Question

The quantity of electricity required to produce 18 g of Al from molten Al\(_2\)O\(_3\) is (Atomic mass of Al = 27):

• A. 2F
• B. 4F
• C. 5F
• D. 6F
• E. 1.5F

Hint:

Remember, each Faraday (F) corresponds to the charge of one mole of electrons, approximately 96500 coulombs. In electrolysis calculations, the charge (in Faradays) directly corresponds to the amount of substance reacted or produced based on stoichiometry.

Answer 1

The Correct Option is A

The reduction reaction at the cathode during the electrolysis of aluminum oxide (Al\(_2\)O\(_3\)) is typically: \[ {Al}^{3+} + 3e^- \rightarrow {Al} (s) \] For every mole of aluminum produced, three moles of electrons are required. To produce 18 g of aluminum: \[ {Moles of Al} = \frac{18 \, {g}}{27 \, {g/mol}} = \frac{2}{3} \, {mol} \] Thus, the number of moles of electrons needed is: \[ 3 \times \frac{2}{3} \, {mol} = 2 \, {mol} \, {of electrons} \] Since 1 mole of electrons corresponds to 1 Faraday (F) of charge, the total charge required is: \[ 2 \, {mol} \times 1F/{mol} = 2F \]