When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO_4 and CuSO_4 connected in series, 2g of cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A?

When a steady current of 2A was passed through two electrolytic cells

1.	When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO_4 and CuSO_4 connected in series, 2g of cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A?
Answer» Solution :Suppose the current flowed for t seconds Total charge =`2A times ts =2t` coulomb Thus, 2 FARADAY or `2 times 96500` coulombs charge is required to deposit 1 mole of copper `2 times 96500` coulombs deposit =63.5g of copper 2t coulombs deposit= `63.5/(2 times 96500) times 2t=(63.5t)/96500 G` Also `(63.5t)/96500 g=2g` or `t=(2 times 96500)/63.5=3039 seconds` Mass of Zn deposited According to Faraday.s second law `(Mass of zi nc)/(Eq. weight of Zn)=(Mass of copper)/(Eq. weight of copper)` or `(Mass of zi nc)/(65//2)=2/(63.5//2)` or Mass of zinc =`(2 times 65)/63.5=2.047g`