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Electrochemistry

Question
CBSEENCH12005976
Wired Faculty App

How much electricity in terms of Faraday is required to produce
40.0 g of Al from molten Al2O3?

Solution
Al2O3 ⇋ 2Al3++3O2-
or       Al3++3e- →     Al      3 mol               1 mol ( = 27 g)                                     (reduction of Al3+)
Charge on Al in Al2O3
2Al  + 3O = 0
Oxygen has –2
2Al +3(–2)   = 0
Al         = 3
Change transfer  n = 3
Charge required for 1 mol of Al    = 3F
Number of moles of Al  = 40 /27  = 1.48
 
The electricity required to produce 1.48 mol of Al = 1.48  × 3 F
                                                                                = 4.44 F

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