to prepare 5.12 kg of aluminium metal by this method,would require
Answer (1)
Student Expert
29th Nov, 2019
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
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