Faraday's Law of Electrolysis

Use $m =\frac{ M\cdot{}I\cdot{}t}{(z\cdot{}F)}$. First calculate the charge: $Q = I\cdot{}t$ with the time in seconds. Divide by the Faraday constant $F =\frac{ 96\,485 C}{mol}$ to obtain the amount of electrons, and by z, the number of electrons per deposited particle. Finally multiply by the molar mass. Example copper: $I = 2.0 A$ for one hour gives $Q = 2.0\cdot{}3600 = 7200 C$. With $z = 2 (Cu^{2+} + 2 e^{-} \to{} Cu)$ it follows that $n =\frac{ 7200}{(2\cdot{}96\,485)} = 0.0373 mol$ and $m = 0.0373\cdot{}63.5 \approx{} 2.37 g$. The most common error is a time in minutes or hours instead of seconds.

The Faraday constant is the electric charge of one mole of electrons: $F = N_{A}\cdot{}e = 6.022\times{}10^{23} mol^{-1} \cdot{} 1.602\times{}10^{-19} C \approx{}\frac{ 96\,485 C}{mol}$. It is thus the bridge between the electrical world (charge in coulombs, measurable via current and time) and the material world (amount of substance in moles). If 96,485 C flow during an electrolysis, exactly one mole of electrons has been transferred; how much substance that deposits depends on z: one mole of silver $(z = 1)$, but only half a mole of copper $(z = 2)$ or a third of a mole of aluminium $(z = 3)$. As a rule of thumb it pays to keep $F \approx{}\frac{ 96\,500 C}{mol}$ in your head.

z is the number of electrons transferred per deposited particle in the electrode reaction. You read it from the half-equation, not from the overall equation: $Cu^{2+} + 2 e^{-} \to{} Cu$ means $z = 2$, Ag⁺ + e⁻ → Ag means $z = 1$, $Al^{3+} + 3 e^{-} \to{} Al$ means $z = 3$. Caution with gases: for hydrogen, 2 H⁺ + 2 e⁻ → H₂ applies, so $z = 2$ per H₂ molecule; setting $z = 1$ here wrongly halves the charge. Rule of thumb: z corresponds to the charge number of the discharged ion, or the change in oxidation number times the number of atoms in the particle formed.

Rearrange Faraday's law for the time: $t =\frac{ m\cdot{}z\cdot{}F}{(M\cdot{}I)}$. Example silver plating: for 1.00 g of silver (M = 107.9 g/mol, z = 1) at $I = 0.50 A$ you need $Q =\frac{ m\cdot{}z\cdot{}F}{M} =\frac{ 1.00\cdot{}96\,485}{107.9} \approx{} 894 C$, so $t =\frac{ Q}{I} =\frac{ 894}{0.50} \approx{} 1790 s$, about 30 minutes. The calculation also shows why industrial electrolysis needs enormous currents: a single mole of aluminium (27 g, z = 3) requires almost 290,000 C; even at 1000 A that takes nearly five minutes. Real processes take longer because the current efficiency is below 100 %, for example due to side reactions such as hydrogen evolution.

Wherever current deposits or decomposes substances. In electroplating they determine the layer thickness in chrome, zinc or gold plating: charge, area and density fix the thickness exactly. Copper refining deposits high-purity copper (over 99.99 %) from anode copper; aluminium production uses molten-salt electrolysis with gigantic current demand. In water electrolysis for green hydrogen the law links charge and gas volume: $2\cdot{}96\,485 C$ produce one mole of H₂, about 24.8 L at room conditions. In analytics, coulometry measures amounts of substance directly via the charge, and when charging batteries the same relation describes the material turnover at the electrodes.