In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of:

The extraction of aluminium by the Hall-Héroult process relies on the electrolytic reduction of molten $$\text{Al}_2\text{O}_3$$ dissolved in molten cryolite ($$\text{Na}_3\text{AlF}_6$$). In any electrolytic cell, we always require two electrodes: the anode (positive) where oxidation occurs and the cathode (negative) where reduction takes place.

In this commercial cell, both the anodes and the cathode are normally made of carbon (graphite). The anodes are large blocks that dip into the electrolyte from the top, while the entire inner lining of the steel container is packed with a thick layer of carbon; this carbon lining itself is connected to the negative terminal of the power supply and therefore serves as the cathode.

At the cathode, the desired reduction reaction is written as $$\text{Al}^{3+} + 3e^- \longrightarrow \text{Al(l)}.$$ We can see that nothing in this reaction would demand an expensive or noble-metal electrode. What is essential is a material that

• conducts electricity well,
• withstands the very high temperature (about 950 °C), and
• is inexpensive because the process is carried out on a huge scale.

Carbon (graphite) meets all these requirements: it is a good electronic conductor, it remains solid and stable at the working temperature, and it is far cheaper than metals like platinum or even copper. For this reason the industry uses carbon both for the cathode lining and for the anode blocks. Platinum or copper would be economically prohibitive and pure aluminium cannot serve as a cathode material because we are actually collecting liquid aluminium at the bottom of the cell.

Therefore, the cathode in the Hall-Héroult process is made of carbon.

Hence, the correct answer is Option B.