Electrolysis in Industrial Metal Extraction and Refining
Faraday's laws of electrolysis are usually first met as a small-scale laboratory calculation — a few amps, a few minutes, a few milligrams of metal deposited on an electrode. The same equation, completely unchanged, is what an aluminium smelter uses to plan how many tonnes of metal a potline will produce in a day, and what a copper refinery uses to plan how pure the copper coming off its cathodes will be. This article scales the calculation up to industrial size and introduces the one correction that real cells always need: current efficiency.
The formula
Current efficiency: η = (actual mass deposited / theoretical mass) × 100%
What each term means
| Term | Meaning | Unit |
|---|---|---|
| m | Mass of metal deposited or dissolved | g |
| I | Current passed through the cell | A |
| t | Time the current is passed | s |
| M | Molar mass of the metal | g/mol |
| n | Number of electrons needed to deposit or dissolve one ion of the metal (its charge) | dimensionless |
| F | Faraday constant | 96 485 C/mol |
Worked example 1 — aluminium production by the Hall–Héroult process
Aluminium is extracted by electrolysing alumina (Al₂O₃) dissolved in molten cryolite — dissolving it lowers the melting point from around 2 050 °C down to roughly 950–980 °C, which is what makes the whole process energy-feasible at all. At the cathode, Al³⁺ + 3e⁻ → Al, so n = 3. Take a plausible industrial cell current of 150 000 A run for 24 hours, and find the mass of aluminium deposited (atomic mass of Al = 26.98).
t = 24 h = 24 × 3 600 = 86 400 s
Charge passed: Q = I × t = 150 000 × 86 400 = 1.296 × 10¹⁰ C
Moles of electrons: Q ÷ F = (1.296 × 10¹⁰) ÷ 96 485 = 134 322 mol e⁻
Moles of Al: 134 322 ÷ 3 = 44 774 mol
Mass: 44 774 × 26.98 = 1 208 000 g
≈ 1 208 kg, about 1.21 tonnes of aluminium in 24 hours from this one cell, at 100% current efficiency
Real smelting cells (called "pots") are wired in long series circuits of hundreds of pots carrying the same current one after another, precisely because Faraday's law says charge — not voltage — is what determines the mass deposited, so putting cells in series multiplies total aluminium output for the same current without multiplying the current itself.
Worked example 2 — current efficiency
Suppose the cell above is actually measured to deposit only 1 112 kg of aluminium in the same 24 hours, rather than the theoretical 1 208 kg calculated from Faraday's law. Find the current efficiency.
η = (actual ÷ theoretical) × 100% = (1 112 ÷ 1 208) × 100%
η ≈ 92.1%
The missing 7.9% is not an error in Faraday's law — the law itself is exact. It represents charge consumed by side reactions (such as some re-oxidation of aluminium already deposited, or parasitic reduction of other species present) instead of by the intended metal-deposition reaction, which is why current efficiency, not just current, is a routine reported figure in electrolytic metal production.
Where this is actually used
The Hall–Héroult process above is how essentially all primary aluminium is produced worldwide; the carbon anodes are slowly consumed as they react with the liberated oxygen to form CO₂, so anode replacement is itself a routine, planned part of running a smelter. Copper is purified by a related but distinct process, electrorefining: impure "blister" copper is cast as the anode and dissolves into a copper sulfate electrolyte, while pure copper deposits at the cathode (Cu²⁺ + 2e⁻ → Cu, n = 2). The valuable difference is what happens to the impurities — more reactive metals stay dissolved in the electrolyte, while less reactive ones (including silver and gold present in trace amounts) do not dissolve at all and fall to the bottom of the cell as anode sludge, which is itself collected and processed for those metals. Electrolysis is used specifically where a metal is too reactive to be reduced by carbon or another chemical reducing agent — sodium, magnesium and aluminium all sit above carbon in reactivity and are extracted electrolytically for exactly that reason.
Common mistakes that cost marks
- Forgetting n, or using the wrong value of n. n is the ionic charge of the metal being deposited (3 for Al³⁺, 2 for Cu²⁺), not the number of atoms in a formula unit — using the wrong n scales the whole answer incorrectly.
- Assuming real cells run at 100% current efficiency. Faraday's law gives the theoretical maximum mass for the charge passed; industrial cells routinely run below 100% because of side reactions, and the actual output must be scaled down accordingly.
- Mixing up units of time. t must be in seconds to match F in coulombs per mole; forgetting to convert hours or days to seconds is the single most common arithmetic slip in this topic.
- Confusing electrolysis (extraction/refining, energy is put in) with a galvanic cell (energy is produced). The same Nernst-family thinking about electrode potentials applies to both, but electrolysis forces a non-spontaneous reaction to occur by supplying external electrical work.
Exam relevance
| Exam | Typical use |
|---|---|
| IIT-JAM / CUET-PG Physical Chemistry | Faraday's law numericals, current efficiency calculations |
| GATE Chemistry | Electrolytic processes, metallurgy and extraction methods |
| CSIR-NET Physical Chemistry | Electrolysis, electrode reactions, quantitative electrochemistry |
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