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प्रश्न
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)
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उत्तर
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
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