Al2O3 is reduced by electrolysis at low potentials and high currents . If 4.0 x 10^4 amperes of current is passed through molten Al2O3 for 6 hours, What mass of aluminium is produced?( Assume 100% current efficiency, At. mass of Al= 27g mol^-1)

1- 1.3 x 10^4g

2- 9 x 10^3 g

3- 8.1 x 10^4 g

4- 2.4 x 10^5 g

According to Faraday's first law of electrolysis :
Amount of material deposited is given as (W ) = z × I × t..........(1)
z = equivalent mass of metal deposited/faraday constant
  = equivalent mass of Al3+/96500 C
   = 27 g/3 / 96500 C
I = 4 x 104 A
t = 6 hr
 = 6 x 3600 s
Putting the values in eq(1) we get :
W = 9 g/96500 C x 4 x104 A  x 6 x 3600 s
   = 8.1 x 104 g


 

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