Electrochemistry
Faraday's Laws of Electrolysis
Charge in, atoms out — exactly
Faraday's first law says the mass deposited at an electrode is proportional to the charge passed. His second says equal charge deposits chemically equivalent amounts of any element. One ampere-hour passes 0.0373 mol of electrons — enough to deposit 1.19 g of copper or 4.02 g of silver.
- Faraday constant F96 485 C/mol
- Electrons per ampere-second6.24 × 10¹⁸
- Moles of e⁻ per A·hr0.0373
- First publicationFaraday, 1834
- Cu plated by 1 A·hr1.19 g
- Al smelted per kWh~83 g
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The two laws, stated cleanly
Michael Faraday published these two laws together in 1834, before electrons were known to exist. He stated them as quantitative observations about how much chemistry one could do with a given quantity of "electricity." A century later, when J.J. Thomson identified the electron, Faraday's two laws became one fact about counting charges.
First law. The mass m of substance liberated at an electrode is proportional to the charge Q passed:
m = Z · Q
where Z is the electrochemical equivalent of the substance.
Second law. For the same charge through different electrolytes, the masses liberated are in the ratio of their chemical equivalents (molar mass / valence):
m₁ : m₂ : m₃ = (M₁/n₁) : (M₂/n₂) : (M₃/n₃)
The two combine into the single working formula:
┌──────────────────────┐
│ Q · M │
│ m = ───── │
│ n · F │
└──────────────────────┘
m = mass liberated (g)
Q = charge passed (coulombs) = I × t
M = molar mass (g/mol)
n = electrons per ion
F = 96 485 C/mol
Equivalently, in moles: moles of substance = Q / (n·F). This is the bookkeeping equation behind every electroplating bath, every aluminum pot line, and every battery capacity rating.
Worked numbers — the canonical 1 A · 1 hr
Pass 1 ampere for 1 hour through any electrolyte. How many electrons cross?
- Charge: Q = I·t = 1 × 3600 = 3600 C.
- Moles of electrons: Q / F = 3600 / 96 485 = 0.03731 mol — almost exactly 1/27 of a mole.
- Number of electrons: 0.03731 × 6.022 × 10²³ = 2.247 × 10²² electrons.
Now feed that charge into different cells:
| Reaction at cathode | n (e⁻ per ion) | M (g/mol) | Mass deposited per A·hr | Volume at STP if a gas |
|---|---|---|---|---|
| H⁺ + e⁻ → ½ H₂ | 1 | 1.008 | 0.038 g (H₂ gas) | 0.418 L H₂ |
| Ag⁺ + e⁻ → Ag | 1 | 107.87 | 4.025 g | — |
| Cu²⁺ + 2e⁻ → Cu | 2 | 63.55 | 1.186 g | — |
| Au³⁺ + 3e⁻ → Au | 3 | 196.97 | 2.450 g | — |
| Al³⁺ + 3e⁻ → Al | 3 | 26.98 | 0.336 g | — |
| 2 H₂O → O₂ + 4H⁺ + 4e⁻ | 4 | 32.00 (O₂) | 0.298 g | 0.209 L O₂ |
The 4:1 mass ratio of Ag to Cu (4.025 / 1.186 ≈ 3.4) reflects both the higher atomic mass of silver and its single-electron reduction. That's Faraday's second law made concrete.
Worked example: a tonne of aluminum
Aluminum is reduced from molten Al₂O₃ in cryolite at ~960 °C in the Hall-Héroult process. Each Al³⁺ needs 3 electrons:
- Moles of Al per tonne: 1 000 000 g / 26.98 = 37 064 mol.
- Charge required: 37 064 × 3 × 96 485 = 1.073 × 10¹⁰ C.
- At 4 V cell voltage (typical industrial): energy = QV = 4.29 × 10¹⁰ J = 11 920 kWh per tonne.
- Theoretical minimum (E° = 1.66 V): 4940 kWh/t. Real plants run at ~13 000 kWh/t, so current efficiency × voltage efficiency ≈ 38%.
Aluminum smelters consume about 3% of global electricity for this single reaction. Pricing power-coupled aluminum on a kWh basis is more accurate than pricing it on an ore basis.
Different ways the law shows up
| Application | What you compute | From | Typical setpoint |
|---|---|---|---|
| Electroplating thickness | plating time | m = QM/(nF), thickness = m/(ρA) | 30 µm Ni in 30 min at 5 A/dm² |
| Electrowinning copper | plant throughput | kg/day = I·t·M/(nF·1000) | 250 kA cell, 7 t Cu/day |
| Hall-Héroult Al smelt | energy per tonne | Q·V where Q from law | ~13 MWh/t, 4.0–4.5 V |
| Chlor-alkali Cl₂ output | cell sizing | n=2 for Cl⁻ → ½Cl₂ | 80–150 kA per cell |
| Battery capacity Q | Coulombs from mass | Q = nF·moles of active material | Li-ion: 3.86 mAh per mg of LiCoO₂ |
| Anodizing Al | oxide thickness | m = QM/(nF) for Al + O | 20 µm @ 1.6 A/dm² for 30 min |
| Coulometric titration | Analyte amount | moles = Q/(nF) | µg of Cu detectable |
Real-world facts and figures
- Hall-Héroult cells. Modern AP60-class pots run at 600 kA. One pot drops a tonne of aluminum every ~10 hours. A line is 300+ pots in series.
- Norsk Hydro 1928. Norway's hydroelectric surplus drove 130 MW of alkaline electrolyzers at Rjukan to make ammonia (Birkeland-Eyde). Faraday's law was the design equation; flooded fjords were the energy source.
- Silver coulometer. Until 1948, the international ampere was defined as the current that deposits 1.118 mg of silver per second from AgNO₃ solution — a literal embodiment of Faraday's first law.
- Coinage and electroforming. The Statue of Liberty's seven-tonne torch flame replicas were electroformed in copper sulfate baths to ~3 mm thickness — Q calculated from the law before any current flowed.
- Spacecraft Cu electrowinning. 1970s NASA studies sized lunar oxygen plants directly from Faraday: each kg of O₂ from molten lunar regolith electrolysis needs 6.7 kWh ideal, ~12 kWh real.
- Lithium-ion battery capacity. A LiCoO₂ cathode stores 1 e⁻ per Co atom (Co⁴⁺ ⇌ Co³⁺). Nameplate 274 mAh/g vs theoretical 274 mAh/g — vendors quote 140–155 mAh/g because cycling stability collapses past 50% delithiation.
Variants and edge cases
- Current efficiency. Real systems lose some current to side reactions. Define η = (actual mass) / (Faraday-predicted mass). Cr plating: 12–25%. Cu electrowinning: 92–96%. Cl₂ from chlor-alkali: 95–98%.
- Coulombic efficiency in batteries. The same idea, applied to charge–discharge cycling. Modern Li-ion: > 99.9% per cycle. Without that figure, 1000-cycle life would be impossible.
- Coulometric titration. Generate a titrant (e.g., Br₂ from KBr) at exactly Faraday-predicted rate. The endpoint timestamp gives moles directly — used for trace-water Karl Fischer assays.
- Pulsed plating. Pulse current at kHz rates with short reverse pulses. Same total Q, finer grain structure. Faraday's law cares only about net Q, not waveform.
- Membrane crossover. In a fuel cell or electrolyzer, gases that diffuse across the membrane don't contribute to current but still consume reactant. Reduces "Faradaic efficiency" below 100%.
Common pitfalls and misconceptions
- Forgetting the n. Treating Cu²⁺ like Ag⁺ doubles your predicted mass. Always identify the half-reaction first: how many electrons per atom?
- Mixing C and A·hr. 1 A·hr = 3600 C. Coursework problems are easy to botch by a factor of 3600 if units slip.
- Assuming 100% current efficiency. Faraday's law gives the maximum mass. Anything that consumes electrons elsewhere (water splitting, side reactions) cuts the actual yield. Always quote a current efficiency.
- Ignoring impurities. Copper electrowinning relies on the deposited metal being purer than the solution. Trace metals more noble than Cu (Ag, Au) plate first; less noble (Fe, Zn) stay in solution. The law speaks to gross mass, not chemistry of selectivity.
- Using F as energy. F is in C/mol, not J. Cell voltage matters separately: energy = Q·V, where V depends on overpotentials and operating conditions. Plenty of textbooks gloss this distinction and confuse students.
- Forgetting it works for both electrodes. If 0.5 mol e⁻ deposited 0.5 mol Cu at the cathode, the same 0.5 mol e⁻ must have done something at the anode — typically 0.125 mol O₂ from water if the metal is inert.
Frequently asked questions
What does the Faraday constant actually represent?
F is the magnitude of charge carried by one mole of electrons. F = NA · e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ = 96 485.33 C/mol. Plug a current and time into Q = It and divide by F, and you immediately get how many moles of electrons crossed the cell.
Why does Faraday's law contain n, the number of electrons?
Different ions need different numbers of electrons to be reduced or oxidized. Cu²⁺ takes 2 e⁻ to become Cu⁰, but Al³⁺ takes 3, and Ag⁺ takes 1. For the same total charge, you get half as many moles of Cu as moles of Ag. The factor n in m = QM/(nF) absorbs that valence difference.
What is Faraday's second law in plain English?
If you pass the same amount of charge through different electrolytes, the masses deposited stand in the ratio of their chemical equivalents (atomic mass / valence). One faraday (96 485 C) liberates 1 g of H, 8 g of O, 23 g of Na, 31.75 g of Cu, 108 g of Ag — each is one mole of charge equivalents.
How accurate is Faraday's law in real plating cells?
Astonishingly accurate, provided current efficiency is ~100%. In well-designed copper electrowinning the figure is 92–96% (a few percent of the current splits water instead of plating). For chromium plating it can be as low as 12% — most of the current makes H₂. Faraday's law gives the upper bound; the rest is bookkeeping.
How is Faraday's law used to size a chlor-alkali plant?
1 ton of Cl₂ needs (1 000 000 / 70.9) × 2 × 96 485 ≈ 2.72 × 10⁹ C, and at 80 kA per cell that's about 9.4 hours per ton per cell. A 1 Mt-Cl₂/year plant needs roughly 130 cells running continuously, plus the corresponding 26 Mt of NaOH and 1.7 Mt of co-product H₂.
Why does aluminum smelting need so much electricity?
Faraday's law: m = QM/(nF). For Al³⁺ + 3e⁻ → Al, n = 3 and M = 26.98, so each kg of aluminum needs Q = 1000 × 3 × 96 485 / 26.98 = 1.07 × 10⁷ coulombs. Times the cell voltage (~4 V industrial): 12 kWh per kg, the textbook number. Aluminum is electricity in a solid form.