Electrolysis of aqueous solutions

Electrolysis of an aqueous solution is more involved than electrolysis of a molten compound because water itself supplies a small amount of hydrogen ions, H⁺, and hydroxide ions, OH⁻. That means two cations reach the cathode and two anions reach the anode, and only one is discharged at each. Choosing correctly is the whole skill this Form 5 Redox topic tests.

The three factors for selective discharge

1. Position in the electrochemical series. The ion positioned lower in the series is discharged more readily. For cations this means a less reactive metal ion is discharged in preference to a more reactive one; for anions, OH⁻ is discharged more readily than sulfate or nitrate.
2. Concentration of the ion. A very concentrated ion may be discharged in preference to what the series alone predicts. This is why concentrated chloride gives chlorine at the anode even though OH⁻ sits lower in the series.
3. Type of electrode. Inert electrodes (carbon, platinum) take no part, but a reactive electrode such as copper can itself dissolve and change the products.

Worked examples with balanced half-equations

Dilute sulfuric acid (or water) with carbon electrodes

• Cathode: 2H⁺(aq) + 2e- → H₂(g)
• Anode: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e-

Hydrogen and oxygen are released in a 2 : 1 volume ratio; this is effectively the electrolysis of water.

Copper(II) sulfate solution with carbon electrodes

• Cathode: Cu²⁺(aq) + 2e- → Cu(s) (copper is lower than hydrogen, so it is discharged)
• Anode: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e-

A brown solid coats the cathode and the blue colour of the solution fades as Cu²⁺ is used up.

Concentrated sodium chloride solution (brine) with carbon electrodes

• Cathode: 2H⁺(aq) + 2e- → H₂(g) (sodium is more reactive than hydrogen, so H⁺ is discharged)
• Anode: 2Cl⁻(aq) → Cl₂(g) + 2e- (chloride wins because it is very concentrated)

If the sodium chloride is dilute, the anode instead gives oxygen: 4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e-. This pair of experiments is the classic demonstration of the concentration factor.

Copper(II) sulfate with copper (reactive) electrodes

• Anode: Cu(s) → Cu²⁺(aq) + 2e- (the copper anode dissolves)
• Cathode: Cu²⁺(aq) + 2e- → Cu(s)

Here the blue colour stays constant, the anode loses mass and the cathode gains mass, the basis of electroplating and purifying copper.

Conditions required

The solution must be aqueous, a direct current is supplied, and the electrodes are inert (carbon or platinum) unless the question is testing reactive electrodes.

Observations

Look for gas bubbles at an electrode, a metal deposit, a fading or unchanging solution colour, and a change in electrode mass. Test any gas: a lighted splint pops with hydrogen, a glowing splint relights with oxygen, and chlorine bleaches moist litmus paper.

How it appears in the SPM exam

In Paper 2 (4541/2) a structured question gives a named solution and electrodes and asks for the half-equation, product and observation at each electrode, plus a reason based on the three factors. The concentration comparison of dilute versus concentrated sodium chloride is a favourite. Balance the electrons and charges every time, and always justify your choice of ion using the factor named in the question.