Electrolysis

· Electrolysis = decomposition of an ionic compound using electricity.
· Requires an electrolyte with mobile ions: either molten ionic compound or aqueous ionic solution.
· Electrolytic cell uses a direct current (d.c.) power supply to force a non-spontaneous redox reaction.
· Cathode = negative electrode in electrolysis; attracts cations.
· Anode = positive electrode in electrolysis; attracts anions.
· Reduction occurs at the cathode: ions gain electrons.
· Oxidation occurs at the anode: ions lose electrons.
· Exam memory aid: OIL RIG = Oxidation Is Loss, Reduction Is Gain of electrons.

This diagram summarises the essential structure of an electrolytic cell. It clearly shows cation movement to the cathode and anion movement to the anode, which is key for predicting electrolysis products. Source

Predicting Products: Molten Electrolytes

· In a molten ionic compound, only the compound’s own ions are present.
· Cation → cathode → reduced to the element.
· Anion → anode → oxidised to the element.
· Example: molten NaCl(l)
· Cathode: Na⁺ + e⁻ → Na
· Anode: 2Cl⁻ → Cl₂ + 2e⁻
· Overall: 2NaCl(l) → 2Na(l) + Cl₂(g)
· For molten salts, product prediction is usually straightforward because water is absent.

This diagram shows how molten sodium chloride is electrolysed to produce sodium metal and chlorine gas. It reinforces that molten electrolytes contain only their own ions, so Na⁺ is reduced and Cl⁻ is oxidised. Source

Predicting Products: Aqueous Electrolytes

· In aqueous solutions, ions from water are also present: H⁺(aq) and OH⁻(aq).
· Possible cathode products: metal or hydrogen gas.
· Possible anode products: halogen, oxygen gas, or another oxidised product depending on the ion.
· Product depends on: state of electrolyte, position in the redox series / electrode potential, and concentration.
· At the cathode, the species more easily reduced is discharged.
· Reactive metals such as Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺ usually remain in solution; H₂ forms instead.
· Less reactive metal ions such as Cu²⁺, Ag⁺ can be reduced to the metal.
· At the anode, halide ions such as Cl⁻, Br⁻, I⁻ may form halogens, especially when concentrated.
· If no halide is preferentially discharged, OH⁻ / H₂O is oxidised to form O₂.
· Concentrated chloride solution often gives Cl₂ at the anode; dilute solutions may give O₂.

Electrode Half-Equations

· Always write balanced half-equations using electrons.
· At the cathode, electrons appear on the left because reduction uses electrons.
· Example: Cu²⁺ + 2e⁻ → Cu
· Example: 2H⁺ + 2e⁻ → H₂
· At the anode, electrons appear on the right because oxidation releases electrons.
· Example: 2Cl⁻ → Cl₂ + 2e⁻
· Example: 4OH⁻ → O₂ + 2H₂O + 4e⁻
· To form the overall equation, make the number of electrons equal, then add the half-equations.
· Include state symbols where useful: (s), (l), (g), (aq).

Charge, Current and Time

· Charge passed during electrolysis: Q = It
· Q = charge in coulombs, C
· I = current in amperes, A
· t = time in seconds, s
· Convert time carefully: minutes × 60 = seconds.
· Larger current or longer time means more charge passes.
· More charge passed means more moles of electrons transferred.
· Exam trap: do not use time in minutes unless converted to seconds.

Faraday Constant and Avogadro Constant

· Faraday constant, F = charge carried by 1 mole of electrons.
· Approximate value: F = 96 500 C mol⁻¹.
· Relationship: F = Le
· L = Avogadro constant, in mol⁻¹.
· e = charge on one electron, usually 1.60 × 10⁻¹⁹ C.
· Rearranged: L = F / e.
· Meaning: if you know the charge carried by 1 mol of electrons, and the charge on one electron, you can calculate how many electrons are in one mole.

Quantitative Electrolysis Calculations

· Step 1: Calculate charge using Q = It.
· Step 2: Calculate moles of electrons: mol e⁻ = Q / F.
· Step 3: Use the half-equation mole ratio to find moles of product.
· Step 4: Convert moles to mass using m = nMr or to gas volume using the molar gas volume given in the question.
· For metal deposition: use the ion charge to find electrons needed.
· Example: Cu²⁺ + 2e⁻ → Cu, so 2 mol e⁻ deposit 1 mol Cu.
· Example: Ag⁺ + e⁻ → Ag, so 1 mol e⁻ deposits 1 mol Ag.
· Example: 2Cl⁻ → Cl₂ + 2e⁻, so 2 mol e⁻ form 1 mol Cl₂.
· For gas volumes, use volume = moles × molar gas volume.
· Keep answers to appropriate significant figures and include units.

Determining Avogadro Constant by Electrolysis

· Use electrolysis of a solution such as CuSO₄(aq) with copper electrodes.
· Measure the current, I, and time, t, then calculate Q = It.
· Measure the mass change of a copper electrode.
· Use Cu²⁺ + 2e⁻ → Cu to connect deposited/dissolved copper to moles of electrons.
· Calculate the charge per mole of electrons to estimate F.
· Use F = Le to calculate L, the Avogadro constant.
· Important practical points: clean and dry electrodes, measure mass accurately, keep current steady, and record time precisely.
· Main sources of error: loss of deposited copper, wet electrodes during weighing, fluctuating current, and side reactions.

Common Exam Product Examples

· Molten NaCl → cathode: Na, anode: Cl₂.
· Aqueous CuSO₄ with inert electrodes → cathode: Cu, anode: O₂.
· Concentrated aqueous NaCl / brine → cathode: H₂, anode: Cl₂, solution becomes NaOH(aq).
· Dilute aqueous NaCl → cathode: H₂, anode often O₂.
· Acidified water / dilute H₂SO₄ → cathode: H₂, anode: O₂.
· CuSO₄ with copper electrodes → cathode gains Cu, anode loses Cu; solution concentration stays approximately constant.
· With inert electrodes, electrodes do not take part in the reaction.
· With reactive electrodes, the electrode material may be oxidised or deposited.

This diagram is useful for remembering the 2:1 volume ratio of hydrogen to oxygen in water electrolysis. It also reinforces that H₂ forms at the cathode and O₂ forms at the anode. Source