The answer is 24.3 h.
Solution:
The half-reaction for the reduction of Al3+ (in Al2O3) is given by
Al3+(aq) + 3e- (aq) --> Al(s)
Faradays 1st law is used
w = (i x t / F) x (M/z)
i is current used = 20.0 A
t is time in seconds which we calculate first
F is Faraday constant = 96485 C/mol
M is atomic mass of Al = 26.98g/mol
z is the number of electrons involved per reduction = 3
w is the weight of Al = weight of Al required for 12 cans
= ( 13.6 g / 1can) x 12 cans = 163.2g
substituting we get
163.2g = (20.0A x t / 96485C/mol) x (26.98g/mol / 3)
t = 87545 s
= (87545s) x ( 1h / 3600s)
= 24.3 h
Thus, the time in hours needed for the production of Aluminium that can make up 12 cans = 24.3.
Electrolysis of molten aluminum oxide at carbon electrodes produces pure metal at the cathode and oxygen at the anode. This oxygen reacts with the positively charged carbon electrode to produce carbon dioxide gas. Extraction is done by electrolysis, but the alumina must first be melted so that electricity can flow.
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