Faraday's Laws of Electrolysis

Pick a bath, set a current and a time, and read off the mass, the thickness and the electricity bill that follow from them.

Worked example — Faraday's Laws of Electrolysis

Charge passed	6,000 C
Moles of electrons	0.0622 mol
Metal per coulomb	0.30416 mg/C
Metal deposited	1.73 g
Metal the lost charge would have made	0.09 g
Average thickness	9.73 µm
Deposition rate	29.19 µm/h
Electricity into the cell	0.0067 kWh

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

A tonne of aluminium is 2,979,980 ampere-hours 🖖

A tonne of aluminium is 37,062 moles of atoms, and Al(III) wants three electrons each. That fixes the charge before anyone builds a smelter: 1.0728 × 10¹⁰ coulombs, or 2,979,980 ampere-hours. It is arithmetic on the periodic table. No plant design touches it.

Press Aluminium pot. Three hundred and fifty thousand amperes for a day is 30,240,000,000 C, and the tool puts 2,677,870.38 g on the cathode, 95% of what the charge could have made. The bill was 36,120.0000 kWh, so 13.4883 kWh for every kilogram.

Halvor Kvande and Per Arne Drabløs give two figures for a modern smelter: cells run at 4.0 to 4.5 V, and the direct-current electricity comes to close to 13 kWh per kilogram in the best plants and close to 14 as a world average. Put 13 and 14 back through this page at 95% current efficiency and they ask for 4.14 V and 4.46 V. The two bands are one band, seen from either side of Faraday's law.

Read the last three rows. The reversible potential of the reaction is about 1.2 V at 960 °C and the pot runs at 4.3, so of that 36,120 kWh only 9,576.0000 is the thermodynamics of pulling aluminium out of alumina. 24,738.0000 kWh is overpotential at the electrodes and resistance in the molten cryolite, which leaves the cell as heat. The charge that made no metal accounts for 1,806.0000, the smallest of the three by a wide margin.

Drop the cell voltage to 4.0 V and the bill falls 6.98%. Push the current efficiency from 95% to a perfect 100 and it falls 5.00%, which is everything current efficiency has left to give, for ever. Above its 1.2 V floor the cell is carrying 3.1 V of overpotential, and that is the 68.49% row.

Change the bath and the same chromium plates at half the rate 🖖

Faraday's second law says a coulomb buys M/(zF) grams, and a real tank gets η of that. Fix the metal and fix the charge, and the bath chooses the two terms that are left: how many electrons the ion carries, and what fraction of the charge ends up as metal at all. The first of those is an integer.

Chromium is where that costs money. A hexavalent bath carries chromium at Cr(VI), six electrons an atom, 0.08982 mg per coulomb. A trivalent bath carries the same chromium at three and gives 0.17963 mg. Press Hard chromium, then move the ion selector to Cr(III) and change nothing else: 20.24 µm against 40.47 µm. The ratio is exactly 2, because the molar mass cancels and 6/3 is all the division has left.

Five elements on the comparison chart appear twice, and the pairs are the whole argument. Gold from a cyanide bath is Au(I) at 2.04141 mg/C and from an acid bath Au(III) at 0.68047, exactly three apart. Copper and tin give 2. Iron gives 3/2.

Silver sits second from the top of the chart and aluminium second from the bottom, for two reasons at once. A silver atom is 3.998 times the mass of an aluminium atom and takes one electron where aluminium takes three, so a coulomb of silver weighs 11.99 times a coulomb of aluminium. Roughly four of that is the mass and exactly three of it is the electron count.

Real baths differ in current efficiency as well, which is why the comparison above holds it fixed. That way the only thing moving is z.

Faraday's constant stopped being measured in 2019 🖖

Every row on this page that turns charge into matter divides by F = 96,485.33212 C/mol. That used to be an experimental quantity with an uncertainty attached. Since 20 May 2019 it has been the product of two exact ones: the SI defines the Avogadro constant as 6.02214076 × 10²³ mol⁻¹ and the elementary charge as 1.602176634 × 10⁻¹⁹ C, and F is what you get by multiplying them. This tool's module multiplies them rather than carrying the answer as a constant.

Faraday reached the law first, and reached it blind. His experiments of 1834 established that the mass deposited is proportional to the charge passed, and that the constant of proportionality is the substance's chemical equivalent. He had no unit of charge worth the name and no Avogadro number to divide by. He measured a ratio, and the ratio held.

He named the furniture in the same year, after writing to William Whewell at Trinity College, Cambridge: electrode, anode, cathode, ion, electrolyte. The words arrived with the law they describe.

The Moles of electrons row is where the count becomes visible. Press Aluminium pot and it reads 313,415.5144 mol, which is 1.89 × 10²⁹ individual electrons crossing the cathode in a day.

Problem solved in full

One pot, one day, and where the 36,120 kWh went 9 steps
A Hall–Héroult cell runs at 350,000 A for twenty-four hours at 4.3 V, and 95% of its charge ends up as metal. Work out how much aluminium comes out, what the electricity cost per kilogram, and how much of that bill the chemistry is actually responsible for. Then find the part of the cost that has nothing to do with electricity at all.

Open this problem in the tool
1. Q = I·t, with the time in seconds: 350,000 A over 86,400 s is 30,240,000,000 C, which is what the tool prints under Charge passed. Written as 3.024 × 10¹⁰ C it is easier to carry through the next four steps, and in the units a rectifier's meter counts it is 8,400,000 ampere-hours.
  
  $Q = I t = 350, 000 A \cdot 86, 400 s = 3.024 \times 1 0^{10} C = 8, 400, 000 A h$
2. Dividing by Faraday's constant turns coulombs into electrons: 313,415.5144 mol of them, under Moles of electrons. F is not a measurement here. It is the Avogadro constant times the elementary charge, both exact since 2019.
  
  $F = N_{A} e = 6.02214076 \times 1 0^{23} \cdot 1.602176634 \times 1 0^{- 19} = 96, 485.33212 C mo l^{- 1} \frac{Q}{F} = 313, 415.5144 mol$
3. Aluminium arrives as Al(III), so three electrons build one atom. Every electron doing its job would give 104,471.8381 mol of aluminium; at 95% current efficiency, 99,248.2462 mol of it does.
  
  $n_{Al} = \frac{1}{z} \cdot \frac{Q}{F} = \frac{313 , 415 . 5144}{3} = 104, 471.8381 mol η n_{Al} = 0.95 \cdot 104, 471.8381 = 99, 248.2462 mol$
4. Weigh that. The tool prints 2,677,870.38 g under Metal deposited, and the 5% that went elsewhere is the 140,940.55 g row: aluminium the charge paid for and the pot never saw, lost to the back reaction in which dissolved metal is re-oxidised by the gas coming off the anode.
  
  $m = η \frac{QM}{z F} = 99, 248.2462 \cdot 26.9815385 = 2, 677, 870.38 g (1 - η) \frac{QM}{z F} = 140, 940.55 g$
5. M/(zF) for aluminium is 0.09321 mg per coulomb, which is 0.3356 g per ampere-hour. Kvande and Drabløs quote 0.335 kg per kilo-ampere-hour for the same quantity, arrived at independently. A tonne therefore needs 2,979,980 ampere-hours, and no smelter design changes that number.
  
  $\frac{M}{z F} = \frac{26 . 9815385}{3 \cdot 96 , 485 . 33212} = 9.3215 \times 1 0^{- 5} g C^{- 1} = 0.3355727 g (A h)^{- 1} \frac{1 0 ^{6}}{0 . 3355727} = 2, 979, 980 A h (t Al)^{- 1}$
6. Energy is charge times voltage whatever the charge goes on to do: 3.024 × 10¹⁰ C at 4.3 V is 1.30032 × 10¹¹ J, or 36,120.0000 kWh. Against the metal that arrived, 13.4883 kWh/kg, which puts this pot between the best plants and the world average.
  
  $E = Q V = 3.024 \times 1 0^{10} \cdot 4.3 = 1.30032 \times 1 0^{11} J = 36, 120 kWh \frac{36 , 120}{2 , 677 . 87} = 13.4883 kWh k g^{- 1}$
7. Al₂O₃ + 3⁄2 C → 2 Al + 3⁄2 CO₂ has a reversible potential of about 1.2 V at 960 °C, so the thermodynamic share is 0.95 × 3.024 × 10¹⁰ × 1.2 J, which is 9,576.0000 kWh or 26.51% of the bill. The other 3.1 V of the cell is overpotential at both electrodes plus the resistance of the bath: 24,738.0000 kWh, 68.49%. Charge that made no metal still crossed the full 4.3 V and cost 1,806.0000 kWh, 5.00%. The three add to 36,120 exactly, because they are QV cut three ways rather than three separate estimates.
  
  $η Q E_{rev} η Q (V - E_{rev}) (1 - η) Q V Q V 0.95 \cdot 3.024 \times 1 0^{10} \cdot 1.2 0.95 \cdot 3.024 \times 1 0^{10} \cdot 3.1 0.05 \cdot 3.024 \times 1 0^{10} \cdot 4.3 9, 576.0 kWh (26 . 51%) 24, 738.0 kWh (68 . 49%) 1, 806.0 kWh (5 . 00%) 36, 120.0 kWh$
8. Which one is worth chasing? Take the cell to 4.0 V and the bill drops 6.98%. Take the current efficiency to a perfect 100% and it drops 5.00%, and then there is nothing left to take. The voltage term has 3.1 V standing on a 1.2 V floor; the efficiency term has five points standing on nothing.
  
  $\frac{13 . 4883 - 12 . 5473}{13 . 4883} = 6 . 98% (V : 4.3 \to 4.0) \frac{13 . 4883 - 12 . 8139}{13 . 4883} = 5 . 00% (η : 0.95 \to 1)$
9. The anode in a Hall–Héroult cell is carbon, and the reaction eats it. Three quarters of a mole of carbon goes per mole of aluminium, which is 0.3339 kg of carbon and 1.2233 kg of CO₂ for every kilogram of metal. This pot therefore emits about 3.28 tonnes of CO₂ a day from the anode alone, before a kilowatt-hour has been accounted for, and moving the smelter onto hydroelectricity leaves that figure exactly where it is: the carbon is a reagent here, not a fuel. In practice it is worse, because carbon also burns in the oxygen and the CO₂ around it; Kvande and Drabløs put net anode consumption at 0.40 to 0.45 kg per kilogram of aluminium, and the emission at about 1.5 kg. An anode that survived would make oxygen gas instead, which is the harder product: the same authors put the carbon anode's contribution at a full volt off the cell voltage, so an inert anode has to find that volt somewhere. None of this is on the page above. Faraday's law contains no anode.
  
  $A l_{2} O_{3} + \frac{3}{2} C \to 2 Al + \frac{3}{2} C O_{2} : \frac{0 . 75 \cdot 12 . 011}{26 . 9815385} = 0.3339, \frac{0 . 75 \cdot 44 . 009}{26 . 9815385} = 1.2233 kg k g^{- 1} \Rightarrow 2, 677.87 \cdot 1.2233 = 3, 276 kg C O_{2} da y^{- 1}$
Answer

2,677,870.38 g of aluminium, 36,120.0000 kWh, and 13.4883 kWh for every kilogram.

The chemistry is 26.51% of that. Overpotential and the resistance of the bath are 68.49%, and charge that made no metal is 5.00%. Perfect current efficiency would save 5.00% of the bill and is by far the smaller prize, because the voltage term has 3.1 V above its floor while the efficiency term has five percentage points above nothing.

And the carbon anode is 1.2233 kg of CO₂ per kilogram of aluminium whatever the electricity is made from, 1.5 kg once the side reactions are counted. This page cannot compute either, because Faraday's law contains no anode.

$m = η \frac{QM}{z F} = 2, 677, 870 g \frac{Q V}{m} = 13.4883 kWh k g^{- 1} \frac{η Q E _{rev}}{Q V} = \frac{0 . 95 \cdot 1 . 2}{4 . 3} = 26 . 51%$

References (7)

The law itself, and the paper that named the electrode: Michael Faraday, "VI. Experimental researches in electricity.-Seventh Series", Philosophical Transactions of the Royal Society of London 124, 77–122, 1834. doi:10.1098/rstl.1834.0008
Where the vocabulary came from, which is a better-documented story than the usual telling: S. Ross, "Faraday consults the scholars: the origins of the terms of electrochemistry", Notes and Records of the Royal Society of London 16(2), 187–220, 1961. doi:10.1098/rsnr.1961.0038
Why the Faraday constant is now arithmetic rather than a measurement: Bureau International des Poids et Mesures, "The International System of Units (SI)", 9th edition, 2019, §2.2 and Resolution 1 of the 26th CGPM. doi:10.59161/AUEZ1291
And the value itself, with its uncertainty printed as the word "exact": "Faraday constant", 2022 CODATA recommended values, NIST Reference on Constants, Units and Uncertainty.
Every smelter figure on this page, and the cross-check insight block 1 is built on: Halvor Kvande and Per Arne Drabløs, "The Aluminum Smelting Process and Innovative Alternative Technologies", Journal of Occupational & Environmental Medicine 56(5 Suppl), S23–S32, 2014. doi:10.1097/JOM.0000000000000062
The reversible potential the aluminium preset carries: "U.S. Energy Requirements for Aluminum Production: Historical Perspective, Theoretical Limits and Current Practices", Industrial Technologies Program, U.S. Department of Energy, February 2007, §5.
Why a hard-chromium tank throws most of its charge away: "12.20 Electroplating", AP-42: Compilation of Air Emissions Factors, U.S. Environmental Protection Agency, revision 10/2021.

By the 42Math team · Reviewed against published sources. Corrections welcome.