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Chemistry Question on Electrochemistry

When a quantity of electricity is passed through CuSO4CuSO_4 solution, 0.16 g0.16\, g of Copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H2H_2 liberated at STPSTP will be [Given At.Wt.Cu=64At. Wt. Cu = 64]

A

4.0 cm34.0 \,cm^3

B

56 cm356\, cm^3

C

604 cm3604 \,cm^3

D

8.0 cm38.0\, cm^3

Answer

56 cm356\, cm^3

Explanation

Solution

 Wt. of Cu deposited  Wt. of H2 produced = E  wt. of Cu E wt. of H\frac{\text { Wt. of Cu deposited }}{\text { Wt. of } H _{2} \text { produced }}=\frac{\text { E } \text { wt. of } Cu }{\text { E wt. of } H }
0.16wt . of H2=64/21=321\frac{0.16}{ wt \text { . of } H _{2}}=\frac{64 / 2}{1}=\frac{32}{1}

wt. of H2=0.1632=5×10−3 g H _{2}=\frac{0.16}{32}=5 \times 10^{-3}\, g

Volume of H2H _{2} liberated at STP

=224002×5×10−3 cc=56 cc=\frac{22400}{2} \times 5 \times 10^{-3} \,cc =56\, cc