How Electrolysis Works

Electrolysis is examined heavily across Papers 2, 4 and 6. The content splits into two levels: molten electrolysis (straightforward) and aqueous electrolysis (requires predicting which ion discharges). Both levels demand half-equations at Extended tier.

The setup

An electrolysis circuit has five components:

1. Power supply providing direct current (d.c.)
2. Two electrodes — conductors dipped into the electrolyte (usually graphite or platinum, because they are inert and do not react)
3. Electrolyte — an ionic compound that is either molten or dissolved in water, containing ions free to move
4. Cathode — the negative electrode (connected to the negative terminal)
5. Anode — the positive electrode (connected to the positive terminal)

Memory aid: Cathode = negative = Cations attracted (positive ions go to the negative electrode). Anode = positive = Anions attracted (negative ions go to the positive electrode).

What happens during electrolysis

When the circuit is switched on:

1. Cations (positive ions) migrate toward the cathode
2. Anions (negative ions) migrate toward the anode
3. At the cathode, cations gain electrons (reduction): M^n+ + ne- -> M
4. At the anode, anions lose electrons (oxidation): X^n- -> X + ne-

Oxidation occurs at the anode. Reduction occurs at the cathode. Use the mnemonic: OA RC (Oxidation at the Anode, Reduction at the Cathode) or AN OX, RED CAT.

The electrons flow through the external circuit from anode to cathode. Inside the electrolyte, ions carry the charge.

Electrolysis of molten compounds

With a molten binary compound, only two ions are present, so the products are predictable.

Example: molten lead(II) bromide, PbBr2

Ions present: Pb2+ and Br-

Electrode	Ion attracted	Half-equation	Product
Cathode (-)	Pb2+	Pb2+ + 2e- -> Pb	Lead metal
Anode (+)	Br-	2Br- -> Br2 + 2e-	Bromine vapour

Observation at cathode: silvery bead of molten lead forms. Observation at anode: brown/orange vapour of bromine produced.

Example: molten aluminium oxide, Al2O3

This is the industrial extraction of aluminium, covered in extraction of metals.

Electrode	Ion attracted	Half-equation	Product
Cathode (-)	Al3+	Al3+ + 3e- -> Al	Aluminium metal
Anode (+)	O2-	2O2- -> O2 + 4e-	Oxygen gas

The carbon anodes burn away because oxygen reacts with the hot carbon electrodes (C + O2 -> CO2), so they must be replaced regularly.

Electrolysis of aqueous solutions

Aqueous electrolysis is more complex because water provides additional ions:

H2O ⇌ H+ + OH-

So the electrolyte contains ions from the dissolved compound AND ions from water. The question becomes: which ion is discharged at each electrode?

Rules for predicting cathode products

The metal’s position in the reactivity series determines what happens:

• Metal more reactive than hydrogen (K, Na, Ca, Mg, Al, Zn): H+ ions are discharged instead, producing hydrogen gas. The reactive metal ions stay in solution.
• Metal less reactive than hydrogen (Cu, Ag): The metal ion is discharged, depositing the metal on the cathode.

Half-equation for hydrogen: 2H+ + 2e- -> H2

Rules for predicting anode products

• Halide ion present (Cl-, Br-, I-): The halogen is produced.
• No halide ion (SO4^2-, NO3-, OH-): Hydroxide ions from water are discharged, producing oxygen.

Half-equation for oxygen from OH-: 4OH- -> 2H2O + O2 + 4e-

Worked example: electrolysis of aqueous copper(II) sulfate with inert electrodes

Ions present: Cu2+, SO4^2-, H+, OH-

Electrode	Ions competing	Discharged	Half-equation	Product
Cathode (-)	Cu2+ vs H+	Cu2+ (Cu less reactive than H)	Cu2+ + 2e- -> Cu	Copper deposited
Anode (+)	SO4^2- vs OH-	OH- (no halide)	4OH- -> 2H2O + O2 + 4e-	Oxygen gas

Observation: pink-brown copper coats the cathode; bubbles of gas at the anode; solution becomes paler blue (Cu2+ ions removed).

Worked example: electrolysis of concentrated aqueous sodium chloride (brine)

Ions present: Na+, Cl-, H+, OH-

Electrode	Ions competing	Discharged	Half-equation	Product
Cathode (-)	Na+ vs H+	H+ (Na more reactive)	2H+ + 2e- -> H2	Hydrogen gas
Anode (+)	Cl- vs OH-	Cl- (halide present)	2Cl- -> Cl2 + 2e-	Chlorine gas

The solution remaining becomes sodium hydroxide (Na+ and OH- left behind). This is the basis of the chlor-alkali industry.

Worked example: electrolysis of dilute sulfuric acid

Ions present: H+, SO4^2-, OH- (from water)

Electrode	Discharged	Half-equation	Product
Cathode (-)	H+	2H+ + 2e- -> H2	Hydrogen gas
Anode (+)	OH- (no halide)	4OH- -> 2H2O + O2 + 4e-	Oxygen gas

Volume of hydrogen is twice the volume of oxygen (2:1 ratio from the overall equation: 2H2O -> 2H2 + O2). The sulfuric acid concentration stays the same because it is not consumed.

Effect of electrode material

When copper electrodes are used in copper(II) sulfate solution, the result changes:

• Cathode: Cu2+ deposited as before (Cu2+ + 2e- -> Cu)
• Anode: The copper anode dissolves (Cu -> Cu2+ + 2e-) instead of oxygen being produced

The anode loses mass, the cathode gains mass, and the concentration of Cu2+ stays constant. This is the principle behind electroplating and copper purification.

Common exam mistakes

1. Confusing anode (+) and cathode (-): remember PANIC — Positive is Anode, Negative Is Cathode.
2. Saying “electrons flow through the electrolyte” — electrons flow through the external circuit; ions carry charge through the electrolyte.
3. Forgetting to balance half-equations: the electrons must balance.
4. Using alternating current — only d.c. works for electrolysis.

Exam-style worked questions

Dilute aqueous copper(II) chloride is electrolysed using graphite electrodes. State the product at each electrode and write the half-equation for each. [4 marks]

• Cathode: copper is deposited [1]. Cu2+ + 2e- -> Cu [1]
• Anode: chlorine gas is produced [1]. 2Cl- -> Cl2 + 2e- [1]

(Cu is less reactive than H, so Cu2+ is discharged at cathode. Cl- is a halide, so chlorine is produced at anode.)

Explain why molten sodium chloride must be used, rather than solid sodium chloride, for electrolysis to occur. [2 marks]

• In solid NaCl the ions are in fixed positions and cannot move [1]
• When molten, the ions are free to move to the electrodes / there are mobile charge carriers [1]