Question

The major reduction half-reaction occurring in the cell in which molten \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and molten aluminum salts are electrolyzed is \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{s})\). The cell operates at \(5.0 \mathrm{~V}\) and \(1.0 \times 10^{5} \mathrm{~A} .\) Calculate the mass \((\mathrm{g})\) of aluminum metal produced in \(8.0 \mathrm{~h}\).

Short answer

The mass of aluminum produced is approximately 268331 grams.

Step-by-step solution

Determine Total Charge

Calculate the total charge (Q) passed through the cell by using the formula \( Q = I \times t \), where \( I \) is the current and \( t \) is the time in seconds. Given that \( I = 1.0 \times 10^5 \) A and \( t = 8.0 \times 60 \times 60 \) s (convert hours to seconds), we find that \( Q = 1.0 \times 10^5 \times 28800 \ = 2.88 \times 10^9 \) C.

Calculate Moles of Electrons

Use Faraday's constant (\( F = 96485 \) C/mol) to find the number of moles of electrons. The formula is \( n = \frac{Q}{F} \). Substituting the known values, we get \( n = \frac{2.88 \times 10^9}{96485} \approx 29846.83 \) moles of electrons.

Determine Moles of Aluminum Produced

Using the reduction half-reaction \( \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al} \), we see that 3 moles of electrons produce 1 mole of aluminum. Therefore, the moles of aluminum produced is \( \frac{29846.83}{3} \approx 9948.94 \) moles of aluminum.

Convert Moles of Aluminum to Mass

Multiply the moles of aluminum by its molar mass (\( M = 26.98 \) g/mol) to calculate the mass. The mass \( m \) is given by \( m = 9948.94 \times 26.98 \approx 268331.05 \) grams.

Key concepts

Electrolysis

Electrolysis is the process of using electricity to drive a chemical reaction that does not occur spontaneously. It plays an essential role in extracting and refining metals, as well as in chemical manufacturing.
In the context of producing aluminum, electrolysis involves passing an electric current through molten cryolite and bauxite to break down the compounds and extract aluminum.
The process takes place in an electrolytic cell, where electrodes are submerged in the molten compound.

• The anode, which is positive, attracts negatively charged ions or anions.
• The cathode, which is negative, attracts positively charged ions or cations. Here, reduction occurs as electrons are gained, helping to produce aluminum.

This entire setup is crucial for transforming electrical energy into chemical energy necessary to extract desired materials from their ore.

Reduction Half-Reaction

In electrochemistry, a reduction half-reaction involves the gain of electrons by atoms or ions. It is one of the two half-reactions in electrolysis. The other half is the oxidation reaction, which involves loss of electrons.
For the production of aluminum, the significant reduction half-reaction is: \[ \text{Al}^{3+} + 3 \text{e}^{-} \rightarrow \text{Al} \]This equation tells us that aluminum ions (\( \text{Al}^{3+} \)) gain three electrons (\( \text{e}^{-} \)) to form solid aluminum metal (\( \text{Al} \)).

• The species that undergoes reduction receives electrons, decreasing its oxidation state.
• This half-reaction takes place at the cathode of the electrolytic cell.

Understanding this reaction is key to knowing how metal ions become metal atoms in processes like electroplating or extracting metals from ores.

Faraday's Law

Faraday's Law of Electrolysis is fundamental to understanding how much substance is altered at an electrode during electrolysis. According to the law, the amount of chemical change is directly proportional to the total electric charge passed through the substance.
The mathematical expression of Faraday’s law is: \[ Q = nF \]Where:

• \( Q \) is the total electric charge in coulombs
• \( n \) is the number of moles of electrons exchanged
• \( F \), Faraday's constant, is approximately 96485 C/mol, representing the charge of one mole of electrons

For example, in calculating the moles of electrons during aluminum production, knowing the electric charge transferred allows you to find out how much aluminum is made by using this vital principle. Faraday's law thus connects electrical energy usage to chemical product yield, making it invaluable in industrial applications.

Moles of Electrons

In the context of electrolysis, the concept of moles of electrons is crucial to quantify the amount of substance being transformed in any given reaction.
This involves using Faraday's constant to determine how many moles of electrons are involved based on the total charge.
In the aluminum production scenario, step two of the solution calculates this by:\[ n = \frac{Q}{F} \]Here, \( n \) represents the number of moles of electrons, \( Q \) is the charge (in coulombs), and \( F \) is Faraday's constant (96485 C/mol).

• This calculation helps to bridge how much electricity is used and how many electrons are participating in the electrochemical reaction.
• Since each atom of aluminum requires three electrons, understanding the number of moles of electrons helps ascertain how much aluminum can be extracted.

Thus, determining the moles of electrons becomes a fundamental step in translating electrical input into chemical output effectively.