Question

4(c). During the electrolysis of molten Al₂O₃, a current of 6A was passed through the electrolyte for 1hr 30mins. Calculate the mass of aluminium deposited at the cathode.

Accepted Answer

Al³⁺(l)+3e⁻→Al(s)
Q=It, I=6A, t=1hr30mins=90mins=5400s.
Q=6×5400=32400C.
From the equation, 1 mole of Al³⁺ requires 3e⁻ (3F). Since 1F=96500C, quantity of electricity required for 1 mole (27g) of aluminium = 3×96500=289500C.
If 289500C deposits 27g of aluminium, 32400C will deposit (27/289500)×32400 = 3.02g of aluminium.

WAEC Chemistry 2014 Theory — Question 32

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