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प्रश्न
How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten Al2O3?
(Given: Molar mass of Aluminium is 27 g mol−1.)
How much electricity in terms of Faraday is required to produce 40.0 g of Al from molten Al2O3?
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उत्तर १
\[\ce{Al2O3 -> 2Al^{3+} + 3O^{2-}}\]
27 g of aluminium needs = 3 mole of electrons
= 3 × 96500 coulombs
∴ 40.0 g of aluminium needs = `(3 xx 96500 xx 40.0)/27`
= 4.28888 × 105 coulombs
= 4.44 F
उत्तर २
\[\ce{1/2 Al2O3 -> Al}\]
or \[\ce{\underset{1 mole}{Al^3+} + \underset{3 moles}{3e-} -> \underset{1 mole}{Al}}\]
26.98 g of aluminium require electricity = 3F
∴ 40 g of Al will require electricity = `(3 xx F)/26.98 xx 40`
= 4.448 F
Notes
Students should refer to the answer according to their questions.
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