Electrolysis

Q: Why do solid ionic compounds not conduct electricity?

Solid ionic compounds have ions held in a fixed giant lattice structure, so there are no free-moving charged particles to carry electric current. Ions only become mobile when the compound is molten or dissolved in water.

Q: How do I tell if an electrode reaction is oxidation or reduction?

Use two mnemonics: 1. OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain of electrons. 2. Red Cat: Reduction occurs at the Cathode, so oxidation always occurs at the anode.

Edexcel International GCSE Chemistry· 1.55C–1.60C· 25 min read

1. Conductivity of Ionic and Covalent Compounds★★☆☆☆⏱ 5 min

Electrical conductivity requires free-moving charged particles (either delocalised electrons or mobile ions) to carry current. This property distinguishes conducting ionic compounds from non-conducting simple covalent compounds.

📘 Definition

Electrolyte

A substance that conducts electricity when molten or dissolved in water, and is chemically broken down during electrolysis. All electrolytes are ionic compounds.

Simple covalent compounds do not conduct electricity in any state, as they are made of neutral molecules with no free ions or delocalised electrons. Giant covalent structures like graphite are an exception, but these are not classified as electrolytes.

Ionic compounds do not conduct electricity when solid, as their ions are locked in a fixed lattice by strong ionic bonds. When molten or dissolved in water, the ionic bonds break, releasing free-moving ions that can carry current.

📐 Worked Example

Explain why solid potassium iodide does not conduct electricity, but aqueous potassium iodide does.

1
Potassium iodide is an ionic compound with a giant ionic lattice in the solid state.
2
In solid form, K⁺ and I⁻ ions are held in fixed positions, so there are no free-moving charged particles to carry current.
3
When dissolved in water, the ionic lattice breaks apart, so K⁺ and I⁻ ions become free to move and carry electric charge, so the solution conducts.

Exam tip:

Always reference "free-moving charged particles" in conductivity questions, as this is a mandatory marking point for full credit.

2. Key Electrolysis Terms and Redox Classification★★★☆☆⏱ 6 min

Cation: Positively charged ion, attracted to the negative cathode
Anion: Negatively charged ion, attracted to the positive anode
Cathode: Negative electrode, where reduction (electron gain) occurs
Anode: Positive electrode, where oxidation (electron loss) occurs

📘 Definition

Half-equation

Ionic equation showing electron transfer at one electrode. Must balance both atoms and total charge on both sides of the equation.

📐 Worked Example

Classify the reaction $2 O^{2 -} \to O_{2} + 4 e^{-}$ as oxidation or reduction, and state which electrode it occurs at.

1
Identify electron transfer: the oxide ions lose 4 electrons total to form oxygen gas.
2
Apply OIL RIG: loss of electrons = oxidation.
3
Apply Red Cat: oxidation occurs at the anode.
4
Final answer: Oxidation, anode.

3. Electrolysis of Molten Ionic Compounds★★★☆☆⏱ 5 min

Molten binary ionic compounds (made of two elements) only contain the ions from the compound itself, so product prediction is straightforward. The metal cation is always discharged at the cathode, and the non-metal anion is discharged at the anode.

📐 Worked Example

Predict the products of electrolysis of molten lead(II) bromide (PbBr₂) using inert electrodes, and write balanced half-equations for each reaction.

1
Identify the ions present: Pb²⁺ (lead cation) and Br⁻ (bromide anion).
2
Cathode reaction: Pb²⁺ ions gain 2 electrons to form lead metal (reduction):
3
$Pb^{2+} + 2e^- \rightarrow Pb$
4
Anode reaction: Two Br⁻ ions each lose 1 electron to form bromine gas (oxidation):
5
$2Br^- \rightarrow Br_2 + 2e^-$
6
Final products: Lead metal at cathode, bromine gas at anode.

Exam tip:

For molten electrolytes, you never have to consider ions from water, so you only need to work with the ions from the pure ionic compound given in the question.

4. Electrolysis of Aqueous Solutions★★★★☆⏱ 7 min

Aqueous solutions contain H⁺ and OH⁻ ions from the partial dissociation of water, plus ions from the dissolved solute, giving four total ions. Product prediction follows two simple rules:

Cathode rule: Hydrogen gas ( $H_{2}$ ) is produced unless the metal cation is less reactive than hydrogen (below H in the reactivity series, e.g. Cu²⁺, Ag⁺), in which case solid metal is deposited.
Anode rule: Oxygen gas ( $O_{2}$ ) is produced from OH⁻ ions unless the solution has a high concentration of halide ions (Cl⁻, Br⁻, I⁻), in which case the halogen gas is produced. Sulfate and nitrate ions are never discharged, so oxygen is always produced for these anions.

📐 Worked Example

Predict the products of electrolysis of concentrated aqueous sodium chloride (brine) using inert electrodes, and write balanced half-equations.

1
Identify ions present: Na⁺, Cl⁻ (from NaCl), H⁺, OH⁻ (from water).
2
Cathode: Na is above H in the reactivity series, so H⁺ is discharged to form hydrogen gas:
3
$2H^+ + 2e^- \rightarrow H_2$
4
Anode: High concentration of Cl⁻ ions, so Cl⁻ is discharged instead of OH⁻ to form chlorine gas:
5
$2Cl^- \rightarrow Cl_2 + 2e^-$
6
Final products: Hydrogen gas at cathode, chlorine gas at anode, sodium hydroxide solution left behind.

📐 Worked Example

Predict the products of electrolysis of dilute aqueous copper(II) sulfate using inert electrodes.

1
Identify ions present: Cu²⁺, SO₄²⁻ (from CuSO₄), H⁺, OH⁻ (from water).
2
Cathode: Cu is below H in the reactivity series, so Cu²⁺ is discharged to form solid copper deposit:
3
$Cu^{2+} + 2e^- \rightarrow Cu$
4
Anode: SO₄²⁻ ions are stable, so OH⁻ is discharged to form oxygen gas:
5
$4OH^- \rightarrow O_2 + 2H_2O + 4e^-$
6
Final products: Copper metal at cathode, oxygen gas at anode.

Exam tip:

Always check if a halide solution is concentrated or dilute: dilute halide solutions produce oxygen instead of the halogen, so this detail changes the anode product.

5. Required Practical: Electrolysis of Aqueous Solutions★★★☆☆⏱ 4 min

You must be able to describe the standard method to investigate electrolysis of aqueous solutions using inert electrodes, and how to test the products formed:

Set up the circuit: connect two inert graphite electrodes to a low-voltage (4-6V) power supply, and place them in a beaker of the aqueous electrolyte.
To collect gaseous products, invert test tubes filled with water over each electrode to capture gas.
Turn on the power supply and leave for 5-10 minutes, recording observations (gas bubbles, solid deposits, colour changes).
Test collected gases: hydrogen gives a squeaky pop with a lit splint, oxygen relights a glowing splint, chlorine bleaches damp blue litmus paper.

📐 Worked Example

Describe how you would test the products of electrolysis of dilute sulfuric acid to confirm their identity.

1
Products of dilute sulfuric acid electrolysis are hydrogen at the cathode and oxygen at the anode.
2
Test cathode gas: Hold a lit splint at the mouth of the test tube of collected gas. A squeaky pop confirms hydrogen is present.
3
Test anode gas: Insert a glowing splint into the test tube of collected gas. If the splint relights, oxygen is present.

Exam tip:

For practical valid comparison questions, control variables include voltage, electrode material, electrolyte volume and concentration, and duration of electrolysis.

6. Common Pitfalls

Wrong move:

Stating solid ionic compounds conduct electricity.

Why:

Ions are locked in a fixed lattice in solid ionic compounds, so there are no free-moving charged particles to carry current.

Correct move:

Specify ionic compounds only conduct when molten or dissolved in water, as ions become free to move.

Wrong move:

Mixing up anode and cathode reactions, e.g. saying oxidation occurs at the cathode.

Why:

The cathode is negatively charged, so it attracts positive cations that gain electrons (reduction). The anode attracts negative anions that lose electrons (oxidation).

Correct move:

Use the mnemonic Red Cat: Reduction at Cathode, so oxidation always occurs at the anode.

Wrong move:

Writing unbalanced half-equations, e.g. $2 C l^{-} \to C l_{2} + e^{-}$ .

Why:

Charge is not balanced: total charge on the left is -2, while total charge on the right is -1.

Correct move:

Balance atoms first, then add electrons to balance charge: $2 C l^{-} \to C l_{2} + 2 e^{-}$ (charge -2 on both sides).

Wrong move:

Predicting sodium metal deposition at the cathode for aqueous sodium chloride.

Why:

Sodium is more reactive than hydrogen, so H⁺ ions from water are discharged instead of Na⁺ ions.

Correct move:

Apply the cathode rule: hydrogen is produced unless the metal is below hydrogen in the reactivity series.

Wrong move:

Predicting chlorine gas from dilute sodium chloride solution.

Why:

Only concentrated halide solutions produce halogens at the anode; dilute solutions produce oxygen from OH⁻ ions.

Correct move:

Check if the halide solution is concentrated: only produce the halogen if concentration is high, otherwise produce oxygen.

Wrong move:

Stating all covalent compounds never conduct electricity.

Why:

Giant covalent structures like graphite have delocalised electrons that conduct, but simple molecular covalent compounds do not.

Correct move:

Specify that simple covalent compounds do not conduct electricity, as they have no free ions or delocalised electrons.

7. Quick Reference Cheatsheet

Electrolyte Type	Cathode Product Rule	Anode Product Rule	Example Half Equations
Molten binary ionic	Metal from compound cation	Non-metal from compound anion	Cathode: $P b^{2 +} + 2 e^{-} \to P b$ Anode: $2 B r^{-} \to B r_{2} + 2 e^{-}$
Aqueous (metal > H in reactivity)	Hydrogen gas ( $H_{2}$ )	O₂ if sulfate/nitrate/dilute halide; halogen if concentrated halide	Cathode: $2 H^{+} + 2 e^{-} \to H_{2}$ Anode (conc NaCl): $2 C l^{-} \to C l_{2} + 2 e^{-}$
Aqueous (metal < H in reactivity)	Solid metal deposit	Oxygen ( $O_{2}$ ) for sulfate/nitrate anions	Cathode: $C u^{2 +} + 2 e^{-} \to C u$ Anode (dilute $C u S O_{4}$ ): $4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}$

Electrolyte Type

Cathode Product Rule

Anode Product Rule

Example Half Equations

Molten binary ionic

Metal from compound cation

Non-metal from compound anion

Cathode: $P b^{2 +} + 2 e^{-} \to P b$

Anode: $2 B r^{-} \to B r_{2} + 2 e^{-}$

Aqueous (metal > H in reactivity)

Hydrogen gas ( $H_{2}$ )

O₂ if sulfate/nitrate/dilute halide; halogen if concentrated halide

Cathode: $2 H^{+} + 2 e^{-} \to H_{2}$

Anode (conc NaCl): $2 C l^{-} \to C l_{2} + 2 e^{-}$

Aqueous (metal < H in reactivity)

Solid metal deposit

Oxygen ( $O_{2}$ ) for sulfate/nitrate anions

Cathode: $C u^{2 +} + 2 e^{-} \to C u$

Anode (dilute $C u S O_{4}$ ): $4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}$

8. Frequently Asked

Why do solid ionic compounds not conduct electricity?

Solid ionic compounds have ions held in a fixed giant lattice structure, so there are no free-moving charged particles to carry electric current. Ions only become mobile when the compound is molten or dissolved in water.

How do I tell if an electrode reaction is oxidation or reduction?

Use two mnemonics: 1. OIL RIG: Oxidation Is Loss of electrons, Reduction Is Gain of electrons. 2. Red Cat: Reduction occurs at the Cathode, so oxidation always occurs at the anode.

What's Next

Now that you have mastered core electrolysis concepts for Paper 2C, you can apply this knowledge to context-specific applications like the electrolytic extraction of aluminium (covered in Section 2 of the syllabus). Make sure you memorise the product prediction rules and half-equation balancing steps, as these are high-frequency marking points in exams. You should also practice past paper questions on electrolysis to familiarise yourself with common question phrasing and practical method requirements, which are often worth 3-4 marks per question.