A disc of copper of area 12 cm² is to be electro‑plated with gold to a thickness of 1.5 µm. Density of gold = 19.3 g cm⁻³, molar mass = 197 g mol⁻¹. A constant current of 0.80 A is passed with a current efficiency of 90 %. For the reduction reaction $ceA{u}^{3+}+3e-->Au$, calculate the time required (in seconds) to obtain the desired coating.

Step-by-step solution

1. Step 1: Convert thickness to cm: $1.5,mumathrmm=1.5\times 10^{-4},\mathrm{cm}$.
2. Step 2: Volume of gold deposited: $V=A\times t_{\text{thick}}=12,\mathrm{c}{\mathrm{m}}^2\times 1.5\times 10^{-4},\mathrm{cm}=1.8\times 10^{-3},\mathrm{c}{\mathrm{m}}^3$.
3. Step 3: Mass needed: $m=\rho V=19.3\times 1.8\times 10^{-3}=0.03474,\mathrm{g}$.
4. Step 4: Moles of gold: $n_{ceAu}=m/M=0.03474/197=1.76\times 10^{-4},\mathrm{mol}$.
5. Step 5: Each gold atom requires 3 electrons, so electrons needed: $n_e=3n_{ceAu}=5.29\times 10^{-4},\mathrm{mol}$.
6. Step 6: Correct for 90 % efficiency: $n_{e,\text{actual}}=5.29\times 10^{-4}/0.90=5.88\times 10^{-4},\mathrm{mol}$.
7. Step 7: Charge: $Q=n_{e,\text{actual}}\times F=5.88\times 10^{-4}\times 96485=56.7,\mathrm{C}$.
8. Step 8: Time: $t=Q/I=56.7/0.80=70.9,\mathrm{s}approx71,\mathrm{s}$.
9. Final Answer: 71 s

Concept tested: Faraday’s law for multi‑electron deposition

Published 10 October 2026 · How this solution was prepared and checked