The same quantity of electricity is passed through solutions of silver nitrate and cupric sulphate connected in series. If the weight of silver deposited is 0.54 g, calculate the weight of copper deposited. (Equivalent mass of Ag = 108, equivalent mass of Cu = 31.75)

Explanation

According to Faraday.s second law
`(W_(Cu))/(W_(Ag)) = (E_(Cu))/(E_(Ag)) rArr W_(Cu) = (E_(Cu) xx W_(Ag))/(E_(Ag))`
Mass of Cu deposited `=(31.75 xx 0.54)/(108) = 0.159g`