In Hall-Heroult process, the following is used for reducing Al$$_2$$O$$_3$$:-

In the Hall-Heroult process for the extraction of aluminium, $$Al_2O_3$$ is dissolved in a molten mixture and electrolyzed.

We have the following setup in the Hall-Heroult process:

- $$Al_2O_3$$ (alumina) is dissolved in molten cryolite ($$Na_3AlF_6$$)

- $$CaF_2$$ (fluorspar) is added to lower the melting point of the mixture

- Graphite electrodes are used (carbon anodes)

- Electrolysis reduces $$Al^{3+}$$ to Al at the cathode

Now, at the cathode: $$Al^{3+} + 3e^- \to Al$$

At the anode: $$C + O^{2-} \to CO_2 + 4e^-$$

So the carbon (graphite) anode acts as the reducing agent in the overall reaction, while $$CaF_2$$ serves as a flux to lower the melting point.

Hence, the correct answer is Option D: $$CaF_2$$.