A watch case of area 15 $c{m}^2$ is to be electro‑plated with gold to a thickness of 2 µm. The density of gold is 19.3 g $c{m}^{-3}$, its molar mass is 197 g $mo{l}^{-1}$. A constant current of 0.50 A is passed and the current efficiency is 85 %. For the reduction reaction $A{u}^{3+}+3e^{-}$ → Au, calculate the time required (in seconds) to obtain the desired coating. Use F = 96485 C $mo{l}^{-1}$.

Step-by-step solution

1. Convert the required thickness to centimetres: 2 µm = 2×$10^{-4}$ cm.
2. Calculate the volume of gold needed: V = area × thickness = 15 $c{m}^2$ × 2×$10^{-4}$ cm = 3.0×$10^{-3}$ $c{m}^3$.
3. Find the mass of gold: m = V × density = 3.0×$10^{-3}$ $c{m}^3$ × 19.3 g $c{m}^{-3}$ = 5.79×$10^{-2}$ g.
4. Determine the amount in moles: $n_{Au}$ = m / M = 5.79×$10^{-2}$ g / 197 g $mo{l}^{-1}$ = 2.94×$10^{-4}$ mol.
5. Calculate the charge required for the reduction (n = 3 electrons per Au atom): $Q_{theoretical}$ = n × F × $n_{Au}$ = 3 × 96485 C $mo{l}^{-1}$ × 2.94×$10^{-4}$ mol = 8.51×$10^1$ C.
6. Adjust for current efficiency: $Q_{actual}=Q_{theoretical}$ / 0.85 = 1.00×$10^2$ C.
7. Find the time using t = Q / I: t = 1.00×$10^2$ C / 0.50 A = 2.00×$10^2$ s.
8. Round to two significant figures.

Concept tested: Faraday’s law with current efficiency

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked