Problem 61 - 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 \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow\) \(\mathrm{Al}(\mathrm{s}) .\) If the electrolysis cell operates at \(5.0 \mathrm{V}\) and \(1.0 \mathrm{X}\) \(10^{5} \mathrm{A},\) what mass of aluminum metal can be produced in a 24-h day?

Step-by-Step Solution

Answer

Approximately 806.81 kg of Al can be produced in 24 hours.

Step 1: Identify the Reaction

The reaction provided is the reduction of aluminum ions to form solid aluminum: \( \mathrm{Al}^{3+} + 3 \mathrm{e}^{-} \longrightarrow \mathrm{Al} \). This means that for every mole of Al produced, 3 moles of electrons are needed.

Step 2: Determine Total Charge Used

To find the total charge, use the formula \( Q = I \times t \), where \( I \) is current and \( t \) is time in seconds. Given \( I = 1.0 \times 10^5 \) A and time = 24 hours (which is 86400 seconds), the calculation is: \( Q = 1.0 \times 10^5 \times 86400 = 8.64 \times 10^9 \) C.

Step 3: Calculate Moles of Electrons

Use Faraday's constant (\( 96500 \) C/mol) to determine moles of electrons. Divide total charge by Faraday's constant: \[ \text{moles of } \mathrm{e}^{-} = \frac{8.64 \times 10^9}{96500} \approx 89534.2 \].

Step 4: Relate Moles of Electrons to Moles of Aluminum

From the reaction, 3 moles of \( \mathrm{e}^{-} \) produce 1 mole of \( \mathrm{Al} \). Therefore, the moles of Al produced is \( \frac{89534.2}{3} \approx 29844.73 \) moles.

Step 5: Calculate Mass of Aluminum Produced

The molar mass of aluminum is approximately 27 g/mol. Multiply the moles of aluminum by its molar mass to find the mass: \[ \text{mass of Al} = 29844.73 \times 27 \approx 806807.71 \text{ grams} \].

Step 6: Convert Mass to Kilograms

Convert the mass from grams to kilograms for practicality: \( 806807.71 \text{ grams} = 806.81 \text{ kg} \).

Key Concepts

Aluminum production

Aluminum production through electrolysis is a key industrial process. The main raw material used is aluminum oxide, also known as alumina ( \( \mathrm{Al}_2\mathrm{O}_3 \) ). Utilizing electrolysis, this compound is broken down to extract pure aluminum metal. The process involves passing a large electric current through the electrolyte, which is a molten mixture usually containing cryolite and alumina. This mixture is heated to a high temperature where alumina becomes conductive.

The reduction of aluminum ions ( \( \mathrm{Al}^{3+} \) ) to elemental aluminum ( \( \mathrm{Al} \) ) occurs at the cathode. This involves the transfer of electrons, facilitated by the high current supplied, producing aluminum as a solid metal. The aluminum is collected at the bottom of the electrolysis cell. This process allows for large-scale production of aluminum, making it a widely available and cost-effective material.

• Essentially, the method to produce aluminum from alumina is energy-intensive but highly efficient.
• Large electrolysis cells are used to produce substantial aluminum mass day-to-day.
• The setup enhances global aluminum supply, vital for many industries.

Faraday's law of electrolysis

Faraday's law of electrolysis is fundamental in understanding how electricity is used to provoke chemical changes, specifically in electrolysis.
This law states that the amount of substance transformed at the electrode is directly proportional to the total electric charge passed through the substance. In mathematical terms, the law can be expressed as:\[ \text{Mass of substance} = \left( \frac{Q}{F} \right) \times n \times M \]where:

• \( Q \) is the total charge in coulombs ( \( C \) ), calculated as \(I \times t\) where \(I\) is current and \(t\) is time.
• \( F \) is Faraday's constant ( \(96500\) C/mol).
• \( n \) is the number of moles of electrons exchanged.
• \( M \) is the molar mass of the substance.

This principle helps us understand how much aluminum can be produced by electrolysis under certain conditions. By using Faraday’s law, it becomes possible to calculate the practical yield of aluminum production based on the charge applied. It provides a precise link between the electricity consumed and the aluminum output, positioning Faraday's law as a cornerstone of electrochemical applications.

Redox reactions

Redox reactions play a central role in electrolysis and aluminum production. These reactions involve the transfer of electrons between substances, where one species is oxidized, and another is reduced.
In the context of aluminum electrolysis, the aluminum ions ( \( \mathrm{Al}^{3+} \)) are reduced at the cathode by gaining electrons to form solid aluminum.The process can be viewed in two half-reactions:

• Reduction Half-Reaction: \[ \mathrm{Al}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Al} \] at the cathode.
• Oxidation Half-Reaction: Occurs at the anode, usually involving the oxidation of oxygen ions to form oxygen gas.

These reactions together constitute the redox process. It involves a continuous exchange of electrons, enabling the transformation of raw materials to pure aluminum metal. Redox reactions exemplify the underlying chemistry in electrolysis, making it possible for electrical energy to drive chemical changes.

By understanding the mechanics of redox reactions, students can appreciate how energy is converted and stored in chemical bonds, a phenomenon that is crucial for both scientific and industrial applications. This knowledge lends insight into how resources are refined efficiently using electrochemical methods.