Introduction:
The Hall-Heroult process is a method used to extract aluminum metal from purified bauxite ore. It involves the electrolysis of molten aluminum oxide (Al2O3) dissolved in a molten cryolite (Na3AlF6) bath. This process takes place in a cell with a carbon anode and a carbon cathode.

The Electrolysis Reaction:
At the anode (positive electrode), oxygen gas (O2) is produced through the oxidation reaction:
2O2- → O2 + 4e-

At the cathode (negative electrode), aluminum metal (Al) is formed through the reduction reaction:
Al3+ + 3e- → Al

Explanation:
According to the stoichiometry of the electrolysis reaction, for every 4 moles of electrons transferred at the anode, 1 mole of O2 is produced. Since the balanced equation for the oxidation reaction involves the transfer of 4 electrons, we can conclude that for every mole of O2 formed, 4 moles of electrons are involved.

Considering the molar ratio between the electrons and aluminum ions in the reduction reaction, we can determine the number of moles of aluminum formed at the cathode. For every 3 moles of electrons transferred, 1 mole of Al is produced.

Since the number of moles of electrons involved in the oxidation reaction is equal to the number of moles of electrons involved in the reduction reaction, we can conclude that for every 1 mole of O2 formed, 1 mole of Al is also formed.

However, the question states that 3 moles of O2 are formed at the anode. As we established earlier, for every 1 mole of O2 formed, 1 mole of Al is formed. Therefore, with 3 moles of O2 formed, the number of moles of Al formed at the cathode would be 3 times the number of moles of O2, which is 3 x 1 = 3 moles of Al.

However, the correct answer given is '4', which contradicts the stoichiometry of the reaction. It is possible that there may be an error in the answer provided.