Question

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?

Answer 1

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 = $\frac{12945.43}{1.5}$ s
= 863.6 s
= 864 s
= 14.40 min
Again,
Cu2+(aq) + 2e- $\rightarrow$ 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+(aq) + 2e- $\rightarrow$ 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