# 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. - Chemistry

Numerical

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)

#### Solution

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

Concept: Electrolytic Cells and Electrolysis - Introduction
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