The time period to coat a metal surface of \(80 \mathrm{~cm}^2\) with \(5 \times 10^{-3} \mathrm{~cm}\)…

The time period to coat a metal surface of \(80 \mathrm{~cm}^2\) with \(5 \times 10^{-3} \mathrm{~cm}\) thick layer of silver (density \(1.05 \mathrm{~g} \mathrm{~cm}^{-3}\) ) with the passage of \(3 \mathrm{~A}\) current through a silver nitrate solution is
  1. \(115 \mathrm{sec}\)
  2. \(125 \mathrm{sec}\)
  3. \(135 \mathrm{sec}\)
  4. \(145 \mathrm{sec}\).

Solution

$\begin{aligned} & \text { Weight of Ag required }=V \times d \\ & =80 \times 5 \times 10^{-3} \times 1.05=0.42 \\ & \because \quad W=\frac{\text { Eit }}{96500} \\ & \therefore \quad 0.42=\frac{108 \times 3 \times t}{96500} \Rightarrow t=125 \mathrm{sec} .\end{aligned}$

Asked in: NEET 2015 (Phase 1)

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