Problem 116 - General Chemistry: Principles and Modern Applications

Question

The quantity of electric charge that will deposit \(4.5 \mathrm{g}\) Al at a cathode will also produce the following volume at STP of \(\mathrm{H}_{2}(\mathrm{g})\) from \(\mathrm{H}^{+}(\) aq) at a cathode: (a) \(44.8 \mathrm{L} ;\) (b) \(22.4 \mathrm{L} ;\) (c) \(11.2 \mathrm{L} ;\) (d) \(5.6 \mathrm{L}\).

Step-by-Step Solution

Answer

The correct answer is (d) 5.6 L.

Step 1: Calculate the moles of Aluminium

Firstly, convert the mass of Aluminium (Al) into moles. The molar mass of Al is approximately 27 g/mol. \( \frac{4.5g}{27 \frac{g}{mol}} = 0.167 mol \) of Al.

Step 2: Use Faraday's Laws of Electrolysis

From Faraday's laws of electrolysis, we know that for Aluminium, it requires 3 moles of electrons for one mole of Al to be deposited at the cathode. Therefore, \( 0.167 mol \) of Al would require \( 3*0.167 = 0.5 mol \) of electrons.

Step 3: Apply the knowledge to Hydrogen gas

The same amount of charge would produce hydrogen at the cathode by reducing \( H+ \) ions to \( H_{2}. \) Since hydrogen requires 2 moles of electrons to produce 1 mole of \( H_{2}, \) the charge required to deposit 0.167 moles of Al can produce \( \frac{0.5}{2} = 0.25 mol \) of \( H_{2}. \)

Step 4: Convert moles of Hydrogen gas to volume at STP

Finally, convert the moles of hydrogen gas into volume at standard temperature and pressure (STP) by using the molar volume of a gas at STP which is 22.4 L/mol. The volume of H2 is \( 0.25 mol * 22.4 \frac{L}{mol} = 5.6 L \).

Step 5: Select the correct answer

So, the quantity of electric charge that will deposit 4.5 g Al at a cathode will also produce 5.6 L of \( H_{2}(g) \) from \( H^{+}(aq) \) at a cathode. The correct answer is (d) 5.6 L.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a substance is fundamental in chemistry, especially when dealing with reactions and processes like electrolysis. Molar mass is defined as the mass of one mole of a substance and is generally expressed in grams per mole (g/mol). To calculate molar mass, you can consult the periodic table to find the atomic mass of each element in a compound and then add them up according to their stoichiometric coefficients.

For instance, if you're dealing with Aluminium (Al), which appears in the given exercise, simply look up Al's atomic mass, approximately 27 g/mol, and since it's a pure element, that's also its molar mass. Calculating the moles of Al from a given mass involves dividing the mass by the molar mass: \[ \text{moles of Al} = \frac{\text{mass of Al (in g)}}{\text{molar mass of Al (in g/mol)}} \].

This step is critical because most chemical calculations require knowledge of either the amount in moles or the mass of the substance in question, and this conversion enables us to connect the tangible mass with the abstract mole concept. When a problem like the one presented is solved, the molar mass calculation serves as a foundation upon which further relationships are built, such as the connection between moles of a substance and moles of electrons in electrolysis.

Mole-to-Electron Conversion

After you've calculated the amount of a substance in moles, you might need to convert it to or relate it to the number of electrons involved in a chemical reaction, such as in the process of electrolysis. This conversion is important in electrochemistry because electric current is essentially the flow of electrons, and understanding this flow allows prediction of the actual chemical change occurring at the electrodes.

Here's a brief explanation: in chemistry, we often use the Faraday constant, approximately \( 96,485 \text{Coulombs/mol} \), which represents the charge of one mole of electrons. If we're given moles of a substance and we need to find out the moles of electrons required for a reaction, like in the provided exercise, we apply the stoichiometry of the reaction. For Aluminium electrolysis, it takes \( 3 \) moles of electrons to deposit \( 1 \) mole of Al. Therefore, \[ \text{moles of electrons} = 3 \times \text{moles of Al} \].

Conversion is equally applicable in reverse. If you start with a known charge or number of moles of electrons, you can determine the corresponding moles of a substance that will be produced or consumed in a chemical reaction.

STP Volume Calculation

The standard temperature and pressure (STP) conditions provide a reference for measuring gases, which is crucial for comparisons and calculations in gas-related exercises. STP is defined as a temperature of 0 degrees Celsius (273.15 K) and 1 atmosphere pressure. Under these conditions, one mole of an ideal gas occupies a volume of 22.4 liters - this is known as the molar volume of a gas at STP.

To find the volume of a gas at STP, simply multiply the amount of the gas (in moles) by the molar volume constant: \[ \text{Volume at STP} = \text{moles of gas} \times 22.4 \text{L/mol} \].

In our exercise, we convert the moles of hydrogen gas produced during electrolysis directly into volume using the molar volume, as shown in the solution steps. Understanding this relationship is vital for predicting how much gas will be produced or consumed in a reaction occurring under STP conditions. Remember, this direct proportional relationship assumes ideal gas behavior, which is a close approximation for many gases under STP but may require corrections under other conditions.