Electrolysis and Faraday's laws

Faraday's laws in plain language

Electrolysis is quantitative: the amount of chemistry that happens is directly proportional to the amount of charge that flows. Faraday's laws state this in two parts:

1. The mass of substance produced at an electrode is proportional to the quantity of charge passed through the cell.
2. One mole of electrons carries a fixed charge — the faraday, F = 96485 C/mol — so moles of electrons = Q / F.

Since charge is current multiplied by time (Q = It, with current in amperes and time in seconds), you can go from a stopwatch and an ammeter reading straight to grams of product. The bridge is always the half-reaction, which tells you how many electrons each formula unit of product requires.

Worked example: plating copper

Problem: What mass of copper plates out when a current of 2.00 A flows for 1.00 hour through a Cu2+ solution?

Step 1 — charge. Convert time to seconds: 1.00 h = 3600 s. Q = It = (2.00 A)(3600 s) = 7200 C.

Step 2 — moles of electrons. mol e− = Q / F = 7200 / 96485 = 0.0746 mol e−.

Step 3 — moles of copper. The cathode half-reaction is Cu2+ + 2e− → Cu, so 2 mol e− make 1 mol Cu: mol Cu = 0.0746 / 2 = 0.0373 mol Cu.

Step 4 — mass. Molar mass of copper is 63.55 g/mol: mass = 0.0373 × 63.55 = 2.37 g Cu.

Check the arithmetic: 7200/96485 = 0.07462 (yes, under 0.1 mol e−); half of that is 0.03731 mol; times 63.55 gives 2.371 g, which rounds to 2.37 g. Every step's units cancel correctly: A·s = C, C/(C/mol) = mol, mol × g/mol = g.

Products: molten NaCl vs aqueous NaCl

Faraday's laws say how much forms, but potentials decide what forms. Compare two cells:

Molten NaCl (no water): Only Na+ and Cl− are present.

• Cathode (reduction): Na+ + e− → Na(l) — sodium metal.
• Anode (oxidation): 2Cl− → Cl2(g) + 2e− — chlorine gas.

Aqueous NaCl (brine): Water competes at both electrodes.

• Cathode: water is easier to reduce than Na+, so 2H2O + 2e− → H2(g) + 2OH− — hydrogen gas, not sodium metal.
• Anode: chloride is easier to oxidize than water under these conditions, so 2Cl− → Cl2(g) + 2e− — chlorine gas.

Net: brine electrolysis produces H2, Cl2, and NaOH — the industrial chlor-alkali process. The presence of water completely changes the cathode product, which is why 'electrolysis of NaCl' is ambiguous until you specify molten or aqueous.

Efficiency and real-world limits

Real cells rarely achieve 100% of the Faraday's-law prediction. The shortfall comes from:

• Competing reactions — a second reduction or oxidation siphons off some current.
• Current efficiency — industrial cells typically run at 90–98% efficiency; the rest of the charge is lost to side reactions or heat.
• Overpotential — extra voltage beyond the theoretical minimum is needed to drive reactions at a practical rate, wasting energy (though not charge).

For exam problems, assume 100% efficiency unless told otherwise, then multiply by the given efficiency if one is provided. And remember the safety context: industrial electrolysis uses large currents and produces gases like H2 and Cl2 — these calculations belong in proper facilities, not home experiments.

Variables and units

Common mistakes

Real-world applications

• Predicting how much metal electroplating will deposit in a given time

• Sizing industrial cells for aluminum and chlorine production

• Calculating hydrogen output of a water electrolyzer

• Determining the charge needed to refine a batch of copper

• Relating laboratory current measurements to moles of product