Problem 178 - A Complete Resource Book in Chemistry for JEE Main

Question

In the context of the Hall-Heroult process for the extraction of \(\mathrm{Al}\), which of the following statement is false? (a) \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) are produced in this process (b) \(\mathrm{Al}_{2} O_{3}\) is mixed with \(\mathrm{CaF}_{2}\) which lowers the melting point of the mixture and brings conductivity (c) \(\mathrm{Al}^{3+}\) is reduced at the cathode to form \(\mathrm{Al}\) (d) \(\mathrm{Na}_{3} \mathrm{~A} / \mathrm{F}_{6}\) serves as the electrolyte

Step-by-Step Solution

Answer

Statement (d) is false.

Step 1: Understanding the Hall-Heroult process

The Hall-Heroult process is employed for extracting aluminum from its oxide, alumina (\(\mathrm{Al}_2\mathrm{O}_3\)). In this process, alumina is dissolved in molten cryolite (\(\mathrm{Na}_3\mathrm{AlF}_6\)) which acts as the solvent.

Step 2: Verify Statement (a)

In the Hall-Heroult process, carbon anodes are used, and they react with the oxygen from the \(\mathrm{Al}_2\mathrm{O}_3\) to form \(\mathrm{CO}\) and \(\mathrm{CO}_2\). Therefore, statement (a) is true as these gases are indeed produced.

Step 3: Verify Statement (b)

While cryolite (\(\mathrm{Na}_3\mathrm{AlF}_6\)), not calcium fluoride (\(\mathrm{CaF}_2\)), is primarily used to reduce the melting point and bring conductivity, \(\mathrm{CaF}_2\) can be used sometimes to further adjust the melting properties. Therefore, statement (b) is true in general practice.

Step 4: Verify Statement (c)

At the cathode of the Hall-Heroult cell, aluminum ions are reduced as \(\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}\). Hence, statement (c) is true, Al is produced at the cathode through reduction.

Step 5: Verify Statement (d)

\(\mathrm{Na}_3\mathrm{AlF}_6\) is commonly known as cryolite and is the solvent, not the electrolyte. The actual electrolyte consists of molten alumina dissolved in cryolite. Therefore, statement (d) is false as it incorrectly states that \(\mathrm{Na}_3\mathrm{AlF}_6\) serves as the electrolyte by itself.

Key Concepts

Aluminum Extraction

Aluminum extraction primarily occurs through the Hall-Heroult process. This method is a highly efficient way to obtain aluminum from its ore, alumina (\(\mathrm{Al}_2\mathrm{O}_3\)). The process begins by dissolving alumina in molten cryolite. Cryolite serves as a solvent, enhancing the efficiency of the extraction process by lowering the melting point of alumina.

Here's how the Hall-Heroult process works:

• Alumina is mixed with cryolite to form a molten solution.
• When electric current passes through this solution, aluminum ions (\(\mathrm{Al}^{3+}\)) migrate towards the cathode, where they gain electrons and become aluminum metal.
• This conversion is made possible by electrolysis, a key step in the process, which involves breaking down compounds using electricity.

This process is carried out in large industrial cells and is widely used in aluminum production due to its effectiveness and cost efficiency. Understanding this process is crucial to comprehend how aluminum is made from its core material, alumina.

Cryolite

Cryolite (\(\mathrm{Na}_3\mathrm{AlF}_6\)) is an essential component in the aluminum extraction process. It plays multiple roles, but its most significant function is to act as a solvent that facilitates the dissolution of alumina.

One of the key benefits of using cryolite is that it significantly lowers the melting point of alumina from about 2072°F (1133°C) to approximately 1832°F (1000°C). This reduction in the melting point:

• Decreases the energy needed to keep the cell at operating temperature
• Enhances the mobility of ions, making the electrolysis process more efficient
• Saves cost by decreasing the overall energy consumption

Sometimes, calcium fluoride (\(\mathrm{CaF}_2\)) is added to further adjust the molten solution's properties, although cryolite remains the primary agent. Therefore, in the Hall-Heroult process, cryolite is indispensable for efficient aluminum extraction.

Electrolysis

Electrolysis is a key chemical process in aluminum extraction. In the context of the Hall-Heroult process, it involves passing an electric current through molten alumina dissolved in cryolite.

This electric current flows between two types of electrodes:

• Anodes: These are made of carbon and positively charged.
• Cathodes: The negatively charged electrodes where reduction takes place.

As electricity flows, it causes the aluminum ions (\(\mathrm{Al}^{3+}\)) in the solution to move towards the cathode. Here, they gain electrons to form pure aluminum.: \[\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}\]At the same time, oxygen ions move towards the anode, where they combine with the carbon to form carbon monoxide (\(\mathrm{CO}\)) and carbon dioxide (\(\mathrm{CO}_2\)). This process, although energy-intensive, is crucial for effectively isolating pure aluminum from its ore.

Cathodic Reduction

Cathodic reduction refers to the gain of electrons by metal ions during electrolysis, an essential step in the Hall-Heroult process. At the cathode, which is the negatively charged electrode, aluminum ions (\(\mathrm{Al}^{3+}\)) undergo reduction.

Here's how this works:

• Aluminum ions are attracted to the cathode due to its opposite charge.
• These ions receive three electrons (\(3e^-\)) to become neutral aluminum atoms.
• The reaction can be expressed as: \(\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}\)

This cathodic reaction is vital to produce aluminum metal from its oxide form. The other side of this process involves the anodes, where oxygen evolves, forming gases that must be managed. By efficiently conducting this reaction, the Hall-Heroult process ensures the pure metal is obtained.