Three electrolytic cells A,B,C containing solutions of ZnSO4,AgNO3 and CuSO3, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Cell B: Ag++e−⇌Ag at cathode.
1 mole (108 g) of Ag is deposited by 96500 C.
1.45 g of Ag will be deposited by 96500×1.45108=1295.6C.
Now,
Q=It
1295.6=1.5×t
t=864s
Cell A: Zn2++2e−→Zn
2 moles of electrons (2×96500 C of current) produces 1 mole (63.5 g) of zinc.
1295.6 C of electricity will deposit 65.32×96500×1295.6=0.438 g of zinc
Cell C: Cu2++2e−→Cu
2 moles of electrons (2×96500 C) of current will produce 1 mole (63.5 g) of Cu.
1295.6 C of current will deposit 63.5×1295.62×96500=0.426g of copper