Electrolysis
Faraday’s laws: mass deposited from current and time, or the current and time a mass needs.
Cu²⁺ + 2e⁻ → Cu · cathode · z = 2 · M = 63.546 g/mol
0.018656 mol of product
1800 s
3600 C of charge
| Charge passed | 3600 C | Q = I × t |
| Moles of electrons | 0.03731137 | Q ÷ F, with F = 96485.33212331 C/mol |
| Moles of product | 0.01865569 | electrons ÷ z, since each unit needs 2 |
| Mass | 1.18549418 g | moles × molar mass |
Electrolysis is counting electrons. The charge that passes fixes how many electrons arrive, and the half-equation fixes how many each ion needs, so the mass follows: m = (I × t × M) ÷ (z × F), with F = 96485 C/mol.
Charge in, atoms out
Every calculation here is the same chain run in whichever direction the question needs. Current times time gives charge in coulombs. Dividing by the Faraday constant converts that to moles of electrons. Dividing again by z — the electrons each ion needs — gives moles of product. Multiplying by the molar mass gives grams.
The constant tying it together is the charge on a mole of electrons. Since the 2019 redefinition of the SI both the elementary charge and the Avogadro constant are exact numbers, so F is exact too: 1.602176634 × 10⁻¹⁹ C multiplied by 6.02214076 × 10²³ per mole, which comes to 96485.33212 C/mol.
Why z matters more than anything else
The same charge deposits very different amounts of different metals, and the reason is the ionic charge rather than the atomic mass. One faraday deposits a full mole of silver from Ag⁺, half a mole of copper from Cu²⁺, and only a third of a mole of aluminium from Al³⁺. That is why aluminium smelting is so electricity-hungry: every atom costs three electrons, and the metal is light, so a great deal of charge buys very little mass.
Q = I·t · moles of electrons = Q ÷ F · moles of product = electrons ÷ z
- 1 Work out the charge. Q = I × t, with the time in seconds. 2 A for 30 minutes is 2 × 1800 = 3600 C.
- 2 Convert to moles of electrons. 3600 ÷ 96485 = 0.03731 mol of electrons.
- 3 Read z off the half-equation. Cu²⁺ + 2e⁻ → Cu needs two electrons per copper atom, so z = 2.
- 4 Divide to get moles of product. 0.03731 ÷ 2 = 0.018656 mol of copper.
- 5 Multiply by the molar mass. 0.018656 × 63.546 = 1.1855 g of copper deposited.
What one faraday deposits
96485 C — one mole of electrons. The ionic charge, not the atomic mass, decides the amount.
| Half-reaction | z | Moles produced | Mass |
|---|---|---|---|
| Ag⁺ + e⁻ → Ag | 1 | 1 | 107.87 g |
| Cu²⁺ + 2e⁻ → Cu | 2 | 0.5 | 31.77 g |
| Zn²⁺ + 2e⁻ → Zn | 2 | 0.5 | 32.69 g |
| Al³⁺ + 3e⁻ → Al | 3 | 0.333 | 8.99 g |
| 2H₂O + 2e⁻ → H₂ + 2OH⁻ | 2 | 0.5 | 1.01 g (11.36 L) |
| 2H₂O → O₂ + 4H⁺ + 4e⁻ | 4 | 0.25 | 8.00 g (5.68 L) |
Two places the arithmetic goes wrong
The first is the time unit. Q = I × t only works in seconds, and questions are almost always set in minutes or hours. Using minutes directly puts the answer out by a factor of sixty, which is the single commonest error on this topic.
The second is z. It is the number of electrons in the half-equation per unit of the product named, and that has to match what you are calculating. For oxygen from water the half-equation gives four electrons per O₂ molecule, so z = 4 and the molar mass is 32 g/mol for the molecule — not 16 for the atom. Mixing the two halves of that pairing is easy and gives an answer out by a factor of two.
The gas volumes here use IUPAC standard conditions, 0 °C and 100 kPa, which give 22.711 L/mol. The older 22.414 L/mol is the molar volume at 101.325 kPa, and using one figure with the other set of conditions puts you 1.3% out. Notice too that the hydrogen and oxygen from water come out in exactly a 2:1 ratio — that falls straight out of the electron counts rather than being put in by hand, which is a useful check that the half-equations are right.