A certain amount of electricity is passed through aqueous silver nitrate and copper sulphate solution connected in series .If the mount of silver deposited at the catchode is 1.08 g , calculate the amount of copper deposited .(Atomic weight of Ag= 108 and atomic weight of Cu = 63 . 53)

Explanation

According to the Faraday .s second law ,
`("Mass of Ag deposited")/("Mass of Cu deposited")=("equivalent mass of Ag")/("equivalent mass of Cu")`
`(1.08)/(W_(cu))=(108//1)/(63.53//2)=(108)/(31.76)`
Therefore , mass of Cu deposited `=(1.08xx31.76)/(108)`
`=0.3176 g`