Question

The overall reaction for the electrolytic production of aluminum by means of the Hall process may be represented as $$ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Al}(l)+3 \mathrm{CO}(g) $$ At \(1000^{\circ} \mathrm{C}\), the standard free-energy change for this process is \(594 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the minimum voltage required to produce 1 mole of aluminum at this temperature. (b) If the actual voltage applied is exactly three times the ideal value, calculate the energy required to produce \(1.00 \mathrm{~kg}\) of the metal.

Short answer

(a) Minimum voltage needed is 1.03 V, (b) Energy required is 11.31 MJ for 1.00 kg of Al.

Step-by-step solution

Understand the Reaction

The given chemical reaction represents the electrolysis of aluminum oxide to produce aluminum and carbon monoxide. The reaction is: \[ \mathrm{Al}_2\mathrm{O}_3(s) + 3\mathrm{C}(s) \longrightarrow 2\mathrm{Al}(l) + 3\mathrm{CO}(g) \] and occurs at \(1000^{\circ} \mathrm{C}\) with a standard free-energy change of \( \Delta G = 594 \text{ kJ/mol} \).

Calculate the Moles of Electrons Transfer

For the overall reaction, every mole of \(\mathrm{Al}_2\mathrm{O}_3\) produces 2 moles of \(\mathrm{Al}\), which involves 6 moles of electrons since each aluminum ion needs 3 electrons to be reduced to aluminum metal. Therefore, the number of moles of electrons, \( n = 6 \).

Use Gibbs Free Energy to Calculate Minimum Voltage

The relationship between the free-energy change and the voltage is given by \(\Delta G = -nFE\) where \(F\) is Faraday's constant \((96485\;C/mol)\) and \(E\) is the voltage. Solving for \(E\), we have \[ E = -\frac{\Delta G}{nF} \] Substitute \( \Delta G = 594,000 \; \text{J/mol} \), \( n = 6 \), and \( F = 96485 \) C/mol to find \(E\).

Calculate the Ideal Voltage

Substituting values into the equation, we have:\[ E = -\frac{594,000}{6 \times 96485} = -1.03 \text{ V}\] (It is a negative sign, indicating the energy required.) Thus, the minimum voltage required is \(1.03\; \text{V}\).

Calculate Actual Voltage Applied

According to the problem, the actual voltage applied is three times the ideal value. Therefore, \[ E_{actual} = 3 \times E_{ideal} = 3 \times 1.03 = 3.09 \text{ V} \]

Determine Moles of Aluminum in 1 kg

To calculate the energy required for producing \(1.00\; \text{kg}\) of aluminum, first find the number of moles of aluminum: \[ \text{moles of Al} = \frac{1000 \text{ g}}{26.98 \text{ g/mol}} \approx 37.08 \text{ mol} \]

Calculate Energy Required

The energy required per mole of aluminum is given by \( E_{actual} \) using the formula \( Energy = nFE \). Total energy for 1 kg of aluminum: \[ Energy = 37.08 \times 3 \times 96485 \times 1.03 \approx 11311823 \text{ J (or 11.31 MJ)} \]

Conclusion

Thus, the energy required to produce \(1.00 \text{ kg}\) of \( \mathrm{Al} \) when applying three times the ideal voltage is approximately \(11.31 \text{ MJ} \).

Key concepts

Hall process

The Hall process is an electrochemical method used to produce pure aluminum metal from aluminum oxide. Named after Charles Martin Hall, who discovered the process in 1886, it revolutionized the aluminum industry by providing an efficient way to extract aluminum. At the core of the Hall process is the electrolysis of alumina (aluminum oxide), where alumina is dissolved in molten cryolite and then subjected to an electric current.
This current facilitates the reduction of aluminum ions to aluminum metal.

• First, aluminum oxide (\(\text{Al}_2\text{O}_3\)) is dissolved in a bath of molten cryolite.
• An electric current is passed through the solution, which results in the electrons reducing the aluminum ions to form aluminum metal.
• Simultaneously, oxygen from the aluminum oxide reacts with carbon from the electrode to form carbon monoxide or carbon dioxide.

Thanks to the Hall process, aluminum, once a rare metal, became available on a large scale, transforming industrial applications.

Gibbs free energy

Gibbs free energy (\(\Delta G\)) is a thermodynamic property that can predict the spontaneity of a chemical reaction. It combines the system's enthalpy (or heat content) and entropy (or disorder) to provide a measure of the maximum reversible work that can be performed by the system at constant temperature and pressure.
The formula is:\[\Delta G = \Delta H - T\Delta S\]Where:

• \(\Delta H\) is the change in enthalpy,
• \(T\) is the temperature in Kelvin, and
• \(\Delta S\) is the change in entropy.

A negative \(\Delta G\) means the reaction proceeds spontaneously, while a positive \(\Delta G\) requires an input of energy. In electrochemistry, the relationship between Gibbs free energy and electrical work is particularly useful. The connection is given by:\[\Delta G = -nFE\]where \(n\) is the number of moles of electrons exchanged, \(F\) is Faraday's constant, and \(E\) is the cell potential. This relation is central for calculating the minimum voltage needed in reactions like the Hall process.

Faraday's constant

Faraday's constant (\(F\)) is a crucial figure in electrochemistry that provides the quantity of electric charge carried by one mole of electrons. Its approximate value is \(96485\) coulombs per mole. Named after Michael Faraday, this constant forms the bridge between macroscopic measurements of electricity and the microscopic activity of electrons.
In practical terms:

• For any electrochemical calculation, \(F\) helps to relate the amount of substance involved in the reaction to the electric charge needed.
• When multiplied by the moles of electrons,\(n\), it gives the total charge transferred in the reaction: \(Q = nF\).
• Using this constant, along with the relation \(\Delta G = -nFE\), one can determine the energy requirements or efficiency of electrochemical processes such as aluminum electrolysis.

Understanding \(F\) is fundamental for students and professionals looking to precisely calculate the energetics of electrochemical transformations.

Aluminum electrolysis

Aluminum electrolysis is the process used to extract aluminum metal from its oxide, primarily using the Hall process. This method relies on passing a large electric current through molten aluminum oxide mixed with cryolite, leading to the formation of pure liquid aluminum at the cathode.
Key steps in the process include:

• The anode reaction, where oxygen ions produce carbon dioxide by reacting with carbon electrodes.
• The cathode reaction, where aluminum ions gain electrons to form metallic aluminum.
• The efficiency of this process depends on carefully managed temperatures and potentials, ensuring maximum yield.

Even though aluminum electrolysis is energy-intensive, advances in technology have improved its efficiency. This progress has made aluminum one of the most recycled and economically feasible metals in the world, essential for everything from aerospace to household items.