A copper strip of area 5\,$c{m}^2$ is to be electro‑plated with silver. A current of 0.85\,A is passed for 30\,minutes. The current efficiency is 95\%. Given: $M_{Ag}$=107.87\,g\,$mo{l}^{-1}$, density of Ag=10.5\,g\,$c{m}^{-3}$, and Faraday constant F=96485\,C\,$mo{l}^{-1}$. (a) Calculate the mass of silver actually deposited. (b) Determine the thickness of the silver layer formed on the copper strip.

Step-by-step solution

1. Calculate total charge passed: $Q_{total}$=I·t=0.85\,A × 30\,min × 60\,s/min = 0.85×1800 = 1530\,C.
2. Account for current efficiency: $Q_{effective}$=0.95×1530=1453.5\,C.
3. For Ag⁺+e⁻→Ag, n=1. Moles of Ag deposited = $Q_{effective}$/F = 1453.5/96485 = 0.01506\,mol.
4. Mass of Ag = moles × molar mass = 0.01506\,mol × 107.87\,g\,$mo{l}^{-1}$=1.62\,g (rounded).
5. Volume of deposited Ag = mass/density = 1.62\,g / 10.5\,g\,$c{m}^{-3}$=0.1543\,$c{m}^3$.
6. Thickness = volume/area = 0.1543\,$c{m}^3$ / 5\,$c{m}^2$=0.0309\,cm = 3.09×$10^{-2}$\,cm (≈310\,µm).

Concept tested: Faraday’s law of electrolysis

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked