Electrochemical Extraction of Aluminum from Bauxite Ore Process

To determine the correct statements regarding the electrochemical extraction of aluminum from bauxite ore, we analyze each option in detail:

1. Option (A): Direct pyrometallurgical reduction of alumina ($\text{Al}_2\text{O}_3$) using carbon (coke) at high temperatures ($> 2000\,^\circ\text{C}$) is thermodynamically difficult and economically unviable because aluminum reacts with carbon at high temperatures to form aluminum carbide ($\text{Al}_4\text{C}_3$) rather than metallic aluminum: $2\text{Al}_2\text{O}_3 + 9\text{C} \rightarrow \text{Al}_4\text{C}_3 + 6\text{CO}$ Hence, carbon reduction is not employed for the extraction of aluminum. Thus, Option (A) is incorrect.

2. Option (C): In the Bayer process (concentration/leaching of bauxite ore), the powdered bauxite ore is digested with a concentrated hot aqueous solution of $\text{NaOH}$ at $473 - 523\text{ K}$ ($200 - 250\,^\circ\text{C}$) and $35 - 36\text{ bar}$ pressure. $\text{Al}_2\text{O}_3$ is amphoteric and dissolves to form soluble sodium aluminate, while impurities like $\text{Fe}_2\text{O}_3$ and $\text{TiO}_2$ remain insoluble: $\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \xrightarrow{\Delta} 2\text{Na}[\text{Al(OH)}_4]\text{(aq)}$ Thus, Option (C) is correct.

3. Option (B): The sodium aluminate solution obtained from the leaching step is neutralized by passing $\text{CO}_2$ gas through it. This precipitates hydrated alumina ($\text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ or $\text{Al}_2\text{O}_3 \cdot 3\text{H}_2\text{O}$): $2\text{Na}[\text{Al(OH)}_4]\text{(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O(s)} \downarrow + 2\text{NaHCO}_3\text{(aq)}$ The precipitated hydrated alumina is then filtered, dried, and heated above $1470\text{ K}$ to yield pure alumina ($\text{Al}_2\text{O}_3$). Thus, Option (B) is correct.

4. Option (D): In the Hall-Héroult process, pure $\text{Al}_2\text{O}_3$ is dissolved in molten cryolite ($\text{Na}_3\text{AlF}_6$) and fluorspar ($\text{CaF}_2$) to lower its melting point ($\sim 950\,^\circ\text{C}$) and increase electrical conductivity. Electrolysis is carried out using a steel vessel lined with carbon as the cathode and graphite rods as the anode:

   • At Cathode: $\text{Al}^{3+}(\text{melt}) + 3e^- \rightarrow \text{Al(l)}$

   • At Anode: $\text{C(s)} + \text{O}^{2-}(\text{melt}) \rightarrow \text{CO(g)} + 2e^-$ $\text{C(s)} + 2\text{O}^{2-}(\text{melt}) \rightarrow \text{CO}_2(\text{g}) + 4e^-$

   • Overall Reaction: $2\text{Al}_2\text{O}_3 + 3\text{C} \rightarrow 4\text{Al} + 3\text{CO}_2$

   This process yields pure aluminum at the cathode and releases $\text{CO}_2$ at the anode. Thus, Option (D) is correct.

Conclusion: The correct options are B, C, and D.