A disc of copper with area 12 cm² is to be electro‑plated with gold to a thickness of 1.5 µm. The density of gold is 19.3 g cm⁻³, its molar mass is 197 g mol⁻¹. A constant current of 0.80 A is passed for 90 % current efficiency. Calculate the time required in seconds.

Step-by-step solution

1. Step 1: Convert thickness to centimetres: $1.5,\mu \text{m}=1.5\times 10^{-4},\text{cm}$.
2. Step 2: Calculate the volume of gold deposited: $V=A\times t=12,{\text{cm}}^2\times 1.5\times 10^{-4},\text{cm}=1.8\times 10^{-3},{\text{cm}}^3$.
3. Step 3: Find the mass: $m=\rho V=19.3,{\text{g cm}}^{-3}\times 1.8\times 10^{-3},{\text{cm}}^3=3.474\times 10^{-2},\text{g}$.
4. Step 4: Compute moles of gold: $n=m/M=0.03474/197=1.764\times 10^{-4},\text{mol}$.
5. Step 5: Determine charge required for the reduction reaction $A{u}^{3+}+3e^{-}\to Au$: $Q_{\text{ideal}}=n,z,F=1.764\times 10^{-4}\times 3\times 96485=5.106\times 10^1,\text{C}$.
6. Step 6: Correct for current efficiency: $Q=Q_{\text{ideal}}/0.90=56.73,\text{C}$.
7. Step 7: Time needed: $t=Q/I=56.73/0.80=70.9,\text{s}\approx 71,\text{s}$.

Concept tested: Faraday’s law and current efficiency in electro‑plating

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked