Question

The Hall process is an important method by which pure aluminum is prepared from its oxide (alumina, \(\mathrm{Al}_{2} \mathrm{O}_{3}\) ) by indirect reaction with graphite (carbon). Balance the following equation, which is a simplified representation of this process.$$\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{Al}(s)+\mathrm{CO}_{2}(g)$$

Short answer

The balanced chemical equation for the Hall process is: \[\mathrm{Al}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \rightarrow 2\mathrm{Al}(s) + 3\mathrm{CO}_{2}(g)\]

Step-by-step solution

Write down the unbalanced equation

The unbalanced equation is given as: $$\mathrm{Al}_{2} \mathrm{O}_{3}(s) + \mathrm{C}(s) \rightarrow \mathrm{Al}(s) + \mathrm{CO}_{2}(g)$$

Count the number of atoms of each element on both sides of the equation

On the reactant side, we have: - 2 atoms of Aluminum (Al) - 3 atoms of Oxygen (O) - 1 atom of Carbon (C) On the product side, we have: - 1 atom of Aluminum (Al) - 2 atoms of Oxygen (O) - 1 atom of Carbon (C)

Balance the Aluminum atoms

Since we have 2 Al atoms on the reactant side and only 1 on the product side, let's multiply the Al on the product side by 2 to balance the Aluminum atoms: $$\mathrm{Al}_{2} \mathrm{O}_{3}(s) + \mathrm{C}(s) \rightarrow 2\mathrm{Al}(s) + \mathrm{CO}_{2}(g)$$ Now on the reactant side and product side, we have: - 2 atoms of Aluminum (Al)

Balance the Carbon and Oxygen atoms

With Aluminum balanced, let's now balance the Carbon and Oxygen atoms. We can do this by multiplying the CO₂ on the product side by 3: $$\mathrm{Al}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \rightarrow 2\mathrm{Al}(s) + 3\mathrm{CO}_{2}(g)$$ Now on both the reactant side and product side, we have: - 3 atoms of Oxygen (O) - 3 atoms of Carbon (C)

Check if the equation is balanced

We verify that the number of atoms for each element is the same on the reactant and product sides of the equation: - 2 atoms of Aluminum (Al) on both sides - 3 atoms of Oxygen (O) on both sides - 3 atoms of Carbon (C) on both sides As all atoms are balanced, the equation is now balanced. The final and balanced chemical equation for the Hall process is: $$\mathrm{Al}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \rightarrow 2\mathrm{Al}(s) + 3\mathrm{CO}_{2}(g)$$

Key concepts

Chemical Equation Balancing

Understanding how to balance chemical equations is crucial for studying any chemical reaction, including the Hall process for aluminum extraction. Balancing chemical equations ensures that the law of conservation of mass is upheld, meaning the matter cannot be created or destroyed during a chemical reaction.

For instance, in the simplified chemical equation of the Hall process, \(\mathrm{Al}_{2} \mathrm{O}_{3}(s) + \mathrm{C}(s) \rightarrow \mathrm{Al}(s) + \mathrm{CO}_{2}(g)\), we initially observe an imbalance in the quantities of atoms. The goal is to adjust coefficients, the numbers placed in front of compounds, to ensure that the number of each type of atom on the reactants side equals the number on the products side.

After a step-by-step balancing process, we find that a coefficient of 3 in front of \(\mathrm{C}(s)\) and \(\mathrm{CO}_{2}(g)\) yields a balanced equation: \(\mathrm{Al}_{2} \mathrm{O}_{3}(s) + 3\mathrm{C}(s) \rightarrow 2\mathrm{Al}(s) + 3\mathrm{CO}_{2}(g)\). This balance reflects the stoichiometry of the Hall process, key to understanding the quantities of reactants needed and products formed.

Aluminum Extraction

Aluminum extraction through the Hall process is an industrially significant application of chemistry that involves converting bauxite ore, primarily composed of alumina (\(\mathrm{Al}_{2} \mathrm{O}_{3}\)), into pure aluminum metal.

The process is historically pivotal as aluminum was once more valuable than gold or silver due to its difficulty to extract. However, with the advent of the Hall-Héroult process in the late 19th century, pure aluminum became much more accessible. This electrolytic method entails dissolving alumina in molten cryolite and reducing it to aluminum at the cathode, while oxygen is formed at the anode, typically made from carbon. The carbon anode is consumed in the process, reacting with the oxygen to form carbon dioxide.

The innovative process requires less energy and has enhanced our ability to mass-produce aluminum, reducing costs dramatically and making it a staple in various industries, from aerospace to food packaging.

Alumina Reduction

Reduction of alumina to aluminum in the Hall process is the critical stage that transforms a non-metal oxide into a pure metal. This is achieved by a redox reaction in an electrolytic cell. Reduction in chemistry means the gaining of electrons, and in the Hall process, alumina receives electrons to become aluminum metal, effectively reducing the aluminum ions present in alumina.

To facilitate this transformation, a high electric current is passed through the molten alumina-cryolite mixture. The cryolite acts as a solvent and lowers the melting point of alumina, making the process more efficient. The aluminum ions are reduced at the cathode, and oxygen ions are oxidized at the anode. This reaction is represented by the equation \(\mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al}(s)\), signifying aluminum ions gaining three electrons to form pure aluminum.

Understanding the principles of redox reactions is essential to comprehend the chemistry behind the Hall process, as it plays a central role in the extraction and production of various metals beyond aluminum.