Hello there!

Greetings!

I am extremely sorry to tell you dear but you haven't completed your questions. But I guess the question is asking about the quantity of electricity required to produce the aluminium.

Now,from Faraday's first law ,we know that

W = ZQ

where w = Weight Z = Electrochemical equivalent and Q = Quantity of electricity

Now, E = Z F

where E = equivalent weight and F = faraday

or W = E/FQ

or Q = WF/E

or Q = WF/A/n

Thus, Q = nwf/A

= 35.121096500/27

= 5.49 10 C

Thus, to prepare 5.12 kg of aluminium metal we need 5.49 10 C of electricity

Thankyou