Question

Silver is deposited on a metallic vessel by passing a current of 0.2 amps. For 3 hrs. Calculate the weight of silver deposited​

Answer 1

Explanation:

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

.‘. 3600 C of charge will deposit Ag = 108x3600/96500 = 4.029 g

Thickness = Mass/AreaX Density

4.029/ 900 x 10.5

= 4.26 x${{10}^{-4}}$ cm

So Have You Got It