Final answer:
Aluminum forms at the cathode during the electrolysis of a molten mixture containing aluminum oxide. The reduction of aluminum(III) ions takes place at the cathode, leading to the production of molten aluminum metal.
Step-by-step explanation:
In the given electrolysis reaction, **aluminum** would form at the **cathode**. The electrolysis involves the reduction of aluminum ions to aluminum metal, which is a fundamental concept in electrochemistry. In processes like the Hall-Héroult process, **aluminum(III) ions** are reduced at the cathode to produce molten aluminum. Specifically, in this widely used industrial method, a molten mixture that includes aluminum oxide (Al2O3) is electrolyzed, resulting in the formation of aluminum at the cathode.
Moreover, the **Al2O3** must be dissolved in a mixture with other components like cryolite and calcium fluoride to lower its melting point for efficient electrolysis. At high temperatures, these substances form a solution that allows the easy movement of ions, facilitating the reduction of aluminum ions to aluminum metal. The half-reaction at the cathode in this scenario would involve the formation of aluminum metal from aluminum(III) ions, represented by the equation Al3+ + 3e- → Al(l).