When electricity is passed through a solution of AlCl3, 13.5 g of Al is deposited. The number of Faradays (charge) must be:
A
1
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B
1.5
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C
0.5
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D
2
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Solution
The correct option is D 2 AlCl3→Al3++3Cl− Using faraday's law of electrolysis w=E.i.tF where, w = amount of Al deposited F= faraday's constant i = current supplied t = total time for which current is supplied E = equivalent weight = Molar mass/ no of electron transfer per mole put up in equation; 13.5=273i.tF Q=i.t=1.5F