Molten aluminim chloride is electrolysed with a current of 0.5 ampere to produce 27.0 g of aluminium. (a) How many gram equivalents of aluminium were producted ? (b) How long did the electrolysis take place ? (c ) How many litres of chlorine were evolved at S.T.P. ?

Molten aluminim chloride is electrolysed with a current of 0.5 ampere

1.	Molten aluminim chloride is electrolysed with a current of 0.5 ampere to produce 27.0 g of aluminium. (a) How many gram equivalents of aluminium were producted ? (b) How long did the electrolysis take place ? (c ) How many litres of chlorine were evolved at S.T.P. ?
Answer» Solution :The electrolysis of MOLTEN aluminium chloride is represented as : `AlCl_(3)hArrAl^(3+)(l)+3Cl^(-)(l)` `Al^(3+)(l)+3e^(-) to Al(s) , 3Cl^(-)(l) to 3//2 Cl_(2)(G)+3e^(-)` (a) Gram equivalent of aluminium deposited `=("Mass of Al deposited")/("Equivalent mass of" Al)=(27g)/((9equiv^(1)))=3 g equiv^(-1)`. (b) TIME for which electrolysis was carried : `Al^(3+)(l)+underset((3 Faraday))3e^(-) to underset((27 g))Al(s)` Let the current be passed for time =t sec. `:.""` Quantity of charge (Q) passed`=(0.5 amp)xxt(s)=0.5xxt(C)` Actual amout of charge passed`=3F=3xx96500" C"` `:.` Time (t) for which electrolysis was carried `=(3xx96500(C))/(0.5(C))=579000 s=(579000)/(60xx60)=160.83` hr. (c ) Litres of CHLORINE EVOLVED : `3Cl^(-) to underset(3//2xx22.4 L) Cl_(2)(g)+underset(3F)(3e^(-))` =33.6 L 3 Faraday of charge will evolve `Cl_(2)`(g)=33.6 L.