The time period to coat a metal surface of 80 cm² with 5 × 10^–3 cm thick layer of silver (density 1.05 g cm^–3) with the passage of 3A current through a silver nitrate solution is

mass of Ag  required to coat the given surface =$\mathrm{d}\times \mathrm{thickness}\times \mathrm{surface} \mathrm{Area}$

=$80\times 5\times {10}^{-3}\times 1.05=0.42\mathrm{gm}$

Number of equivalents of Ag = $\frac{0.42}{108}$

Number of coulombs required =$96500\times \frac{0.42}{108}=375.28 \mathrm{Coul}$

$$\begin{gathered} \mathrm{Amp}\times \sec =\mathrm{coulomb} \\ \mathrm{t} \left(\sec \right) =\frac{375.28}{3}\cong 125 \end{gathered}$$