3.16     Three electrolytic cells A,B,C containing solutions of ZnSO_{4}AgNO_{3} and CuSO_{4}, 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?

Answers (1)

Since the cells are connected in series so the current passing through each cell will be equal.(1.5 A)

Now we are given that 1.45 g of silver is deposited. So firstly we will consider the cell containing silver.

                                                   Ag^+\ +\ e^-\ \rightarrow Ag

Since for deposition of 108 g silver 96487 C charge is required, thus for 1.45 g deposition of silver charge required will be:- 

                                                              = \frac{96487}{108}\times1.45        = 1295.43\ C

Now we can find the time taken by 1.5 A current to deposit 1.45 g silver.

                                                         Time\ taken = \frac{1295.43}{1.5} \approx 864\ sec.

For copper:- 

                                Cu^{+2}\ + 2e^-\ =\ Cu

Since 2F charge will deposit 63.5 g of Cu,  then deposition by 1295.43 C will be:-

                                                                                                 = \frac{63.5}{2\times96487}\times1295.43         = 0.426\ g

 Hence 0.426 g of copper will be deposited.   

 For zinc:-  

                               Zn^{+2}\ +\ 2e^-\ \rightarrow\ Zn

Since 2F charge will deposit 65.4 g of Zn,  then deposition by 1295.43 C will be:-

                                                                                                 = \frac{65.4}{2\times96487}\times1295.43         = 0.439\ g

 Hence 0.439 g of zinc will be deposited.   

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