The incorrect statement about the Hall-Heroult process is:
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- Carbon anode is oxidised to CO and CO2
- Na3AlF6 helps to decrease the melting point of the electrolyte
- CaF2 helps to increase the conductivity of the electrolyte
- Oxidation state of oxygen changes in the overall cell reaction
The Correct Option is D
Approach Solution - 1
The Hall-Heroult process is used for the extraction of aluminum from its ore, bauxite. It involves the electrolysis of molten aluminum oxide (Al$_2$O$_3$), and the following points are true for this process:
- Option A: Carbon anode is oxidized to CO and CO$_2$ — This is correct. In the Hall-Héroult process, the carbon anode undergoes oxidation, releasing carbon dioxide (CO$_2$) and carbon monoxide (CO) gases.
Option B: Na$_3$AlF$_6$ helps to decrease the melting point of the electrolyte — This is also correct. Sodium hexafluoroaluminate (Na$_3$AlF$_6$), commonly known as cryolite, is added to the electrolyte to lower its melting point and make the process more efficient.
Option C: CaF$_2$ helps to increase the conductivity of the electrolyte — This is correct. Calcium fluoride (CaF$_2$) is added to increase the conductivity of the molten electrolyte.
Option D: Oxidation state of oxygen changes in the overall cell reaction — This is incorrect. In the overall cell reaction, oxygen remains in the same oxidation state. Oxygen ions are reduced at the cathode to form aluminum, and at the anode, oxygen ions are oxidized to produce oxygen gas, but the oxidation state of oxygen remains -2 in both cases.
Thus, the incorrect statement is (D). The oxidation state of oxygen does not change in the overall cell reaction.
Therefore, the correct answer is (D).
Approach Solution -2
- In the Hall-Heroult process, the carbon anode is oxidized to produce CO and CO₂.
- Na₃AlF₆ (sodium cryolite) is used to lower the melting point of the electrolyte, making the process more efficient.
- CaF₂ (calcium fluoride) increases the conductivity of the electrolyte.
- The incorrect statement is (D) because in the Hall-Heroult process, oxygen in the electrolyte is reduced at the anode to form oxygen gas (O₂) without a change in its oxidation state. Oxygen is reduced (not oxidized) at the cathode in the overall reaction.
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