Copper Plating Calculator
2024
This calculator determines the amount of copper deposited with acid copper electroplating.
Calculator
Assumptions
It takes two electrons to reduce a Cu²⁺ ion (from CuSO₄) to copper metal. There are 3600 Coulombs in an Amp Hour and \(6.2415 \cdot 10^{18}\) electrons in a Coulomb meaning there are \(3600 \cdot 6.2415 \cdot 10^{18} = 2.247 \cdot 10^{22}\) electrons in per Amp Hour. The number of moles of copper deposited can be determined by dividing the number of electrons by Avogadro's number (\(6.022 \cdot 10^{23}\) mol⁻¹) and by the number of electrons required to reduce each ion. This results in \(2.247 \cdot 10^{22} / (2 \cdot 6.022 \cdot 10^{23}) = 0.0186\) moles of copper deposited per Amp Hour. The number of grams of copper deposited can be determined by multiplying the number of moles by the molar mass of copper (63.546 g/mol) which results in \(0.0186 \cdot 63.546 = 1.182\) grams of copper deposited per Amp Hour at 100% cathode efficiency.
Another way of arriving at the same answer is to use Faraday's law: \(Q = n \cdot F \cdot \frac{m}{M}\), where:
- Q is charge in coulombs
- n is the number of electrons (2 for Cu²⁺)
- F is Faraday's constant (96485 C/mol)
- m is mass deposited
- M is the molar mass of copper (63.546 g/mol)
For acid copper electroplating, the cathode efficiency is typically near 100%, so this theoretical number is not far from the truth.
Ammusingly, Google's impossible to disable AI overview gives an entirely different (wrong) number:
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