Electrolytic Cells
Key Terms
- Anode: positive electrode. Anions migrate here and are oxidised (lose electrons). OIL: Oxidation Is Loss.
- Cathode: negative electrode. Cations migrate here and are reduced (gain electrons). RIG: Reduction Is Gain.
- Electrolyte: ionic conductor (molten salt or aqueous solution).
- Current (I): flow of charge; I = Q/t (amps = coulombs/second).
- Charge (Q): Q = I × t (coulombs = amps × seconds).
Electrolytic cell: DC supply drives non-spontaneous redox reactions
Galvanic (voltaic) cell: generates electrical energy from spontaneous reaction; anode = negative, cathode = positive.
Factors Affecting Discharge of Ions
Factor 1: Position in the Electrochemical Series
Ions are discharged in order of their standard electrode potential (E°).
- At cathode (reduction): ions with the highest (most positive) E° are preferentially reduced. Easy to reduce = low in activity series / high in electrochemical series.
- At anode (oxidation): ions with the lowest (most negative) E°) are preferentially oxidised — i.e. easiest to oxidise = highest in activity series.
| Ion | Electrode Reaction | E° (V) | Ease of Discharge at Cathode |
|---|---|---|---|
| Ag+ | Ag+ + e− → Ag | +0.80 | Easiest (most positive E°) |
| Cu2+ | Cu2+ + 2e− → Cu | +0.34 | Easy |
| H+ | 2H+ + 2e− → H2 | 0.00 | Moderate |
| Zn2+ | Zn2+ + 2e− → Zn | −0.76 | Difficult |
| Na+ | Na+ + e− → Na | −2.71 | Very difficult |
Factor 2: Concentration of Ions
Higher concentration of an ion increases its tendency to be discharged. In concentrated solutions, an ion may be preferentially discharged over one that would normally be favoured thermodynamically.
Example: in dilute HCl, O2 is evolved at the anode (OH−/H2O oxidised). In concentrated HCl, Cl2 is evolved at the anode (high [Cl−] overcomes the thermodynamic preference for O2).
Factor 3: Nature of the Electrode (Overpotential)
Inert electrodes (Pt, graphite/carbon): do not participate in electrode reactions. Products are solely from solution ions.
Active electrodes (Cu, Ag, Ni): the electrode itself dissolves at the anode and deposits at the cathode. Example: Cu anode in CuSO4 solution → Cu oxidised at anode, Cu2+ reduced at cathode.
Overpotential: additional voltage needed beyond the theoretical E° to actually drive the reaction at a given electrode. H2 and O2 have high overpotentials on most electrodes, explaining why Cl2 can be preferentially discharged in concentrated Cl−.
Electrolysis of Water (Dilute H₂SO₄)
Pure water is a very poor conductor. A small amount of H2SO4 (or NaOH) is added to provide ions. The SO42− ions are not discharged; H2O molecules are oxidised at the anode.
Hydrogen is produced at the cathode in twice the volume of oxygen at the anode.
O2: relights a glowing splint.
Electrolysis of Copper(II) Sulfate Solution
Case A: Inert Platinum or Carbon Electrodes
Case B: Copper Electrodes (Active Electrodes)
Active (Cu): Cu deposited at cathode; Cu anode dissolves; [Cu²⁺] stays constant; electrodes change mass.
Predicting Electrode Products
Predict the products at each electrode when dilute CuSO₄ solution is electrolysed using: (a) graphite electrodes; (b) copper electrodes.
Ions present: Cu²⁺, SO₄²⁻, H⁺, OH⁻ (from water).
Cathode: Cu²⁺ discharged in preference to H⁺ (E°(Cu²⁺) = +0.34 V > E°(H⁺) = 0.00 V) → copper metal deposited.
Anode: OH⁻/H₂O discharged in preference to SO₄²⁻ → oxygen gas evolved.
Cathode: Cu²⁺ + 2e⁻ → Cu → copper deposited (cathode gains mass).
Anode: Cu → Cu²⁺ + 2e⁻ → copper anode dissolves (anode loses mass).
Electrolysis of Brine (Concentrated NaCl Solution)
| Product | Electrode | Test | Industrial Use |
|---|---|---|---|
| Chlorine (Cl2) | Anode (+) | Turns damp starch-iodide paper blue; bleaches damp litmus paper | PVC, disinfectants, bleach, HCl manufacture |
| Hydrogen (H2) | Cathode (−) | Squeaky pop with lit splint | Fuel, Haber process (NH3), hydrogenation |
| Sodium hydroxide (NaOH) | Remains in solution | Red litmus → blue; pH > 7 | Soap, paper pulp, aluminium production, rayon |
Faraday's First Law of Electrolysis
m ∝ Q or m = Z × Q
where m = mass deposited (g), Q = charge (coulombs = A×s), Z = electrochemical equivalent (g/C).
Working Formula
Faraday's First Law Calculation
A current of 2.00 A is passed through CuSO₄ solution for 30 minutes. Calculate the mass of copper deposited. (Mr(Cu) = 63.5, z = 2, F = 96,500 C/mol)
Volume of Gas Produced
Calculate the volume of O₂ produced at STP when 0.500 A flows through acidified water for 2.00 hours. (Molar volume at STP = 22.4 L/mol)
n(O₂) = 0.03731/4 = 9.327×10⁻³ mol
Faraday's Second Law of Electrolysis
m1/m2 = (M1/z1) / (M2/z2)
where M/z is the equivalent mass (or chemical equivalent) of the substance.
Equivalence of the Faraday
One Faraday (96,500 C) deposits one mole of equivalents of any substance:
Faraday's Second Law — Two Cells in Series
The same charge is passed through a silver coulometer and a copper coulometer. 5.40 g of Ag is deposited. Find the mass of Cu deposited. (Mr(Ag)=107.9, z=1; Mr(Cu)=63.5, z=2)
Industrial Applications of Electrolysis
| Application | Electrolyte | Cathode Product | Anode Product | Significance |
|---|---|---|---|---|
| Chlor-alkali industry | Conc. NaCl (brine) | H2 | Cl2 | Also NaOH in solution; key industrial chemicals |
| Copper refining | CuSO4/H2SO4 | Pure Cu (99.99%) | Impure Cu anode dissolves | Precious metals (Au, Ag) collect as anode sludge |
| Aluminium extraction | Molten Al2O3/cryolite | Al metal | CO2 (C anodes burn) | Hall-Héroult process; energy-intensive |
| Electroplating | Salt of plating metal | Metal deposited on object | Plating metal dissolves | Protective coatings (Ni, Cr, Ag, Au on base metals) |
| Electroforming | Metal salt solution | Thick metal deposit | Metal anode | Making moulds, gramophone records, printing plates |
| Electrolytic H₂ production | Alkaline water or PEM | H2 | O2 | Green H₂ from renewable electricity; fuel cells |
| Sodium production | Molten NaCl (Downs cell) | Na metal | Cl2 | Na used in making TiCl₄ → Ti metal |
| Anodising aluminium | H2SO4 | — | Al oxidised to Al2O3 | Protective, dyeable oxide layer on Al surface |
Copper Electrorefining — Detail
Impure copper (from smelting, ~98% Cu) is cast as the anode. A thin sheet of pure copper is the cathode. Electrolyte: CuSO4/H2SO4.
- At anode: impure Cu dissolves along with impurities (Fe, Zn, Ni dissolve as ions; Au, Ag, Pt don't dissolve → fall as anode sludge).
- At cathode: only Cu²⁺ is reduced (more noble impurity ions like Fe²⁺, Ni²⁺ remain in solution and are not deposited).
- Result: 99.99% pure copper at cathode; precious metals recovered from anode sludge.
Aluminium Extraction (Hall-Héroult Process)
This process is very energy-intensive (~15 kWh/kg Al), which is why aluminium recycling (using only 5% of the energy) is so important.
Electroplating Calculation
A steel object is nickel-plated using a NiSO₄ solution. A current of 3.00 A flows for 45 minutes. Calculate: (a) the mass of Ni deposited; (b) the thickness of the deposit if the plated area is 50 cm². (Mr(Ni)=58.7, z=2, density of Ni=8.9 g/cm³)
Thickness = V/area = 0.2764/50 = 5.53×10⁻³ cm = 0.0553 mm
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Exercises
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State the products at each electrode when: (a) dilute H₂SO₄ is electrolysed with platinum electrodes; (b) molten NaCl is electrolysed; (c) concentrated NaCl solution is electrolysed with carbon electrodes; (d) dilute CuSO₄ is electrolysed with copper electrodes.
(a) Cathode: H₂; Anode: O₂. (2H⁺ + 2e⁻ → H₂; 2H₂O → O₂ + 4H⁺ + 4e⁻)
(b) Cathode: Na metal; Anode: Cl₂. (Na⁺ + e⁻ → Na; 2Cl⁻ → Cl₂ + 2e⁻)
(c) Cathode: H₂; Anode: Cl₂. (2H⁺ + 2e⁻ → H₂; 2Cl⁻ → Cl₂ + 2e⁻)
(d) Cathode: Cu deposited; Anode: Cu dissolves. ([Cu²⁺] remains constant) -
A current of 1.50 A is passed through silver nitrate solution for 20 minutes. Calculate the mass of silver deposited. (Mr(Ag)=108, z=1, F=96500 C/mol)
Q = 1.50 × 20 × 60 = 1800 C
n(e⁻) = 1800/96500 = 0.01865 mol
Ag⁺ + e⁻ → Ag (z=1): n(Ag) = 0.01865 mol
m(Ag) = 0.01865 × 108 = 2.01 g -
The same quantity of electricity deposits 1.080 g of silver and a mass of zinc. Calculate the mass of zinc deposited. (Mr(Ag)=108, z=1; Mr(Zn)=65.4, z=2)
n(Ag) = 1.080/108 = 0.01000 mol; n(e⁻) = 0.01000 mol
Zn²⁺ + 2e⁻ → Zn: n(Zn) = 0.01000/2 = 0.005000 mol
m(Zn) = 0.005000 × 65.4 = 0.327 g
[Or using 2nd law: m(Zn)/m(Ag) = (Mr(Zn)/2)/(Mr(Ag)/1) = 32.7/108 = 0.3028; m(Zn) = 1.080 × 0.3028 = 0.327 g] -
Explain why aluminium cannot be extracted by the electrolysis of aqueous AlCl₃ solution, but molten Al₂O₃/cryolite is used instead.
In aqueous AlCl₃, water is present. At the cathode, H⁺ (from water) is much easier to reduce than Al³⁺ (E°(Al³⁺/Al) = −1.66 V vs E°(H⁺/H₂) = 0.00 V). So H₂ would be produced instead of Al metal — aluminium cannot be deposited from aqueous solution.
In the Hall-Héroult process, Al₂O₃ is dissolved in molten cryolite (Na₃AlF₆) at ~960°C. There is no water present, so only Al³⁺ ions are available for reduction at the cathode. Despite the very negative E°, the absence of competing H⁺ allows Al to be deposited. -
How long must a current of 0.800 A flow to deposit 3.18 g of copper from CuSO₄ solution? (Mr(Cu)=63.5, z=2, F=96500 C/mol)
n(Cu) = 3.18/63.5 = 0.05008 mol
n(e⁻) = 2 × 0.05008 = 0.10016 mol
Q = n(e⁻) × F = 0.10016 × 96500 = 9665 C
t = Q/I = 9665/0.800 = 12081 s = 201 minutes (3 hours 21 min) -
Write all electrode half-equations for the industrial electrolysis of brine. State three products and give one industrial use for each.
Cathode: 2H⁺ + 2e⁻ → H₂ (or 2H₂O + 2e⁻ → H₂ + 2OH⁻)
Anode: 2Cl⁻ → Cl₂ + 2e⁻
Net: 2NaCl + 2H₂O → Cl₂ + H₂ + 2NaOH
Products and uses:
1. Chlorine: manufacture of PVC; disinfection of water; making bleach (NaOCl).
2. Hydrogen: Haber process (making NH₃); fuel for fuel cells; hydrogenation of oils.
3. Sodium hydroxide: making soap; processing paper pulp; manufacturing aluminium (dissolving bauxite).
Interactive Quiz
Unit 13 Quiz — Electrolysis
25 QuestionsIn an electrolytic cell, oxidation occurs at the:
The Faraday constant (96,500 C/mol) represents:
When dilute H₂SO₄ is electrolysed, which gas is produced at the cathode?
Faraday's First Law states that the mass deposited at an electrode is proportional to:
A current of 2A passes for 1930 seconds through CuSO₄. Moles of Cu deposited (z=2, F=96500):
In the electrolysis of concentrated NaCl solution, the product at the anode is:
When copper electrodes are used in the electrolysis of CuSO₄, the concentration of Cu²⁺ in solution:
Faraday's Second Law states that for the same quantity of electricity, the masses of different substances deposited are proportional to:
The Hall-Héroult process extracts aluminium from:
Why is AC current not suitable for electrolysis?
The volume ratio of H₂:O₂ produced in the electrolysis of water (dilute H₂SO₄) is:
In copper electrorefining: impure copper is used as the:
How much charge is needed to deposit 3.175 g of copper (Mr=63.5, z=2, F=96500)?
Why must cryolite be added to Al₂O₃ in the Hall-Héroult process?
The electrochemical equivalent Z (g/C) for silver (Mr=108, z=1, F=96500) is:
During electrolysis of dilute CuSO₄ with platinum electrodes, the solution gradually:
Electroplating a steel spoon with silver uses the spoon as:
In the electrolysis of molten NaCl, sodium is produced at the cathode because:
The test for chlorine gas involves:
The equivalent mass of Al (Mr=27, z=3) is:
Which ion is preferentially discharged at the cathode from a solution containing both Cu²⁺ and Zn²⁺?
Anodising aluminium uses H₂SO₄ as electrolyte and aluminium as the:
A current of 3A deposits 1.98 g of metal X (z=2) in 30 minutes. The molar mass of X is:
The three products of the industrial electrolysis of brine are:
Which of the following correctly states Faraday's Second Law?
Unit Test
Section A — Short Answer
30 marksFor each electrolysis below, state the product at each electrode and write the half-equation:
(a) Molten PbBr₂ with carbon electrodes
(b) Aqueous NiSO₄ with nickel electrodes
(c) Dilute H₂SO₄ with platinum electrodes
(d) Concentrated HCl with graphite electrodes
(b) NiSO₄/Ni electrodes: Cathode: Ni²⁺ + 2e⁻ → Ni(s) (nickel deposited); Anode: Ni(s) → Ni²⁺ + 2e⁻ (nickel dissolves) — [Ni²⁺] remains constant
(c) Dilute H₂SO₄/Pt: Cathode: 2H⁺ + 2e⁻ → H₂(g); Anode: 2H₂O → O₂(g) + 4H⁺ + 4e⁻
(d) Conc. HCl/graphite: Cathode: 2H⁺ + 2e⁻ → H₂(g); Anode: 2Cl⁻ → Cl₂(g) + 2e⁻
A current of 0.500 A is passed through molten AlCl₃ for 2.00 hours. Calculate:
(a) Charge passed; (b) Moles of electrons; (c) Moles of Al deposited; (d) Mass of Al deposited (Mr=27.0, z=3); (e) Volume of Cl₂ evolved at STP (molar vol = 22.4 L/mol)
(b) n(e⁻) = 3600/96500 = 0.03731 mol
(c) Al³⁺ + 3e⁻ → Al; n(Al) = 0.03731/3 = 0.01244 mol
(d) m(Al) = 0.01244 × 27.0 = 0.336 g
(e) 2Cl⁻ → Cl₂ + 2e⁻; n(Cl₂) = 0.03731/2 = 0.01866 mol; V(Cl₂) = 0.01866 × 22.4 = 0.418 L = 418 cm³
In an electrolysis experiment, three cells connected in series contain: AgNO₃ solution, CuSO₄ solution, and ZnSO₄ solution. A total of 96,500 C of electricity is passed through all three.
(a) Moles of Ag deposited (z=1); (b) Moles of Cu deposited (z=2); (c) Moles of Zn deposited (z=2);
(d) Mass of each metal deposited (Mr: Ag=108, Cu=63.5, Zn=65.4);
(e) State which law is illustrated and explain it using your results.
(a) n(Ag) = 1.000/1 = 1.000 mol
(b) n(Cu) = 1.000/2 = 0.500 mol
(c) n(Zn) = 1.000/2 = 0.500 mol
(d) m(Ag) = 1.000×108 = 108 g; m(Cu) = 0.500×63.5 = 31.75 g; m(Zn) = 0.500×65.4 = 32.7 g
(e) Faraday's Second Law: the same charge (1F) deposits different masses of different metals, proportional to their equivalent masses (Mr/z). Equivalent masses: Ag=108/1=108; Cu=63.5/2=31.75; Zn=65.4/2=32.7. The masses deposited (108:31.75:32.7) are exactly in the ratio of their equivalent masses.
Describe the industrial electrolysis of brine (the chlor-alkali process). Include: the electrode reactions, the three products, how each product is collected/separated, and three industrial uses for each product.
Cathode: 2H⁺(aq) + 2e⁻ → H₂(g)
Anode: 2Cl⁻(aq) → Cl₂(g) + 2e⁻
Net: 2NaCl + 2H₂O → Cl₂ + H₂ + 2NaOH
Products and collection:
1. Chlorine (Cl₂): rises from anode as gas, collected and dried. Industrial uses: (i) PVC manufacture; (ii) water disinfection; (iii) making bleach (NaOCl); (iv) solvents; (v) pesticides.
2. Hydrogen (H₂): rises from cathode as gas, collected. Industrial uses: (i) Haber process for NH₃; (ii) fuel for fuel cells; (iii) hydrogenation of vegetable oils; (iv) refinery processes.
3. Sodium hydroxide (NaOH): remains in solution; concentrated by evaporation. Industrial uses: (i) soap and detergent manufacture; (ii) paper pulp processing; (iii) aluminium production (dissolving bauxite Al₂O₃); (iv) rayon fibre production.
Separation: The cell must be designed to keep Cl₂ and H₂ separate (explosive if mixed) and to prevent NaOH mixing with Cl₂ (forms hypochlorite). Membrane cells or mercury cells are used.
Explain the factors that determine which ion is preferentially discharged at each electrode during electrolysis. Illustrate with the example of aqueous CuSO₄ electrolysis using: (a) platinum electrodes; (b) copper electrodes. Explain the differences observed.
1. Position in electrochemical series (E°): at cathode, ion with highest E° is preferentially reduced; at anode, ion with lowest E° (easiest to oxidise) is preferentially oxidised.
2. Concentration: higher [ion] favours that ion's discharge.
3. Nature of electrode: inert electrodes don't react; active electrodes (same metal as deposit) dissolve preferentially at anode.
4. Overpotential: some reactions require extra voltage; e.g. O₂/H₂ have high overpotentials.
(a) Platinum electrodes:
Cathode: E°(Cu²⁺/Cu)=+0.34V > E°(H⁺/H₂)=0.00V → Cu²⁺ preferentially discharged → copper metal deposited (blue solution fades). Anode: H₂O more easily oxidised than SO₄²⁻ → O₂ evolved. [Cu²⁺] decreases; solution becomes more acidic.
(b) Copper electrodes:
Cathode: Cu²⁺ + 2e⁻ → Cu (as before). Anode: copper electrode itself preferentially oxidises (Cu → Cu²⁺ + 2e⁻) rather than water/SO₄²⁻ — because the active Cu anode has lower activation energy for dissolution than oxidising water. Result: Cu transfers from anode to cathode; [Cu²⁺] stays constant. Anode loses mass; cathode gains mass equally.
A student electroplates a steel key with chromium using a CrSO₄ solution. The key has a surface area of 30.0 cm² and the desired coating thickness is 0.020 mm. Density of Cr = 7.15 g/cm³; Mr(Cr) = 52.0; z = 3; F = 96,500 C/mol.
(a) Calculate the volume of Cr needed; (b) mass of Cr needed; (c) moles of Cr; (d) charge required; (e) time needed at 1.50 A; (f) write the cathode half-equation.
(b) m = V × density = 0.0600 × 7.15 = 0.429 g
(c) n(Cr) = 0.429/52.0 = 8.25×10⁻³ mol
(d) n(e⁻) = 3 × n(Cr) = 3 × 8.25×10⁻³ = 0.02475 mol; Q = 0.02475 × 96500 = 2389 C ≈ 2390 C
(e) t = Q/I = 2389/1.50 = 1593 s ≈ 26.5 minutes
(f) Cr³⁺ + 3e⁻ → Cr(s)
Section B — Extended Response
20 marks(a) State Faraday's First and Second Laws of Electrolysis. For each law, write the relevant formula and give a numerical example. [6 marks]
(b) A silver coulometer and a copper coulometer are connected in series. The silver coulometer deposits 5.394 g of Ag. Calculate: the charge passed; the mass of Cu deposited; and verify Faraday's Second Law using the equivalent masses. (Mr: Ag=107.9, Cu=63.5; z: Ag=1, Cu=2; F=96500) [4 marks]
The mass of substance deposited is directly proportional to the charge passed: m = (I×t×Mr)/(z×F)
Example: 2A for 30 min through CuSO₄: Q=3600C; m(Cu) = (3600×63.5)/(2×96500) = 228600/193000 = 1.185 g
Faraday's Second Law:
For the same charge, masses deposited ∝ equivalent masses (Mr/z): m₁/m₂ = (M₁/z₁)/(M₂/z₂)
Example: 1F deposits 108 g Ag (equivalent mass 108) and 31.75 g Cu (equivalent mass 31.75). Ratio = 108:31.75 = 3.40:1.
(b)
n(Ag) = 5.394/107.9 = 0.05000 mol
Q = n(Ag) × F / z(Ag) = 0.05000 × 96500 / 1 = 4825 C
n(Cu) = Q/(z×F) = 4825/(2×96500) = 0.02500 mol
m(Cu) = 0.02500 × 63.5 = 1.588 g
Verification: m(Ag)/m(Cu) = 5.394/1.588 = 3.398; (Mr(Ag)/z(Ag))/(Mr(Cu)/z(Cu)) = 107.9/31.75 = 3.399 ✓ — Faraday's Second Law confirmed.
(a) Describe the Hall-Héroult process for the extraction of aluminium. Include: raw materials, why cryolite is used, electrode reactions with half-equations, why carbon anodes must be regularly replaced, and why recycling aluminium saves energy. [6 marks]
(b) Compare the electrolysis of concentrated NaCl(aq) with the electrolysis of molten NaCl. State the products at each electrode for both cases and explain any differences, particularly why different products form at the cathode. [4 marks]
Raw materials: Al₂O₃ (alumina, from bauxite ore by Bayer process) dissolved in molten cryolite (Na₃AlF₆).
Why cryolite: Pure Al₂O₃ melts at ~2050°C (impractical). Dissolving in cryolite reduces the melting point to ~960°C, which is achievable with current industrial furnaces and reduces energy costs enormously.
Electrode reactions:
Cathode: Al³⁺ + 3e⁻ → Al(l) [aluminium collects as liquid at bottom of cell]
Anode: 2O²⁻ → O₂ + 4e⁻ [oxygen produced]; C + O₂ → CO₂ [carbon anodes react with O₂]
Net overall: 2Al₂O₃ + 3C → 4Al + 3CO₂
Why anodes replaced: Carbon anodes react with the O₂ produced: C + O₂ → CO₂ and 2C + O₂ → 2CO. The anodes gradually burn away and must be periodically replaced. This increases operating costs and produces CO₂/CO greenhouse gases.
Recycling saves energy: Primary production requires ~15 kWh per kg of Al (enormous electrical energy). Recycling molten aluminium scrap requires only ~0.75 kWh/kg (~5% of primary production energy) — because the energy-intensive electrolytic reduction step is bypassed.
(b) Comparison:
Molten NaCl: Contains only Na⁺ and Cl⁻ ions. Cathode: Na⁺ + e⁻ → Na(l) [sodium metal — only cation available]. Anode: 2Cl⁻ → Cl₂ + 2e⁻.
Concentrated NaCl(aq): Contains Na⁺, Cl⁻, H⁺, and OH⁻ (from water ionisation). Cathode: 2H⁺ + 2e⁻ → H₂ [H⁺ preferentially reduced, E°=0.00V, while E°(Na⁺/Na)=−2.71V — Na⁺ far harder to reduce]. Anode: 2Cl⁻ → Cl₂ + 2e⁻ (in conc. solution).
Key difference at cathode: In aqueous solution, H⁺ (from water) is reduced instead of Na⁺ because E°(H⁺/H₂) = 0.00 V is far more positive than E°(Na⁺/Na) = −2.71 V. Without water (molten), only Na⁺ is available, so Na metal is produced. This is why molten salt electrolysis is essential for producing reactive metals like Na, K, Li.