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:
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
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 16: Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are conn....
Comments
Thanks
it helped me thanks but not still know the current direction
Very better elaborated than my refresher
Ty for ur help