Problem 105 - Chemistry Principles and Practice

Question

An electrolytic cell produces aluminum from \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at the rate of \(10 \mathrm{~kg} /\) day. Assuming a yield of \(100 \%\), (a) how many moles of electrons must pass through the cell in one day? (b) how many coulombs are passing through the cell? (c) how many moles of oxygen \(\left(\mathrm{O}_{2}\right)\) are being produced simultaneously?

Step-by-Step Solution

Answer

1111.2 moles of electrons, 1.072 x 10^8 coulombs, 277.8 moles of oxygen.

Step 1: Calculate moles of aluminum produced

First, determine the number of moles of aluminum produced in one day. The molar mass of aluminum (\(\mathrm{Al}\)) is \(26.98 \, \text{g/mol}\). Since 10 kg of aluminum are produced, we convert this to grams: \(10,000 \, \text{g}\). The number of moles is calculated as:\[\text{moles of } \mathrm{Al} = \frac{10,000 \, \text{g}}{26.98 \, \text{g/mol}} \approx 370.4 \, \text{moles}\]

Step 2: Use stoichiometry to calculate moles of electrons

The balanced reaction for the production of aluminum from \(\mathrm{Al}_2\mathrm{O}_3\) is \(2 \mathrm{Al}_2\mathrm{O}_3 \rightarrow 4 \mathrm{Al} + 3 \mathrm{O}_2\). Each \(\mathrm{Al}_2\mathrm{O}_3\) provides two \(\mathrm{Al}\) atoms and requires 6 moles of electrons (3 for each \(\mathrm{Al}\) atom). Thus, producing \(4 \mathrm{Al}\) involves 12 moles of electrons. Hence, for 370.4 moles of \(\mathrm{Al}\):\[\text{moles of electrons} = 370.4 \times \frac{3}{1} = 1111.2 \, \text{moles}\]

Step 3: Convert moles of electrons to coulombs

Next, convert the moles of electrons into coulombs. The charge of one mole of electrons (1 Faraday) is \(96485 \, \text{C/mol}\). Thus, the total charge is:\[\text{coulombs} = 1111.2 \, \text{moles} \times 96485 \, \text{C/mol} \approx 1.072 \times 10^8 \, \text{C}\]

Step 4: Calculate moles of oxygen produced

Finally, determine how many moles of \(\mathrm{O}_2\) are simultaneously produced. According to the balanced reaction, 4 moles of \(\mathrm{Al}\) generate 3 moles of \(\mathrm{O}_2\). Thus, for 370.4 moles of \(\mathrm{Al}\):\[\text{moles of } \mathrm{O}_2 = 370.4 \times \frac{3}{4} = 277.8 \, \text{moles}\]

Key Concepts

Electrolytic Cells

Electrolytic cells are fascinating components of electrochemistry. They convert electrical energy into chemical energy, allowing chemical reactions that are non-spontaneous to occur. These cells consist of two electrodes, an anode and a cathode, placed in an electrolyte solution.
When a voltage is applied, electrons are compelled to move through the external circuit from the anode (positive electrode) to the cathode (negative electrode). In the case of producing aluminum from alumina (\(\mathrm{Al}_2\mathrm{O}_3\)), as discussed in the exercise, an external electric current is necessary to reduce the oxide to its metallic form \(\mathrm{Al}\).

• The cathode facilitates the reduction of aluminum ions to aluminum metal.
• The anode reaction typically involves the oxidation of oxygen ions, \(\mathrm{O}^{2-}\), releasing \(\mathrm{O}_2\) gas.

The systematic inclusion of electrolytic cells in industries helps in large-scale production of elements like aluminum, provided the right conditions such as electrolyte composition and electrical requirements are maintained.

Stoichiometry

Stoichiometry is the heart of calculating chemical reactions. It involves using balanced chemical equations to determine the relationships between reactants and products.
Understanding stoichiometry is crucial, as it lays the foundation for calculating quantities of substances involved in reactions.

Application in Aluminum Production

In the process of extracting aluminum from \(\mathrm{Al}_2\mathrm{O}_3\), stoichiometry helps us understand the ratios of materials used and produced. The equation: \(2 \mathrm{Al}_2\mathrm{O}_3 \rightarrow 4 \mathrm{Al} + 3 \mathrm{O}_2\) tells us that:

• 2 moles of alumina yield 4 moles of aluminum.
• 3 moles of oxygen gas are produced alongside electrons required in a \(4:6\) proton-electron ratio.

This stoichiometric relationship allows us to calculate events like the amount of oxygen produced when producing a set amount of aluminum, integrating chemical theory with quantitative analysis.

Moles and Molar Mass

Moles and molar mass are basic yet elemental concepts in chemistry. A "mole" is a unit that measures an amount of a substance, whilst "molar mass" is the mass of one mole of a substance, expressed in grams per mole.
For aluminum (\(\mathrm{Al}\)), the molar mass is \(26.98 \, \text{g/mol}\).

Calculation Process

Given a production rate of \(10 \mathrm{~kg}\) of aluminum per day, converting this into moles requires us to:

• Convert \(10 \mathrm{~kg}\) to \(10,000 \, \text{g}\).
• Calculate moles: \(\text{Moles of } \mathrm{Al} = \frac{10,000 \, \text{g}}{26.98 \, \text{g/mol}} \approx 370.4 \, \text{moles}\).

Understanding and using moles and molar mass is vital in scaling laboratory findings to industrial applications or vice versa, validating how much material and energy is necessary for setting up processes like electrolysis.

Coulombs and Charge Calculations

In electrochemistry, coulombs are used to express the amount of electrical charge. A charge of one coulomb is carried by approximately \(6.242 \times 10^{18}\) electrons.
For practical applications like determining charge in electrolysis, we use Faraday's constant (\(96485 \, \text{C/mol of electrons}\)) to convert moles of electrons to charge in coulombs.

• In the given exercise, producing \(370.4\) moles of \(\mathrm{Al}\) consumes roughly \(1111.2\) moles of electrons based on stoichiometry.
• Using \(96485 \, \text{C/mol}\), the total charge is \(1111.2 \times 96485 \approx 1.072 \times 10^8 \, \text{C}\).

Mastering the conversion of moles of electrons to coulombs is essential in evaluating the power requirements and efficiency of electrochemical processes. It informs the design and control of cells in industrial manufacturing, optimizing the use of electricity.