Faraday’s Laws
Part of Electrochemistry
The two quantitative laws that govern the amount of material deposited or dissolved in any electrolytic process — the foundation of all electrochemical calculations.
Why This Matters
Michael Faraday discovered in the 1830s that the mass of material deposited or dissolved at an electrode is precisely proportional to the quantity of electricity passed. This is not an approximation — it is an exact relationship derived from the quantized nature of electric charge and the discrete valence of ions.
These laws are the engineer’s tool for electrolytic processes: they tell you exactly how much copper you will plate per hour, how much chlorine you will generate per kilowatt-hour, how long to run an anodizing bath to reach a target thickness, and how to calculate the energy efficiency of any process. Without them, electrochemical operations require constant empirical adjustment. With them, you design with predictable outcomes.
Faraday’s laws apply to every electrochemical process discussed in this section, without exception.
Faraday’s First Law
The mass of substance deposited at an electrode is directly proportional to the quantity of electricity passed.
Mathematically: m ∝ Q m = (M × Q) / (F × z)
Where:
- m = mass deposited or dissolved (grams)
- M = molar mass of the substance (g/mol)
- Q = charge passed (coulombs) = Current (A) × Time (s)
- F = Faraday’s constant = 96,485 C/mol
- z = valence of the ion (number of electrons transferred per ion)
Example: How much copper deposits if 10 A flows for 1 hour?
- Q = 10 A × 3,600 s = 36,000 C
- M(Cu) = 63.5 g/mol
- z(Cu²⁺) = 2
- m = (63.5 × 36,000) / (96,485 × 2) = 2,286,000 / 192,970 = 11.85 g copper
Faraday’s Second Law
For the same quantity of electricity, the masses of different substances deposited are proportional to their equivalent weights (M/z).
This means that 96,485 C (one Faraday of charge) always deposits exactly one equivalent (M/z grams) of any substance.
Equivalents of common materials per Faraday:
| Element | Molar Mass (g/mol) | Valence z | Equivalent weight (g/F) |
|---|---|---|---|
| Copper (Cu²⁺) | 63.5 | 2 | 31.75 |
| Nickel (Ni²⁺) | 58.7 | 2 | 29.35 |
| Zinc (Zn²⁺) | 65.4 | 2 | 32.7 |
| Iron (Fe²⁺) | 55.8 | 2 | 27.9 |
| Iron (Fe³⁺) | 55.8 | 3 | 18.6 |
| Chromium (Cr³⁺) | 52.0 | 3 | 17.3 |
| Silver (Ag⁺) | 107.9 | 1 | 107.9 |
| Gold (Au³⁺) | 197.0 | 3 | 65.7 |
| Aluminum (Al³⁺) | 27.0 | 3 | 9.0 |
| Hydrogen (H⁺) | 1.0 | 1 | 1.0 |
| Chlorine (Cl⁻) | 35.5 | 1 | 35.5 |
Observation: Silver requires one electron per ion (z=1) and has high atomic mass, so each Faraday deposits 107.9 g of silver. Aluminum requires three electrons (z=3) and has low mass — each Faraday deposits only 9.0 g of aluminum. The Hall-Héroult process consumes enormous electricity per kilogram of aluminum partly for this reason.
Current Efficiency
Real processes often do not achieve 100% of the theoretical Faraday prediction. Some of the current drives side reactions:
- Hydrogen evolution at the cathode (instead of metal deposition)
- Oxygen evolution at the anode (instead of metal dissolution)
- Decomposition of organic additives
Current efficiency (CE) is defined as: CE = (actual mass deposited / theoretical mass by Faraday) × 100%
Typical current efficiencies:
| Process | Current Efficiency |
|---|---|
| Copper acid sulfate plating | 95–98% |
| Nickel sulfamate plating | 96–98% |
| Zinc alkaline plating | 70–85% |
| Chrome (hexavalent) plating | 12–25% |
| Water electrolysis (H₂) | 75–85% |
| Chlor-alkali (Cl₂) | 96–98% |
Chrome plating’s low efficiency (12–25%) is why it requires such extreme current densities — only 15–20% of the charge deposited chromium; the rest evolved hydrogen. This makes chrome plating the most energy-intensive common plating process.
Practical Calculations
Calculating Plating Time for Target Thickness
Convert thickness target to mass per unit area, then apply Faraday’s Law:
Mass per unit area (g/m²) = thickness (m) × density (g/m³)
For 25 μm copper (density 8,960 kg/m³): m/A = 25×10⁻⁶ m × 8,960×10³ g/m³ = 224 g/m²
Charge required (accounting for current efficiency CE = 0.97): Q/A = (m × F × z) / (M × CE) = (224 × 96,485 × 2) / (63.5 × 0.97) = 43,296,320 / 61.6 = 702,859 C/m²
Time at 300 A/m²: t = 702,859 / 300 = 2,343 s = 39 minutes
Calculating Anodizing Thickness
Aluminum oxide grows at approximately 1 μm per minute at typical current densities. The Faraday approach gives more precision:
Al oxidizes with z=3: Al → Al³⁺ + 3e⁻
For standard anodizing at 1.5 A/dm² (150 A/m²), time for 15 μm oxide: Oxide thickness (μm) ≈ current density (A/dm²) × time (min) × 0.3
For 15 μm: time = 15 / (1.5 × 0.3) = 33 minutes
Calculating Gas Production
For hydrogen at the cathode during water electrolysis:
At 100% efficiency, 1 Faraday (96,485 C) produces 1 g (half a mole) of H₂ = 5.6 liters at STP.
At 80% efficiency and 100 A for 1 hour: Q = 100 × 3,600 = 360,000 C Theoretical H₂ = 360,000 × 0.80 / 96,485 × 1.0 g = 2.98 g = 33.1 liters H₂
These calculations allow you to size gas storage vessels, estimate ventilation requirements, and assess the scale of a hydrogen production installation before building it.