Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, 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?
According to the reaction:
Ag+(aq) + e- → Ag(s) (108 g)
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = \(\frac{96487 \times 1.45}{108}\)C
= 1295.43 C
Given, Current = 1.5 A
∴ Time = \(\frac{1295.43}{1.5}\)s
= 863.6 s
= 864 s
= 14.40 min
Again,
Cu2+ + 2e- → Cu(s) (63.5 g)
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = \(\frac{63.5 \times 1295.43}{2 \times 96487}\)g
= 0.426 g of Cu
Zn2+ + 2e- → Zn(s) (65.4 g)
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit = \(\frac{65.4 \times 1295.43}{2 \times 96487}\)g
= 0.439 g of Zn