Question

Calculate the volume of Cl₂ at STP produced during electrolysis of MgCl₂ which produces 6.50 g of Mg. (At. wt. of Mg = 24.3)

Answer 1

According to Faraday’s second law of electrolysis,

`(W_(Mg))/W_(Cl_2) = (E_(Mg))/E_(Cl_2)`

The reactions involved are:

\[\ce{MgCl2 <=> Mg^2+ + 2Cl-}\]

At anode: \[\ce{Cl- - e -> \frac{1}{2} Cl2}\]

At cathode: \[\ce{Mg^2+ + 2e- -> Mg}\]

∴ Equivalent mass of Mg (E_Mg) = `24.3/2`

= 12.15

Equivalent mass of Cl (E_Cl) = `35.5/1`

= 35.5

Thus, we have

`6.50/(W_(Cl_2)) = 12.15/35.5`

or, `W_(Cl_2) = (6.50 xx 35.5)/12.15`

= 18.99 g

∵ One mole (71 g) of Cl₂ has a volume of 22.4 L at STP.

∴ Volume of 18.99 g of Cl₂ at STP

= `22.4/71 xx 18.99`

= 5.99 L