Silver is electrodeposited on a metallic vessel of surface area 800 cm2 by passing
current of 0.2 A for 3 hours. Calculate the thickness of silver deposited.Given:
Element = Ag = at mass = 107.92
Current = 0.2A
Time= 3h = 10800 secs
In coulombs = 0.2 x 10800 = 2160 C
We know that 1A = 96500 C will deposit 107.92 g of Ag
Therefore total amount of Ag deposited from 2160 C can be found out as
(107.92/96500) x 2160 = 2.4156 gms
density of Ag =10.47 g/cc
Volume of Ag deposited = Mass / Density = 2.4156 / 10.47
= 0.2307 cc
thickness deposited = Vol / area
= 0.2307/ 800
thickness= 2.88x10-4 cm