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Q. When the same quantity of current is passed through $CuSO_4$ and $AgNO_3$ solution $2.7 \,g$ of silver is deposited. The amount of copper deposited is

COMEDKCOMEDK 2011Electrochemistry

Solution:

According to Faraday's second law :
$ \frac{ Wt. of \,Cu \, deposited}{ Wt. of\, Ag \,deposited} = \frac{Eq. wt.of \,Cu}{Eq. wt. \,of\, Ag} $
Wt. of Cu deposited
$ = \frac{Eq. wt. of\, Cu \times Wt. of \,Ag \, deposited}{ Eq. wt. of\, Ag} $
$ = \frac{63.5 / 2 \times 2.7}{108} = 0.793 = 0.80 \, g$