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 

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