Question

A current of 3 A has to be passed through a solution of silver nitrate for ${}^{\prime }{t}^{\prime }sec$ to coat a metal surface of 80 $c{m}^2$ with a 0.005-mm-thick layer, then $t$ is: (density of silver is 10.5 g $c{m}^{-3}$) (write your answer to nearest integer)

Answer 1

Volume of the surface (V) = Area $\times$ Thickness
Given,
Area = 80 $c{m}^2$, thickness = 0.005 mm = 0.00005 cm
Mass of Ag(W) = V $\times$ Density = $0.04\times 10.5$ = 0.42 g
$A{g}^{⨁}+e^{-}\to Ag$
$\therefore W_{Ag}=\frac{ZIt}{96500}$
$0.42=\frac{108\times 3\times t}{96500}$
$\therefore t=\frac{0.42\times 96500}{108\times 3}=\frac{40530}{324}$ = 125.09 s