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  1. moles Al = 78 g/ 26.9815 g/mol= 2.89
    The cathode half-reaction is
    Al3+ + 3e- >> Al
    it states that for evey 3 faradays consumed onemole Al is produced

    this means to get 2.89 moles Al we need to use

    2.89 mol ( 3 faradays / 1mole) = 8.67 faradays

    1 F = 96500 Coulombs

    8.67 faradays = 8.67 faradays ( 96500 Coulomb/F)= 836655 coulombs

    836655 Coulomb / 2.0 Coulomb/s = 418328 s

How long will it take to produce 78 g of Al metal by the reduction of Al3+ in an electrolytic cell with a..?

current of 2.0 A..
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