Silver is electrodeposited on a metallic vessel of surface area 800 cm2 by passing a current of 0.20 A for 3.0 hours. What is the thickness of the silver deposited? (Density of silver = 10.47 g cm−3, molar mass of Ag = 108 g mol−1, 1 F = 96500 C mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.89 × 10−4 cm
By Faraday’s first law, the mass deposited is fixed by the charge passed. Each silver ion needs one electron, so 1 F deposits 1 mol (108 g) of Ag:
\( \displaystyle \ce{Ag^+(aq) + e^- -> Ag(s)} \)
Charge passed (t = 3.0 h = 10800 s):
\( \displaystyle Q=It=0.20\,\mathrm{A}\times10800\,\mathrm{s}=2160\,\mathrm{C} \)
Mass of silver deposited:
\( \displaystyle m=\frac{Q}{F}\times M=\frac{2160\,\mathrm{C}}{96500\,\mathrm{C\,mol^{-1}}}\times108\,\mathrm{g\,mol^{-1}}=2.417\,\mathrm{g} \)
Volume of silver:
\( \displaystyle V=\frac{m}{d}=\frac{2.417\,\mathrm{g}}{10.47\,\mathrm{g\,cm^{-3}}}=0.2309\,\mathrm{cm^{3}} \)
Thickness (volume spread over the surface area):
\( \displaystyle \text{thickness}=\frac{V}{A}=\frac{0.2309\,\mathrm{cm^{3}}}{800\,\mathrm{cm^{2}}} \)
\( \displaystyle \text{thickness}=2.886\times10^{-4}\,\mathrm{cm}=2.89\times10^{-4}\,\mathrm{cm} \)