Question

Solve the following question.
A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO₄ and ZnSO₄ until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass : Fe = 56 g mol^–1, Zn = 65.3 g mol^–1, 1F = 96500 C mol^–1)

Answer 1

Given:  I = 2A     

(i)
Fe²⁺ 2e^– → Fe

56 g of Fe requires = 2 × 96500 C charge

2.8 g of Fe requires = `(2 xx 96500)/(56)` x 2.8 C charge

Q = 9650
Q = It

`"t" = "Q"/"I" = (9650)/(2)` = 4825 sec

(ii)
Zn²⁺ + 2e^– → Zn

2 x 96500 C of electricity deposit Zn = 65.3 g

1 C of elecrticity deposit Zn = `(65.3)/(2 xx 96500)`

9650 C  of elecrticity deposit Zn = `(65.3)/(2 xx 96500) xx 9650`

= 3.265 g