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A Steady Current of 2 Amperes Was Passed Through Two Electrolytic Cells X and Y Connected in Series Containing Electrolytes Feso4and Znso4 Until 2.8g of Fe Deposited at

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Question

A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4and ZnSO4 until 2.8g 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=56g mol-1,Zn=65.3g mol-1,1F=96500C mol-1)

Answer in Brief
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Solution

I=2 A
W1= 2.8 g
Fe2+(aq)+2e-→Fe(s)
96500×2 C of charge is required to deposit=56 g of Fe
9650 C of charge is required to deposit=2.8 g of Fe
∵ Q `=1"t"  "or"  "t" 9650/2=4825"s"`
Using Faraday's second law of electrolysis,
`("W"_1("Weight of Fe deposited"))/("W"_2("Weight of Zn deposited")`
`=("E"_1("Equivalent weight of Fe"))/("E"_2("Equivalent weight of Fe")`
`2.8/"W"_2 =(56/5)/((65.3)/2)=56/65.3`
Or W2 = 3.265 g

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2018-2019 (March) 56/1/1

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