How to Use Faraday's Law of Electrolysis

Electrons are reagents you can count

Faraday’s law of electrolysis is the bridge between an ammeter reading and a balance reading. Every coulomb of charge you push through a cell delivers exactly 1/96485 mole of electrons to an electrode, and those electrons reduce or oxidize ions stoichiometrically. If you know the current, the time, and which half-reaction is running, you can predict how many grams of copper plate out, how many liters of chlorine gas leave the anode, or how long an electroplating bath needs to run to deposit a 25 µm coating.

The equation:

m = (I × t × M) / (n × F)

• m: mass deposited or dissolved (g)
• I: current (A, which is C/s)
• t: time (s)
• M: molar mass of the species being deposited (g/mol)
• n: electrons transferred per ion in the half-reaction
• F: Faraday’s constant, 96,485 C/mol e⁻

The structure is intuitive once you parse it: I × t gives total coulombs, dividing by F gives moles of electrons, dividing by n gives moles of product, multiplying by M gives grams.

How to apply it

Get n from the half-reaction, not from the ion’s charge in your head. Cu²⁺ + 2e⁻ → Cu has n = 2. Ag⁺ + e⁻ → Ag has n = 1. Al³⁺ + 3e⁻ → Al has n = 3. For gas-evolution at the anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ has n = 4 per mole of O₂; 2Cl⁻ → Cl₂ + 2e⁻ has n = 2 per mole of Cl₂.

Time in seconds. Always. If your problem says “2.0 hours,” that’s 7200 s before you touch the equation.

Cathode vs. anode matters. Reduction (deposition of metal, production of H₂) happens at the cathode. Oxidation (dissolution of metal, evolution of O₂ or Cl₂) happens at the anode. If you’re being asked about copper deposition you want the cathode half-reaction; if you’re being asked about chlorine production you want the anode.

Worked examples

1. Copper plating. A 3.00 A current passes through CuSO₄(aq) for 2.00 h. How much copper deposits?

t = 7200 s, M(Cu) = 63.55 g/mol, n = 2. m = (3.00 × 7200 × 63.55) / (2 × 96485) = 7.11 g of Cu.

2. Solving for time (silver). How long does a 5.00 A current need to deposit 10.8 g of Ag?

n = 1, M = 107.87. Rearrange: t = (m × n × F) / (I × M) = (10.8 × 1 × 96485) / (5.00 × 107.87) = 1932 s ≈ 32.2 minutes.

3. Gas volume at the anode. Volume of O₂ at STP from 2.00 A through water for 1.00 h?

Moles of e⁻ = (I × t) / F = 7200 / 96485 = 0.0746 mol e⁻. Since 4 e⁻ make 1 O₂: moles O₂ = 0.0746 / 4 = 0.01865. At STP (22.4 L/mol): 0.418 L of O₂.

4. Cells in series. When you wire two electrolytic cells in series, the same current flows through both. The same number of moles of electrons reaches each cathode — but because n and M differ, the deposited masses differ. Pass 0.500 A for 1.00 h through CuSO₄ and AgNO₃ in series: each gets 0.01865 mol e⁻. Cu (n=2): 0.00932 mol × 63.55 = 0.593 g. Ag (n=1): 0.01865 mol × 107.87 = 2.012 g. Same charge, very different mass — because Ag uses one electron per atom while Cu uses two.

Traps to watch for

Wrong n. Cu²⁺ → Cu uses 2 electrons. Au³⁺ → Au uses 3. The charge of the ion is the n, but only after you’ve actually written and balanced the half-reaction — guessing here costs you a factor of 2 or 3 immediately.

Forgetting current efficiency. The equation gives the theoretical maximum. Real plating baths have efficiencies of 50–99% depending on chemistry. If the problem assumes 100%, fine; if it gives you a current efficiency, multiply.

Anode mass loss. Faraday’s law works for dissolution too. A copper anode in CuSO₄ loses mass at exactly the same rate the cathode gains it, because the electrons leaving the anode (Cu → Cu²⁺ + 2e⁻) feed the cathode reaction. If you measure copper consumption at the anode, use the same equation.

Mixing up coulombs and amperes. A coulomb is a quantity of charge; an ampere is a rate (C/s). I × t gives charge. Don’t put hours into the formula and expect coulombs to come out.

Try these, then verify with the Faraday’s Law Calculator:

1. How many grams of Ni (M = 58.69) deposit from NiSO₄ at 4.00 A for 1.50 h? (n = 2)
2. What current deposits 5.00 g of Au (M = 196.97) from Au³⁺ in 30.0 min?
3. How long must 10.0 A flow through molten NaCl to produce 2.30 g of Na?
4. Volume of Cl₂ at STP from 3.00 A through molten NaCl for 45.0 min?
5. In two cells in series (CuSO₄ and AgNO₃) at 0.500 A for 1.00 h, what mass deposits at each cathode?