A quantity of electrical charge that brings about the deposition of `4.5` g Al from `Al^(3+)` at the cathode will also produce the following volume at STP of `H_(2)(g)` from `H^(+)` at the cathode:

Options

A. `44.8 L`
B. `22.4 L`
C. `11.2 L`
D. `2.6 L`

Answer

To solve the problem, we need to determine the volume of hydrogen gas produced at STP when a certain quantity of electrical charge is used to deposit 4.5 g of aluminum from Al³⁺ ions at the cathode.

### Step-by-Step Solution:

1. **Determine the moles of Aluminum deposited:**
- The molar mass of aluminum (Al) is approximately 27 g/mol.
- To find the number of moles of aluminum deposited, we use the formula:
\[ \text{Number of moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{4.5 \, \text{g}}{27 \, \text{g/mol}} = 0.1667 \, \text{mol} \]

2. **Calculate the charge required for the deposition of Aluminum:**
- The reaction for the deposition of aluminum from Al³⁺ is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- This means that 3 moles of electrons are required to deposit 1 mole of aluminum.
- Therefore, for 0.1667 moles of aluminum, the moles of electrons required are:
\[ \text{Moles of electrons} = 0.1667 \, \text{mol Al} \times 3 \, \frac{\text{mol e}^-}{\text{mol Al}} = 0.5001 \, \text{mol e}^- \]

3. **Calculate the total charge (Q) using Faraday's constant:**
- Faraday's constant (F) is approximately 96500 C/mol.
- The total charge required can be calculated as:
\[ Q = \text{moles of electrons} \times F = 0.5001 \, \text{mol} \times 96500 \, \text{C/mol} \approx 48200 \, \text{C} \]

4. **Determine the volume of Hydrogen gas produced:**
- The reaction for the production of hydrogen gas from H⁺ is:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]
- This means that 2 moles of electrons are required to produce 1 mole of hydrogen gas.
- Therefore, the moles of hydrogen produced from the electrons used can be calculated as:
\[ \text{Moles of H}_2 = \frac{0.5001 \, \text{mol e}^-}{2} = 0.25005 \, \text{mol H}_2 \]

5. **Calculate the volume of Hydrogen gas at STP:**
- At STP, 1 mole of gas occupies 22.4 L.
- Thus, the volume of hydrogen gas produced is:
\[ \text{Volume of H}_2 = 0.25005 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 5.6 \, \text{L} \]

### Final Answer:
The volume of hydrogen gas produced at STP is approximately **5.6 L**.