Problem 2 - Chemistry: Principles and Reactions

Question

Write a balanced equation to represent the electrolysis of aluminum oxide. If \(2.00 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(751 \mathrm{~mm} \mathrm{Hg}\) is formed at the anode, what mass of \(\mathrm{Al}\) is formed at the cathode?

Step-by-Step Solution

Answer

Answer: The mass of aluminum formed at the cathode is approximately 2.89 grams.

Step 1: Write the balanced equation for electrolysis of aluminum oxide

The balanced equation for the electrolysis of aluminum oxide is: \[2\mathrm{Al}_2\mathrm{O}_3 \rightarrow 4\mathrm{Al} + 3\mathrm{O}_2\]

Step 2: Convert given volume of oxygen to moles

First, we need to convert the given volume of oxygen gas at given temperature and pressure to the number of moles. For this, we can use the Ideal Gas Law equation: \[PV = nRT\] where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹) and T is the temperature in Kelvin. We need to convert the given temperature and pressure to respective units: Temperature in Kelvin: \(25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15\) K Pressure in atm: \(751 \mathrm{~mm} \mathrm{Hg} \times \frac{1 \mathrm{~atm}}{760 \mathrm{~mm} \mathrm{Hg}} = 0.988 \mathrm{~atm}\) Let's solve for n: \[n_{\mathrm{O}_{2}} = \frac{PV}{RT} = \frac{(0.988 \, \mathrm{atm})(2.00 \, \mathrm{L})}{(0.0821 \, \mathrm{L\, atm \, K^{-1}\, mol^{-1}})(298.15 \, \mathrm{K})} = 0.0805 \, \mathrm{mol}\]

Step 3: Use stoichiometric relationships to find moles of aluminum formed

From the balanced equation, we can see that for every 3 moles of oxygen produced, 4 moles of aluminum are formed. So, we will find the moles of aluminum produced using this relationship: \[n_{\mathrm{Al}} = n_{\mathrm{O}_{2}} \times \frac{4 \, \mathrm{mol\, Al}}{3 \, \mathrm{mol\, O}_{2}} = 0.0805 \, \mathrm{mol} \times \frac{4}{3} = 0.107 \ \mathrm{mol}\]

Step 4: Calculate the mass of aluminum formed

To find the mass of aluminum produced, we will use the molar mass of aluminum (\(26.98 \, \mathrm{g/mol}\)): \[mass_{\mathrm{Al}} = n_{\mathrm{Al}} \times M_{\mathrm{Al}} = 0.107 \, \mathrm{mol} \times 26.98 \, \mathrm{g/mol} = 2.89 \, \mathrm{g}\] So, the mass of aluminum formed at the cathode is approximately 2.89 grams.

Key Concepts

The Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics, used to relate different properties of gases. The ideal gas law is expressed as: \[PV = nRT\]

• P stands for the pressure of the gas, usually measured in atmospheres (atm).
• V represents the volume of the gas, typically recorded in liters (L).
• n is the number of moles of the gas.
• R is the ideal gas constant, with a value of 0.0821 L atm K⁻¹ mol⁻¹.
• T is the temperature, measured in Kelvin (K).

When solving problems involving gases, such as determining the mass of aluminum formed during electrolysis, converting measurements to appropriate units is crucial. For example, pressure should be converted from mm Hg to atm, and the temperature from Celsius to Kelvin. This ensures that all units are consistent to accurately apply the ideal gas law. By rearranging the ideal gas law equation, we can solve for different variables. For instance, solving for the number of moles (n) with given pressure, volume, and temperature allows us to predict chemical quantities in reactions like electrolysis.

Stoichiometry

Stoichiometry is the part of chemistry that helps us understand the quantitative relationships, or ratios, between reactants and products in a chemical reaction. In a balanced chemical equation, the stoichiometric coefficients reveal these relationships. In the electrolysis of aluminum oxide, the reaction is: \[2\mathrm{Al}_2\mathrm{O}_3 \rightarrow 4\mathrm{Al} + 3\mathrm{O}_2\]Here, stoichiometry tells us that:

• 2 moles of aluminum oxide produce 4 moles of aluminum.
• 3 moles of oxygen gas are released in the process.

By understanding these relationships, we can use the moles of one reactant or product to predict the moles of another. For the problem at hand, knowing the moles of oxygen allows us to calculate the moles of aluminum formed using the stoichiometric ratio (4 mol Al : 3 mol O₂). This concept is crucial when moving from theoretical calculations to practical applications.

Molar Mass

Molar mass is a measure of the mass of a given substance (chemical element or chemical compound) divided by the amount of substance, usually expressed in grams per mole (g/mol). It acts as a bridge between the microscopic world of atoms and molecules, and the macroscopic world of grams and liters we handle in laboratories. For elements, the molar mass is equal to the atomic mass found on the periodic table. For compounds, it is the sum of the molar masses of its constituent atoms. For example, the molar mass of aluminum is 26.98 g/mol.When it comes to electrolysis calculations, once we have the number of moles of aluminum formed, we employ the molar mass to convert from moles to grams. The equation used is:\[\text{mass}_{\mathrm{Al}} = n_{\mathrm{Al}} \times M_{\mathrm{Al}}\]This conversion provides a tangible measure of how much aluminum is formed, crucial for practical applications in industrial processes or laboratory experiments.