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Question

Al2O3 is reduced by electrolysis at low potential and high currents. If 4.0×104 amperes of current is passed through molten Al2O3 for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, At. mass of Al=27gmol1)

A
1.3×104g
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B
9.0×103g
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C
8.1×104g
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D
2.4×105g
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Solution

The correct option is D 8.1×104g
Al2O3(l)2Al3+(l)+3O2(l)
Reduction equation of Al3+:
Al3+(aq)+3eAl(s)
As,
Ampere = Coulomb / sec
Time unit should be second.
6hr×60minhr×60secmin=21600sec
Coulomb = Ampere x second =40000×21600=8.64×108
1 mol e carries a charge of 1 faraday, or 96485 coulombs.
Then the mole of e used in the electrolysis will be:
8.64×108coulombs96485coulombs/molofe=8954.8mol ofe
According to the reduction equation, 3 moles of e- are needed to reduce 1 mol of Al^3+ to Al metal.
Then , the mole of Al to be deposited at the cathode will be;
8954.8mol e-3mol of e-mol Al=2985 mol Al
Molar mass of Al:26.98g/mol
Mass of Al produced:2985mol×26.98g/mol=80535g
Al=8.1×104g

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