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Electrolysis is non-spontaneous redox forced uphill by an external electromotive force. In a galvanic cell (lesson 3 of this course) the natural EMF of the cell drives current through an external circuit and the cell does electrical work on the surroundings; in an electrolytic cell the polarity is reversed and the surroundings do electrical work on the cell, driving a reaction whose ΔG° is positive. The pioneering quantitative work was done by Michael Faraday in 1833–34, who established that the mass of substance deposited at an electrode is proportional to the charge passed (first law) and inversely proportional to the number of electrons per ion when comparing different species (second law). This lesson develops the quantitative relations m = MQ/(nF) and Q = It, applies them to deposition and gas-evolution calculations, predicts products of aqueous electrolysis from competing standard electrode potentials, and surveys the industrial-scale processes — Hall-Héroult aluminium extraction, the chlor-alkali plant, electroplating and electrorefining — that consume an appreciable fraction of national electricity supplies.
Spec mapping (AQA 7405): This lesson maps to §3.1.11 (electrode potentials and electrochemical cells, electrolytic extension). The galvanic-cell formalism from lesson 3 (electrochemical cells, cell diagrams, E°cell) is the comparator throughout; the feasibility analysis of lesson 4 (ΔG° = −nFE°cell) is what electrolysis circumvents by supplying external work. Cross-references run to §3.2.5 (transition-metal extraction and electrochemistry), §3.1.2 (stoichiometry — the n = Q/F bridge from coulombs to moles is the same mole concept developed in the AS quantitative-chemistry topic), and §3.1.8 (thermodynamics — the energy economics of Hall-Héroult). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall items include statement of Faraday's first and second laws, definition of the Faraday constant (the charge on one mole of electrons, F = 96 485 C mol⁻¹), and identification of cathode and anode in an electrolytic cell. AO2 calculations dominate the topic: mass of metal deposited from current and time, volume of gas evolved at an electrode at RTP, time required to deposit a target mass. AO3 reasoning is tested by prediction of products in aqueous electrolysis from competing E° values, rationalisation of overpotential effects (notably why chlor-alkali cells evolve Cl₂ rather than O₂ at the anode despite E° suggesting otherwise), and comparison of molten-salt versus aqueous electrolysis.
A galvanic (voltaic) cell consists of two half-cells connected by a salt bridge and an external wire. The half-cell with the more positive E° (the stronger oxidising couple) is the cathode; the half-cell with the more negative E° is the anode; the natural EMF of the cell, E°cell = E°(cathode) − E°(anode), is positive, and the cell does work on the external circuit by driving electrons from anode to cathode through the wire. The reaction is spontaneous (ΔG° = −nFE°cell < 0).
An electrolytic cell takes the same physical setup and reverses the energetic flow. A DC power supply is connected across the electrodes with EMF exceeding the decomposition voltage of the electrolyte. The applied voltage drives current opposite to the spontaneous direction; the surroundings now do work on the cell, and ΔG of the forced reaction is positive. The cathode is defined by where reduction occurs (connected to the negative terminal of the supply); the anode is where oxidation occurs (positive terminal). The reduction-at-cathode convention is invariant — what changes between galvanic and electrolytic modes is the polarity. A lead-acid car battery on discharge is a galvanic cell; on recharge it is an electrolytic cell with the same chemistry running in reverse — the basis of all secondary batteries.
The fundamental observable in electrolysis is the charge passed, Q, measured in coulombs (C). Charge equals current multiplied by time:
Q = I × t
where I is in amperes (A) and t in seconds (s). A current of 1.00 A flowing for one hour passes 3600 C. Industrial Hall-Héroult pots passing 350 kA for 24 hours handle roughly 3 × 10¹⁰ C — about 300 000 mol of electrons per pot per day.
The first law, established by Faraday in 1834, states that the mass of substance deposited at (or dissolved from) an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte. Mathematically:
m ∝ Q
The proportionality is made quantitative by introducing the Faraday constant, F, the magnitude of charge on one mole of electrons:
F = N_A × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ = 96 485 C mol⁻¹
For a half-equation involving the transfer of n electrons per ion, the moles of substance deposited are related to the charge by:
moles deposited = Q / (n × F)
and the mass deposited is:
m = (M × Q) / (n × F)
where M is the molar mass and n the number of electrons per formula unit of the half-equation — 1 for Ag⁺ + e⁻ → Ag, 2 for Cu²⁺ + 2e⁻ → Cu, 3 for Al³⁺ + 3e⁻ → Al, 4 for the anode half-equation 2H₂O → O₂ + 4H⁺ + 4e⁻. The law is exact provided current efficiency is 100% (every electron passed reduces or oxidises the intended species); deviations occur when competing reactions consume charge. The first law is the operational basis of the coulometer, which measures total charge by weighing an electrolytic deposit.
The second law compares deposition in different electrolytic cells through which the same charge has passed. It states that the masses of different substances deposited (or liberated) by the same quantity of charge are proportional to their molar masses divided by the number of electrons per ion in the relevant half-equation:
m₁ / m₂ = (M₁ / n₁) / (M₂ / n₂)
The quantity M/n is the equivalent weight — the mass of substance equivalent to one mole of electrons. One faraday (96 485 C) deposits 107.9 g of Ag from Ag⁺, 31.75 g of Cu from Cu²⁺, 9.0 g of Al from Al³⁺, or liberates 8.0 g (0.5 mol) of O₂ at an aqueous anode. If three coulometers containing these ions are wired in series, the mass ratio of deposits is fixed at 107.9 : 31.75 : 9.0. The two laws combined give the master equation m = MQ/(nF) which underpins every quantitative electrolysis calculation.
A current of 5.00 A is passed through aqueous copper(II) sulfate solution between inert electrodes for 60.0 minutes. Calculate the mass of copper deposited at the cathode.
Half-equation: Cu²⁺(aq) + 2e⁻ → Cu(s), so n = 2.
Step 1. Charge passed. Q = I × t = 5.00 × (60.0 × 60) = 5.00 × 3600 = 18 000 C.
Step 2. Moles of electrons passed. n(e⁻) = Q/F = 18 000 / 96 485 = 0.1866 mol.
Step 3. Moles of copper. n(Cu) = n(e⁻)/2 = 0.0933 mol.
Step 4. Mass of copper. m(Cu) = n × M = 0.0933 × 63.5 = 5.92 g.
The single-line shortcut is m = MQ/(nF) = (63.5 × 18 000) / (2 × 96 485) = 1 143 000 / 192 970 = 5.92 g, identical to within rounding.
In the same electrolysis (Worked Example 1, inert platinum anode), calculate the volume of oxygen gas evolved at the anode at RTP (24.0 dm³ mol⁻¹).
Anode half-equation: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻, so n = 4.
The charge passed is the same — 18 000 C — because the cathode and anode are wired in series and pass the same current for the same time.
Step 1. Moles of O₂. n(O₂) = Q/(nF) = 18 000 / (4 × 96 485) = 18 000 / 385 940 = 0.04663 mol.
Step 2. Volume of O₂ at RTP. V(O₂) = n × 24.0 = 0.04663 × 24.0 = 1.12 dm³.
Note that n(O₂) is exactly half of n(Cu): for every two electrons reducing one Cu²⁺ at the cathode, the anode must release two electrons, and four electrons release one O₂. Hence n(Cu)/n(O₂) = 2.
Calculate the time, in minutes, required to deposit 1.00 g of silver from aqueous silver nitrate at a current of 0.500 A.
Half-equation: Ag⁺(aq) + e⁻ → Ag(s), so n = 1.
Step 1. Moles of Ag required. n(Ag) = m/M = 1.00 / 107.9 = 9.268 × 10⁻³ mol.
Step 2. Charge required. Q = n × n(e⁻ per ion) × F = 9.268 × 10⁻³ × 1 × 96 485 = 894.2 C.
Step 3. Time. t = Q/I = 894.2 / 0.500 = 1788 s = 29.8 minutes.
This is the calculation a coulometer performs in reverse — a known charge deposits a known mass, so weighing the deposit reveals the charge.
Calculate the mass of aluminium produced by a Hall-Héroult cell operating at 350 kA for 24 hours, assuming 100% current efficiency.
Half-equation: Al³⁺ + 3e⁻ → Al, so n = 3.
Q = I × t = 350 000 × (24 × 3600) = 350 000 × 86 400 = 3.024 × 10¹⁰ C.
m(Al) = MQ/(nF) = (27.0 × 3.024 × 10¹⁰) / (3 × 96 485) = 8.165 × 10¹¹ / 2.895 × 10⁵ = 2 820 000 g ≈ 2.82 tonnes.
Real cells achieve 90–95% current efficiency, giving 2.5–2.7 tonnes per pot per day; a smelter runs several hundred pots in line. Aluminium smelting requires 13–15 MWh per tonne — about 4–5% of global electricity supply.
In aqueous electrolyte, water itself can be oxidised or reduced, and there is always competition with the dissolved ions. The thermodynamic guide is E°: the half-reaction with the most positive E° wins at the cathode (most readily reduced); the most negative E° wins at the anode (most readily oxidised).
At the cathode, M^n+ + ne⁻ → M competes with 2H₂O + 2e⁻ → H₂ + 2OH⁻ (E° ≈ −0.41 V at pH 7 after Nernst correction). For Cu²⁺/Cu (E° = +0.34 V) and Ag⁺/Ag (E° = +0.80 V), the metal is deposited preferentially. For Zn²⁺/Zn (−0.76 V) and Na⁺/Na (−2.71 V), H₂ is evolved instead. The d-block cations (Cu, Ag, Au, Ni, Cr) plate cleanly from aqueous solution; the s-block and active-metal cations (Na, K, Mg, Ca, Al) do not.
At the anode, 2X⁻ → X₂ + 2e⁻ competes with 2H₂O → O₂ + 4H⁺ + 4e⁻ (E° ≈ +0.82 V at pH 7). For dilute NaCl(aq) with inert electrodes, O₂ is liberated because the thermodynamic gap to Cl₂/Cl⁻ (+1.36 V) is not bridged. For concentrated brine (chlor-alkali), the high O₂ overpotential on commercial electrodes inverts the prediction and Cl₂ is liberated. SO₄²⁻ and NO₃⁻ are never oxidised in aqueous electrolysis because the relevant couples (S₂O₈²⁻/SO₄²⁻ at +2.01 V) lie well above water oxidation.
Predictive rules for inert-electrode aqueous electrolysis: at the cathode, deposit the metal if E°(M^n+/M) > E°(H₂O/H₂); otherwise evolve H₂. At the anode, evolve the halogen at concentrated halide; otherwise O₂. If the anode is not inert (e.g. a copper electrode in CuSO₄), the electrode itself is oxidised in preference: Cu → Cu²⁺ + 2e⁻. This is the basis of electrorefining and the classroom CuSO₄/Cu demonstration.
The thermodynamic minimum voltage required to drive an electrolysis at equilibrium is E°cell of the forced reaction. In practice, achieving an industrially useful current density requires substantially more — the excess is the overpotential, η:
V(applied) = E°cell(forced) + η(cathode) + η(anode) + IR(electrolyte)
Activation overpotential (kinetic barrier to electron transfer) is usually the largest contributor and depends strongly on electrode material. Two overpotentials dominate aqueous-electrolysis engineering: hydrogen overpotential is large on mercury (~1.0 V), lead (~0.8 V), zinc and iron (0.4–0.5 V), small on platinum — large enough on Hg that sodium can be reduced from aqueous brine (historical mercury-cell chlor-alkali). Oxygen overpotential is ~0.6–0.8 V on most electrode materials, sufficient to invert the E° prediction in concentrated brine and liberate Cl₂ rather than O₂. The full kinetic description is the Butler-Volmer equation; A-Level students need only recognise that the E° prediction for aqueous-anode chemistry can be reversed by kinetics.
Aluminium cannot be extracted from Al₂O₃ by carbon reduction, so all primary aluminium is produced electrolytically. The process invented independently by Charles Hall and Paul Héroult in 1886 dissolves alumina in molten cryolite (Na₃AlF₆) at about 950 °C, well below alumina's own melting point (2070 °C). The cell uses a carbon-lined steel pot as the cathode and consumable carbon anodes; the half-reactions are Al³⁺ + 3e⁻ → Al(l) at the cathode and 2O²⁻ → O₂ + 4e⁻ at the anode, where the O₂ immediately oxidises the carbon to CO₂. The cell runs at 4–5 V and 150–350 kA, consuming 13–15 MWh per tonne of Al — roughly five to seven times the thermodynamic minimum (~6.3 kWh per tonne from ΔG_f of Al₂O₃). The gap is overpotential plus ohmic losses plus the energy to maintain operating temperature. Smelters cluster near low-cost hydroelectric or geothermal generation (Iceland, Quebec, Norway).
Electrolysis of saturated NaCl(aq) produces three commodity chemicals simultaneously: Cl₂, H₂, NaOH. The modern membrane cell uses a sodium-selective ion-exchange membrane (Nafion) to separate anolyte from catholyte. Anode: 2Cl⁻ → Cl₂ + 2e⁻ on coated titanium; cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ on nickel. Na⁺ migrates through the membrane to give 30–35% NaOH product. World production exceeds 80 million tonnes of Cl₂ per year. Cl₂ rather than O₂ is liberated entirely because of the high O₂ overpotential on industrial anodes — a textbook example of kinetics defeating thermodynamics. Mercury-cathode plants (which produced Na amalgam directly) have been phased out for environmental reasons.
Electroplating deposits a thin metal coating on an article serving as the cathode. Typical baths include NiSO₄/NiCl₂/H₃BO₃ (Watts) for nickel, CuSO₄/H₂SO₄ for copper, CrO₃/H₂SO₄ for chromium, K[Au(CN)₂] for gold. The anode is usually the same metal as the deposit, acting sacrificially to maintain bath composition. Plating thickness is set by Faraday's law: for 25 µm Ni on a 100 cm² article, m = 2.23 g, n = 0.0380 mol, Q = 7330 C — about 24 min at 5 A.
Electrorefining uses an impure copper anode and a thin pure-copper cathode in acidified CuSO₄(aq). The applied voltage is held just above 0.3 V, so only Cu²⁺/Cu chemistry is active. Anode: Cu(impure) → Cu²⁺ + 2e⁻ (active impurities Fe, Ni, Zn also dissolve but stay in solution because the voltage is too low to redeposit them); less active impurities (Au, Ag, Pt) are not oxidised and fall as anode slime. The slime is commercially valuable — many copper refineries recover their entire annual gold and silver output from it.
The canonical A-Level demonstration uses two cleaned copper foil electrodes in 0.5 mol dm⁻³ CuSO₄(aq), connected through an ammeter to a low-voltage DC supply, with each electrode weighed before and after.
Procedure: clean each electrode with emery paper, wash, dry, and weigh to ±0.001 g. Electrolyse at 0.2–0.5 A for 15–30 minutes. Rinse, dry, and reweigh both.
Observation: the cathode gains mass (Cu²⁺ + 2e⁻ → Cu); the anode loses mass (Cu → Cu²⁺ + 2e⁻). The mass changes are equal to within 1–3% because the same number of electrons flows through both. The blue colour of the solution is unchanged because anodic dissolution replaces cathodic deposition.
Quantitative check: at 0.300 A for 1800 s, Q = 540 C; n(Cu) = 540/(2 × 96 485) = 2.80 × 10⁻³ mol; expected mass change = 0.178 g. Real values lie in 0.170–0.185 g; deviations arise from non-adherent deposits, parasitic side reactions, and current drift.
Question 1. [13 marks total]
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