A stainless‑steel spoon of area 25 cm² is to be electro‑plated with silver to a thickness of 2 µm. Density of Ag = 10.5 g cm⁻³ and molar mass = 107.87 g mol⁻¹. A constant current of 0.55 A is passed and the current efficiency is 90 %. For the reduction reaction Ag⁺ + e⁻ → Ag, calculate the time required (in seconds) to obtain the desired coating.

Step-by-step solution

1. Step 1: Convert the coating thickness to cm: $2,mumathrmm=2\times 10^{-4},\mathrm{cm}$.
2. Step 2: Calculate the volume of silver deposited: $V=A\times t_{\text{thick}}=25,\mathrm{c}{\mathrm{m}}^2\times 2\times 10^{-4},\mathrm{cm}=5\times 10^{-3},\mathrm{c}{\mathrm{m}}^3$.
3. Step 3: Find the mass of silver needed: $m=\rho V=10.5,\mathrm{g},\mathrm{c}{\mathrm{m}}^{-3}\times 5\times 10^{-3},\mathrm{c}{\mathrm{m}}^3=0.0525,\mathrm{g}$.
4. Step 4: Convert this mass to moles: $n_{\mathrm{Ag}}=m/M=0.0525/107.87=4.86\times 10^{-4},\mathrm{mol}$.
5. Step 5: Because one electron is needed per Ag atom, the theoretical electrons required are $4.86\times 10^{-4},\mathrm{mol};{\mathrm{e}}^{-}$. With 90 % efficiency, the actual electrons needed are $n_{\mathrm{e}}=4.86\times 10^{-4}/0.90=5.40\times 10^{-4},\mathrm{mol}$.
6. Step 6: Calculate the charge: $Q=n_{\mathrm{e}}\times F=5.40\times 10^{-4}\times 96485=52.2,\mathrm{C}$.
7. Step 7: Time required: $t=Q/I=52.2/0.55=94.9,\mathrm{s}approx95,\mathrm{s}$.
8. Final Answer: 95 s

Concept tested: Faraday's laws of electroplating

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