Apply Faraday's Law to electroplating, water electrolysis and electrorefining. Instantly compute deposited mass, film thickness, current density and energy consumption.
Parameters
Process
Metal
Current I
A
Time t
h
Current efficiency η
%
Electrode area A
cm²
Cell voltage V
V
Results
—
Deposited mass m (g)
—
Film thickness δ (μm)
—
Charge Q (Ah)
—
Power P (W)
—
Current density J (A/cm²)
—
Energy consumed (Wh)
Deposited Mass vs Time
Film Thickness vs Current Density (fixed t)
Theory & Key Formulas
$$m = \frac{M \cdot I \cdot t \cdot \eta}{n \cdot F}$$
$F = 96485$ C/mol (Faraday constant)
$\delta = m / (\rho \cdot A)$ — thickness
$J = I / A$ — current density
$P = V \cdot I$ — power
What is Electrolysis & Faraday's Law?
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What exactly is Faraday's Law, and how does it let us predict how much metal gets plated?
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Basically, it's the fundamental link between electricity and chemical change. It says the mass of a substance deposited or dissolved at an electrode is directly proportional to the total electric charge passed through. In practice, if you double the current or time, you double the mass deposited. Try moving the "Current (I)" or "Time (t)" sliders in the simulator above to see this direct relationship instantly.
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Wait, really? So if I set the current to 5 Amps for 1 hour, I get the same mass as 1 Amp for 5 hours?
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Exactly! Because both give you the same total charge (in Amp-hours). The simulator calculates that charge as $I \cdot t$. But there's a catch: not all current goes into plating. That's the "Current Efficiency (η)" slider. For instance, in a real copper plating bath, some current is wasted producing hydrogen bubbles. If η is 90%, only 90% of your charge does useful work.
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Okay, that makes sense. But how do we get from charge to an actual thickness on a part, like chrome on a car bumper?
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Great question! First, Faraday's Law gives you the mass. Then, thickness depends on how that mass spreads over the surface area. That's where the "Surface Area (A)" and "Density (ρ)" parameters come in. A common case is plating a flat sheet: a small mass spread over a large area gives a very thin coating. Try reducing the area in the simulator while keeping mass constant—you'll see the calculated thickness shoot up.
Physical Model & Key Equations
The core of the calculation is Faraday's First Law of Electrolysis, which relates the mass of substance altered at an electrode to the electric charge passed.
$$m = \frac{M \cdot I \cdot t \cdot \eta}{n \cdot F}$$
$m$ = mass deposited (g) $M$ = molar mass of the substance (g/mol) — select from the dropdown. $I$ = current (A) $t$ = time (s) $\eta$ = current efficiency (unitless, as a decimal) $n$ = number of electrons transferred per ion (e.g., $n=2$ for $Cu^{2+}$) $F$ = Faraday constant, 96485 C/mol (charge of one mole of electrons)
For engineering applications, the deposited mass is often converted into a plating thickness, which is a critical design specification.
$$\delta = \frac{m}{\rho \cdot A}$$
$\delta$ = plating thickness (cm) $\rho$ = density of the plated material (g/cm³) — selected automatically based on material. $A$ = surface area of the substrate (cm²)
This equation shows thickness is inversely proportional to area: plating the same mass onto a smaller object yields a thicker layer.
Frequently Asked Questions
Current efficiency is the proportion of current actually used for plating deposition. Due to losses from side reactions (such as hydrogen evolution), the actual deposition amount is less than the theoretical value. Please input an experimental or literature value (e.g., 90–95% for copper plating) within the range of 0 to 1.
In addition to the deposition mass, the film thickness calculation requires the 'plating area' and the 'density of the metal.' Density varies by element (e.g., copper: 8.96 g/cm³). For the area, please input the surface area of the part to be plated in cm².
Yes. As long as the atomic weight, valence, and density are set correctly, it can be applied to any metal. For example, gold (Au) has an atomic weight of 196.97, a valence of 1 (or 3), and a density of 19.32 g/cm³. Please look up and input the physical properties of each metal.
Yes. It is calculated from the power consumption (kWh) and voltage. However, actual electricity costs include rectifier efficiency and wiring losses, so please use this as a reference. For the voltage, input the tank voltage of the plating bath (typically 2–6 V).
Real-World Applications
Decorative & Protective Electroplating: This is the most common use. A thin layer of a desirable metal (like chromium, gold, or nickel) is deposited onto a cheaper base metal (like steel or brass) for corrosion resistance, wear resistance, and appearance. For instance, the shiny, rust-resistant finish on car bumpers, bathroom faucets, and jewelry is achieved this way.
Electrorefining of Metals: Impure metal anodes (e.g., blister copper) are dissolved, and pure metal is deposited at the cathode. This process is how we obtain high-purity copper (99.99%+) for electrical wiring. The simulator's current efficiency parameter is crucial here, as side reactions directly impact yield and cost.
Electroforming: This is additive manufacturing via electroplating. Metal is deposited onto a mandrel (mold) to build up a solid, freestanding object, which is later separated. Common applications include producing intricate nickel meshes for filters, waveguide components for aerospace, and even some musical instrument bells.
Anodizing: While not a plating process, it's a key electrochemical surface treatment. Here, the workpiece is the anode, and an oxide layer (like on aluminum) is grown to enhance corrosion resistance and provide a base for dye. Controlling current density (the $J = I/A$ calculation in the tool) is critical for achieving a uniform, hard anodized layer.
Common Misconceptions and Points to Note
When you start using this calculation tool, there are several pitfalls that beginners on the shop floor often encounter. First and foremost, understand that current efficiency is not a fixed value. While you input it as a constant in the tool, in an actual plating bath, the current efficiency changes with current density (current per unit area), temperature, and bath composition. For example, in nickel plating, if you raise the current density too high, hydrogen evolution can become vigorous, causing efficiency to drop from around 90% to near 70%. When results don't match your calculations, re-evaluating the efficiency is your first step.
Next, there is the fundamental fact that "coating thickness will not be uniform". The result from this calculation is only the "average thickness". On an actual electrode (especially parts with complex shapes), corners and protrusions experience current concentration leading to thicker plating (over-plating), while recessed areas become thinner (under-plating). You use the total surface area to be plated for area "A", but considering uniformity requires a separate dimension of analysis: "current distribution".
Finally, be very careful about unit confusion. The calculation formula uses [cm] and [g/cm³], but on the shop floor, thickness is commonly in [μm] and area in [dm²] (the unit "1 square decimeter" is frequently used, especially in decorative plating). Even if the tool handles conversions internally, ensure unit consistency when doing manual calculations. For instance, mistakenly inputting an area of 10 cm² (=0.1 dm²) as 1 dm² would make the calculated thickness ten times different—a potential disaster.
Enter current (amperes) in field vI and select time unit from sI dropdown (seconds/minutes/hours)
Input time duration in vT and confirm unit from sT selector
Set Faraday constant (96485 C/mol for standard conditions) or accept default, then input charge efficiency (%) in vEta field
Click Calculate to determine moles of substance deposited and mass using M = (I × t × η) / (n × F), where n is electron transfer number
Worked Example
Copper electroplating: 5 amperes applied for 120 minutes with 95% efficiency. Cu atomic mass = 63.5 g/mol, n = 2 electrons. Moles deposited = (5 × 7200 s × 0.95) / (2 × 96485) = 0.177 mol. Mass = 0.177 × 63.5 = 11.24 grams of copper on cathode. For industrial nickel electrorefining at 1000 A, 8 hours, 98% efficiency: Ni (58.7 g/mol, n=2) yields 86.3 kg refined nickel per batch.
Practical Notes
Water electrolysis produces hydrogen at cathode (2H+ + 2e−) and oxygen at anode (2H2O − 4e−); use n=2 for H2, n=4 for O2 stoichiometry
Efficiency accounts for competing side-reactions (hydrogen evolution in copper plating, chlorine formation in brine cells); typical industrial ranges: 85–99%
Temperature and electrolyte concentration affect conductivity; recalculate if operating conditions shift >10°C from calibration baseline
For precious metal recovery (gold, silver), small current fluctuations cause significant mass variance; verify ammeter accuracy to ±2% before calculation