Question

When a steady current of 2 A was passed through two electrolytic cells A and B containing electrolytes ZnSO₄ and CuSO₄ connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow?

What mass of Zn was deposited at cathode of cell A?

[Atomic mass : Cu = 63.5 g mol^–1, Zn = 65 g mol^–1 ; 1 F = 96500 C mol^–1]

Answer 1

From Faraday's laws of electrolysis,

equivalents of Cu deposited = equivalents of Zn deposited

                                             = Faraday's of charge passed.

Now, \(\text{Faradays of charge} = \frac{i \times t}{96500} = \frac{2t}{96500} \) 

 \(\text{and equivalents of Cu} = \frac{2}{\left(\frac{63.5}{2}\right)} = \frac{4}{63.5}\) 

\(\therefore \ \frac{2t}{96500} = \frac{4}{63.5}\)

\(\Rightarrow t \simeq 3039 \ sec \simeq 50.65 \text{ min}\) 

Also, equivalents of Zn = equivalents of Cu.

\(\frac{w \times 2}{65} = \frac{4}{63.5}\)

\(\Rightarrow w = 2.05 \ g\) 

Hence, the current was passed for 3039 sec or 50.65 min and the weight of zinc deposited is approximately 2.05 g