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three electrolytic cells a b c containing solution...

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:

According to the reaction:

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by = 96487 X 1.45 / 108 C

= 1295.43 C

Given,

Current = 1.5 A

Time =1295.43 /1.5s

= 863.6 s

= 864 s

= 864/ 60

= 14.40 min

Again,

i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu

Therefore, 1295.43 C of charge will deposit = (63.5x1295.43) / (2x96487) g

= 0.426 g of Cu

i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn

Therefore, 1295.43 C of charge will deposit

= (65.4x1295.43) / (2x96487) g

= 0.439 g of Zn

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