Question

Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm2 by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited.

[Given : the density of silver is 10.5 g cm-3 and atomic mass of Ag = 108 amu.]

Answer 1

Calculation of mass of Ag deposited :

The electrode reaction is Ag+ + e- → Ag

The quantity of electricity passed

= Current x Time

= 0.5 (amp.) x 2 x 60 x 60 (sec)

= 3600 C.

From the electrode reaction, it is clear that 96500 C of electricity *** Ag = 108 g

Calculation of thickness:

Let the thickness of silver deposited be x cm.

Mass = Volume x Density

= Area x Thickness x Density

(Volume = Area x thickness)