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