Question

In the diagram of the electrolytic cell given below, A, B and C are connected in series having electrolytes of ZnSO₄, AgNO₃ and CuSO₄, respectively.

A steady current of 1.5 A was passed until 1.45 g of Ag was deposited at the cathode of cell B.

(Atomic mass of Ag = 108, Cu = 63.5, Zn = 65.3)

Answer the following questions:

1. How long did the current flow?
2. What weight of Cu and Zn was deposited at the cathode?

Answer 1

(1) Cell B

\[\ce{At Cathode - Ag+ + e- -> Ag(s)}\]

96500 C of current deposits 1 mole (108 g) of Ag.

1295.6 C of current will deposit 1.45 g of Ag.

Now, Q = it

1295.6 = 1.5 × t

t = 863.7 s

The current will flow for 863.7 seconds.

(2) Cell A

\[\ce{Zn^{+2} + 2e- -> Zn(s)}\]

2 moles of current deposits 65.3 g of Zn and 1295.6 C of current will deposit 0.438 g of Zn.

Cell C

\[\ce{Cu^{+2} + 2e- -> Cu(s)}\]

2 moles of current deposit 63.5 g of Cu, while 1295.6 C deposits 0.426 g of Cu.