Electrolysis

Why Covalent Compounds Do Not Conduct Electricity

Covalent compounds do not conduct electricity because they contain no ions or free electrons. In a covalent substance, atoms are bonded by shared pairs of electrons. These shared electrons are fixed between specific atoms, so they cannot move through the substance to carry charge.

Even when a covalent compound is melted or dissolved, no charged particles are released, because the molecules remain intact. Therefore, covalent compounds cannot conduct electricity in any state — solid, liquid, or in solution.

MisconceptionStudents sometimes state that covalent compounds "have no electrons." They do have electrons, but these electrons are localised in shared pairs between atoms and are not free to move. The correct reason is the absence of mobile ions or free electrons.

Exam TipAlways state "no ions or free electrons" rather than just "no charged particles."

Why Ionic Compounds Conduct Only When Molten or in Solution

Ionic compounds conduct electricity only when molten or dissolved in water because the ions must be free to move. In the solid state, ions are held in a fixed, regular lattice [a repeating three-dimensional arrangement]. The ions vibrate but cannot move from their positions, so they cannot carry charge.

When the ionic compound melts or dissolves in water, the lattice breaks apart. The ions become free to move towards the electrodes, so an electric current flows.

This means solid ionic compounds are insulators, but molten ionic compounds and aqueous ionic solutions are electrolytes [substances that conduct electricity because they contain free-moving ions].

Examiner InsightExaminers expect three linked points: (1) ions are present, (2) ions are free to move, and (3) ions carry charge to the electrodes. Omitting "free to move" is the most common reason for lost marks.

Exam TipWrite "ions are free to move and carry charge" as a complete phrase every time.

Anions and Cations

An anion is a negatively charged ion. A cation is a positively charged ion.

During electrolysis, cations are attracted to the cathode (the negative electrode) because opposite charges attract. Anions are attracted to the anode (the positive electrode) for the same reason.

A simple way to remember this: the word "cation" contains a "t" that looks like a + sign, so cations are positive.

MnemonicPANIC — Positive Anode, Negative Is Cathode.

Electrolysis of Molten Compounds — Lead(II) Bromide

Electrolysis is the decomposition of a compound using an electric current. The substance being decomposed is the electrolyte, and the electrodes must be inert [they do not react with the electrolyte or the products] — carbon (graphite) or platinum electrodes are commonly used.

Reading convention for half-equations: the arrow (→) means "forms" or "produces." The electrons (e⁻) appear on the left side if gained (reduction) or on the right side if lost (oxidation).

Lead(II) bromide (PbBr₂) must be molten to electrolyse because the ions must be free to move. When molten PbBr₂ is electrolysed:

At the cathode, lead ions gain electrons:

This is reduction because the ion gains electrons. At the anode, bromide ions lose electrons:

This is oxidation because the ions lose electrons.

The overall decomposition is:

MisconceptionStudents often write that "ions move to the electrode" to describe what happens. The examiners require that you state ions are discharged at the electrode — meaning they gain or lose electrons and become atoms or molecules.

Exam TipUse the phrase "ions are discharged at the electrode" in every electrolysis answer.

Electrolysis of Aqueous Solutions

Electrolysing an aqueous solution is more complex than electrolysing a molten compound. Water itself partially dissociates into hydrogen ions (H⁺) and hydroxide ions (OH⁻), so there are always at least four types of ion present: the cation from the dissolved compound, the anion from the dissolved compound, H⁺ from water, and OH⁻ from water. The product at each electrode depends on which ion is preferentially discharged.

Two rules predict which ion is discharged:

At the cathode (choosing between cations): If the metal is more reactive than hydrogen, hydrogen gas is produced. If the metal is less reactive than hydrogen, the metal is deposited. This is because more reactive metals hold onto their electrons more strongly, so their ions are harder to reduce.

At the anode (choosing between anions): If a halide ion (Cl⁻, Br⁻, or I⁻) is present, the halogen is produced. If no halide is present, oxygen gas is produced from the discharge of OH⁻ ions.

Sodium chloride solution (NaCl):

Ions present: Na⁺, H⁺, Cl⁻, OH⁻

At the cathode, sodium is more reactive than hydrogen, so H⁺ ions are discharged:

At the anode, chloride is a halide ion, so it is preferentially discharged:

Sodium hydroxide (NaOH) remains in solution as the Na⁺ and OH⁻ ions are not discharged.

Dilute sulfuric acid (H₂SO₄):

Ions present: H⁺, SO₄²⁻, OH⁻ (from water)

At the cathode:

At the anode, sulfate is not a halide, so OH⁻ is discharged:

Copper(II) sulfate solution (CuSO₄):

Ions present: Cu²⁺, H⁺, SO₄²⁻, OH⁻

At the cathode, copper is less reactive than hydrogen, so Cu²⁺ ions are discharged:

A pink-brown/orange-brown solid deposits on the cathode.

At the anode, sulfate is not a halide, so OH⁻ is discharged:

The blue colour of the solution fades as Cu²⁺ ions are removed.

Examiner InsightQuestions frequently ask students to predict products and then justify the prediction. Always state the rule (e.g. "copper is less reactive than hydrogen, so copper is deposited") and then write the half-equation. The justification and the equation together earn full marks.

Exam TipFor each electrode, name the product, write the half-equation with state symbols, and state whether the reaction is oxidation or reduction.

MisconceptionStudents often think that the metal in solution is always deposited at the cathode. This only happens when the metal is less reactive than hydrogen. If the metal is more reactive (e.g. sodium), hydrogen gas is produced instead.

Exam TipAlways check the reactivity of the metal relative to hydrogen before predicting the cathode product.

Ionic Half-Equations and Oxidation or Reduction

An ionic half-equation shows what happens at one electrode during electrolysis. It includes only the ion involved, the electrons transferred, and the product formed.

Oxidation occurs at the anode — anions lose electrons. The word OILRIG captures this: Oxidation Is Loss, Reduction Is Gain (of electrons). Reduction occurs at the cathode — cations gain electrons.

To identify whether a half-equation represents oxidation or reduction, check the side on which electrons appear:

Summary of all key half-equations for this section:

MnemonicOILRIG — Oxidation Is Loss, Reduction Is Gain.

SafetyChlorine gas (produced at the anode from NaCl solution) is toxic. Conduct the experiment in a well-ventilated area or fume cupboard and avoid inhaling the gas.

PRACTICAL: Investigate the Electrolysis of Aqueous Solutions

Aim: To investigate the products formed at the electrodes during the electrolysis of different aqueous solutions using inert electrodes, and to identify each product by observation and testing.

Method

1. Pour approximately 100 cm³ of copper(II) sulfate solution into a beaker.

2. Place two graphite (carbon) electrodes into the solution and connect them to a d.c. power supply.

3. Switch on the power supply and allow the current to flow for 5–10 minutes.

4. Observe and record any changes at each electrode (colour changes, gas bubbles, solid deposits).

5. Test any gas produced: hold a glowing splint near the cathode to test for hydrogen (squeaky pop); hold damp litmus paper near the anode to test for chlorine (bleaches litmus) or a glowing splint to test for oxygen (relights).

6. Repeat steps 1–5 with sodium chloride solution and dilute sulfuric acid, using a fresh pair of graphite electrodes each time.

Variables

Independent variable (IV): the aqueous solution used (copper(II) sulfate, sodium chloride, dilute sulfuric acid)

Dependent variable (DV): the products formed at each electrode, identified by observation and gas tests

Control variables (CV): type of electrode (graphite — inert); volume and concentration of solution kept the same; voltage/current from the d.c. power supply kept the same; duration of electrolysis kept the same

Expected results

Copper(II) sulfate produces a copper deposit at the cathode and oxygen gas at the anode, because copper is less reactive than hydrogen. Sodium chloride solution produces hydrogen at the cathode and chlorine at the anode. Dilute sulfuric acid produces hydrogen at the cathode and oxygen at the anode.

Precaution

Chlorine gas is toxic, so electrolysis of sodium chloride must be performed in a well-ventilated room or fume cupboard. Using electrodes that are not inert (e.g. copper electrodes) would introduce a source of error because the anode would dissolve instead of producing gas.