Question

Silver is electrodeposited on a metallic vessel of total surface area 900 cm² by passing a current of 0.5 A for 2 h. Calculate the thickness of silver deposited. Given, density of silver = 10.5 g Atomic mass of silver = 108 u,F = 96500 C.

Answer 1

Q = it = 0.5 x 2 x 60 x 60 = 3600 C
Ag⁺ + e^- → Ag

1F =96500 C 108 g

96500 C of charge deposits 108 g Ag

\(\therefore\) 3600 C of charge will deposit Ag 

= 108 x 3600/96500 = 4.029 g

Thickness = Mass/Area x Density

4.029/ 900 x 10.5

= 4.26 x 10^-4 cm