the cathode, the choice is between the metal ion and the hydrogen ion from water. The rule depends on reactivity:

If the metal is more reactive than hydrogen (for example sodium, potassium, calcium, magnesium, aluminium), then hydrogen is discharged in preference, so hydrogen gas is produced.
If the metal is less reactive than hydrogen (for example copper, silver), then the metal is discharged, so the metal is deposited at the cathode.

So electrolysing sodium chloride solution gives hydrogen at the cathode (sodium is too reactive), while copper(II) sulfate solution gives copper.

Discharge at the anode (ion type and concentration)

At the anode, the choice is between the negative ion of the compound and the hydroxide ion from water:

If a halide ion (chloride, bromide, iodide) is present in high concentration, the halogen is discharged (for example chlorine from concentrated sodium chloride).
Otherwise (for sulfates, nitrates, or dilute halides), the hydroxide is discharged, giving oxygen and water.

So concentration matters at the anode: concentrated chloride gives chlorine, but dilute chloride gives mostly oxygen.

Writing the half-equations

The electrode reactions are written as half-equations showing electrons:

Cathode (reduction): $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ (hydrogen) or $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (a metal).
Anode (oxidation): $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ (chlorine) or $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-$ (oxygen).

Effect on the solution

Removing ions changes the solution. For copper(II) sulfate with inert electrodes, the blue colour fades as copper(II) ions are removed, and the solution becomes acidic as hydroxide is discharged at the anode leaving hydrogen ions and sulfate behind.

Worked example

Predict the products of electrolysing aqueous potassium iodide (a moderately concentrated solution) with inert electrodes, and write the half-equation at each electrode.

Step 1: List the ions present

From potassium iodide: $\text{K}^+$ and $\text{I}^-$ . From water: $\text{H}^+$ and $\text{OH}^-$ . All four are present.

Step 2: Decide the cathode product

The competing cations are $\text{K}^+$ and $\text{H}^+$ . Potassium is more reactive than hydrogen, so hydrogen is discharged in preference. Hydrogen gas forms at the cathode:

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

Step 3: Decide the anode product

The competing anions are $\text{I}^-$ and $\text{OH}^-$ . A halide ion (iodide) present in reasonable concentration is discharged in preference, so iodine forms at the anode:

2\text{I}^- \rightarrow \text{I}_2 + 2e^-

Step 4: State the products

Hydrogen gas is given off at the cathode and iodine forms at the anode (the solution darkens as iodine appears). The remaining ions, potassium and hydroxide, leave a solution of potassium hydroxide.

Examples in context

Example 1. Making chlorine and sodium hydroxide. Electrolysing concentrated sodium chloride solution (brine) produces chlorine at the anode, hydrogen at the cathode, and leaves sodium hydroxide in solution. This is an important industrial process supplying chlorine for water treatment and sodium hydroxide for soap and paper, all from common salt and water.

Example 2. Why dilute and concentrated salt differ. A school experiment electrolysing dilute then concentrated sodium chloride shows oxygen at the anode in the dilute case but chlorine in the concentrated case, while hydrogen always forms at the cathode. This demonstrates the concentration effect on selective discharge in a single comparison.

Try this

Q1. State the product at the cathode when aqueous sodium sulfate is electrolysed, and explain why. [2 marks]

Cue. Hydrogen; sodium is more reactive than hydrogen, so hydrogen ions are discharged in preference to sodium ions.

Q2. State the product at the anode when concentrated sodium chloride solution is electrolysed. [1 mark]

Cue. Chlorine (the halide is concentrated, so the halogen is discharged).

Q3. Write the half-equation for the production of oxygen at the anode. [1 mark]

Cue. $4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4e^-$ .

Exam-style practice questions

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

Original6 marksDilute sodium chloride solution and concentrated sodium chloride solution are each electrolysed with inert electrodes. (a) Name the product at the anode in each case. (b) Explain the difference. (c) Name the product at the cathode in both cases and explain why it is the same.

Show worked answer →

(a) From dilute sodium chloride: oxygen at the anode. From concentrated sodium chloride: chlorine at the anode.

(b) At the anode, both chloride ions and hydroxide ions (from water) are present. Hydroxide tends to be discharged in preference, giving oxygen, when the solution is dilute. When the chloride is concentrated, the high concentration of chloride ions means chlorine is discharged instead. So concentration changes which ion is discharged at the anode.

(c) The cathode product is hydrogen in both cases. The cations present are sodium ions and hydrogen ions (from water). Sodium is more reactive than hydrogen, so hydrogen ions are discharged in preference, giving hydrogen gas, regardless of concentration.

Markers reward oxygen (dilute) and chlorine (concentrated) at the anode, the concentration effect on chloride discharge, and hydrogen at the cathode because the less reactive hydrogen ion is discharged in preference to sodium.

Original4 marksAqueous copper(II) sulfate is electrolysed with inert carbon electrodes. (a) Name the product at the cathode and at the anode. (b) Write the half-equation at the cathode. (c) State how the colour of the solution changes and why.

Show worked answer →

(a) Cathode: copper. Anode: oxygen.

(b) $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ .

(c) The blue colour of the solution fades. The blue is due to copper(II) ions, which are removed from solution as they are discharged and deposited as copper at the cathode, so the solution becomes paler.

Markers reward copper at the cathode and oxygen at the anode, the correct cathode half-equation, and the fading blue colour explained by removal of copper(II) ions.