Problem 59 - Chemistry and Chemical Reactivity

Question

The reaction occurring in the cell in which \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and aluminum salts are electrolyzed is \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow\) Al(s). If the electrolysis cell operates at \(5.0 \mathrm{V}\) and \(1.0 \times 10^{5} \mathrm{A},\) what mass of aluminum metal can be produced in a 24 -hour day?

Step-by-Step Solution

Answer

The mass of aluminum produced in 24 hours is approximately 806.823 kg.

Step 1: Calculate Total Charge

To find the total charge passed through the electrolytic cell in 24 hours, use the formula: \( \text{Total Charge} = \text{Current} \times \text{Time} \). The current is \( 1.0 \times 10^5 \) A, and time is 24 hours, which needs to be converted to seconds (\( 24 \times 60 \times 60 \)). So, \[ \text{Total Charge} = 1.0 \times 10^5 \times 24 \times 60 \times 60 = 8.64 \times 10^9 \text{ C} \].

Step 2: Convert Charge to Moles of Electrons

Use Faraday's constant to convert the charge to moles of electrons. Faraday's constant is \( 96485 \text{ C/mol} \). The number of moles of electrons is calculated as \[ \text{Moles of electrons} = \frac{8.64 \times 10^9}{96485} \approx 89547 \text{ mol} \].

Step 3: Determine Moles of Aluminum

From the reaction, \( \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al} \), 3 moles of electrons produce 1 mole of \( \text{Al} \). Calculate moles of \( \text{Al} \) using the ratio of electrons to moles of aluminum: \[ \text{Moles of Al} = \frac{89547}{3} \approx 29849 \text{ mol} \].

Step 4: Convert Moles of Aluminum to Mass

Use the molar mass of aluminum (27 g/mol) to convert moles to mass. The mass is given by \( \text{Mass} = \text{Moles} \times \text{Molar Mass} \). Thus, \[ \text{Mass of Al} = 29849 \times 27 \approx 806823 \text{ grams} \].

Step 5: Convert Mass to Kilograms

Finally, convert the mass from grams to kilograms by dividing by 1000: \( 806823 \text{ grams} = 806.823 \text{ kg} \).

Key Concepts

Faraday's Constant

Faraday's constant is a crucial figure in the world of electrochemistry. It represents the charge of one mole of electrons. The value is approximately 96485 coulombs per mole (C/mol). This means that when you have a mole of electrons, they carry a total charge of 96485 C.

This constant links the electrical charge to chemical quantities in reactions. Specifically, it's used to calculate the number of moles of electrons transferred in a reaction. Whether it’s for electroplating or aluminum production, this constant comes in handy.

• Named after Michael Faraday, a pioneer in the field of electromagnetism and electrochemistry.
• Helps convert electrical measurements to chemical amounts.
• Always expressed in terms of coulombs per mole (C/mol).

Without Faraday's constant, relating the electricity used in electrolysis processes to chemical changes would be nearly impossible.

Electrolytic Cell

An electrolytic cell is a type of chemical cell used in electrolysis. It uses electrical energy to drive a non-spontaneous chemical reaction. In an electrolytic cell, there's a unique setup designed to make this process possible.

The key components of an electrolytic cell include:

• Two electrodes: An anode and a cathode, each placed in an electrolyte solution.
• An external power source: This supplies the electrical energy needed to drive the reaction.
• An electrolyte: A substance that contains free ions and can conduct electricity.

When current flows through the cell, the reaction progresses with the reduction occurring at the cathode (where the ions gain electrons). At the same time, oxidation happens at the anode (where electrons are lost).

In the case of aluminum production, electrolytic cells play a vital role in the extraction of aluminum from its ore.

Aluminum Production

Aluminum production via electrolysis is a fascinating and essential process in industry. The Bayer and Hall-Héroult processes are the primary methods used to extract aluminum from bauxite ore and to refine it.

Here's a simple breakdown of the Hall-Héroult process used in electrolytic cells:

• Bauxite is refined to produce alumina (\(\mathrm{Al}_2\mathrm{O}_3\)).
• Alumina is dissolved in molten cryolite, which lowers the melting point, making the electrolysis process feasible at lower temperatures.
• The electrolytic reduction occurs, where aluminum ions (\(\mathrm{Al}^{3+}\)) are reduced to aluminum metal (\(\mathrm{Al}\)) at the cathode.

This process demands a significant amount of electrical energy, highlighting the importance of efficient electrolytic cells and the role of Faraday's constant in calculating the charge requirements.

The result is pure aluminum, which is incredibly versatile and used in a range of products from soda cans to aerospace components.