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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.)

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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?

Numerical
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Solution

\[\ce{Al2O3 -> 2Al^{3+} + 3O^{2-}}\]

\[\ce{\underset{1 mole}{Al^3+} + \underset{3 moles}{3e-} -> \underset{1 mole}{Al}}\]

27 g of aluminium needs = 3 mole of electrons

∴ 40.0 g of aluminium needs = `(3 xx 96500 xx 40.0)/27`

= 4.28888 × 105 coulombs

= 4.44 F

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Chapter 2: Electrochemistry - Exercises [Page 60]

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