Advertisement Remove all ads

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


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)

Advertisement Remove all ads


Given:  I = 2A     

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

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
  Is there an error in this question or solution?
Advertisement Remove all ads


Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×