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Chemistry Question on Electrochemistry

When the same quantity of current is passed through CuSO4 {CuSO_4} and AgNO3 {AgNO_3} solution 2.7 g of silver is deposited. The amount of copper deposited is

A

0.4 g

B

3.2 g

C

1.6 g

D

0.8 g

Answer

0.8 g

Explanation

Solution

According to Faraday's second law :
Wt.of Cu depositedWt.of Ag deposited=Ewt.of CuEwt. of Ag{ \frac{ Wt. of \,Cu \, deposited}{ Wt. of\, Ag \,deposited} = \frac{E wt.of \,Cu}{E wt. \,of\, Ag} }
Wt. of Cu deposited
=Ewt.of Cu×Wt.of Ag depositedEwt.of Ag= { \frac{E wt. of\, Cu \times Wt. of \,Ag \, deposited}{ E wt. of\, Ag} }
=63.5/2×2.7108=0.793=0.80 g= \frac{63.5 / 2 \times 2.7}{108} = 0.793 = 0.80 \, g