How long a current of 3 amperes has to be passed through a solution of silver nitrate to coat a metal surface of 80 `cm^2` with 0.005 mm thick layer ?
[Density of Ag is 10.5 g `cm^(-3)`, At wt.of silver = 108.0 u]

Explanation

`Ag^(+) (aq) + e^(-) to Ag (s)`
Volume = Area x Thickness `=80 cm^(2) xx 0.05 xx 1/10 cm = 0.04 cm^(3)`
Mass of Ag = Volume x Density `=0.04 cm^(3) xx 10.5 g cm^(-3) = 0.42 g`
Let the time required be t seconds. Total charge passed - 3t coulombs.
96500 coulombs of charge deposits = 108 g
3t coulombs of charge deposits `=108/96500 xx 3t g`
Thus, `108/96500 xx 3t = 0.42`
or `t = (0.42 xx 96500)/(108 xx 3) = 125.09 s`