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

Explanation

Calculations of mass of Ag deposited:
The electrode reaction is `Ag^(+)+e^(-) to Ag`
The quantity of electricity passed
`= Current times time`
`=0.5 (amp.) times 2 times 60 times 60 (sec)`
=3600 C.
From the electrode reaction, it is clear that 96500C of electricity deposit Ag=108g
360 C of electricity will deposit Ag=`108/96500 times 3600`
=4.03g
Calculation of thickness:
Let the thickness of deposited be x cm.
Mass= Volume `times` density
= Area `times` thickness `times` density
(Volume= Area` times` thickness)
`4.03g=900(cm^2) times x (cm) times 10.5 (g cm^-3)`
`x=4.03/(900 times 10.5)=4.26 times 10^-4 cm`