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

Options

A. `44.8L`
B. `22.4L`
C. `11.2L`
D. `5.6L`

Explanation

To solve the problem, we need to determine the volume of hydrogen gas (H₂) produced when a certain quantity of electrical charge deposits 4.5 g of aluminum (Al) from Al³⁺ ions at the cathode. We will use the concepts of stoichiometry and electrochemistry to find the answer.

### Step-by-Step Solution:

1. **Determine the Molar Mass of Aluminum (Al)**:
- The molar mass of aluminum (Al) is approximately 27 g/mol.

2. **Calculate the Number of Moles of Aluminum Deposited**:
\[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{4.5 \text{ g}}{27 \text{ g/mol}} = 0.1667 \text{ mol} \]

3. **Determine the Number of Electrons Required for the Reduction of Al³⁺**:
- The half-reaction for the reduction of Al³⁺ to Al is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- Therefore, for every mole of Al deposited, 3 moles of electrons are required.
- Thus, the total moles of electrons (n) needed for 0.1667 moles of Al is:
\[ n = 0.1667 \text{ mol Al} \times 3 \text{ mol e}^- = 0.5001 \text{ mol e}^- \]

4. **Determine the Equivalent Weight of Hydrogen (H)**:
- The equivalent weight of hydrogen is 1 g/equiv (since the molar mass of H is 1 g/mol).

5. **Calculate the Moles of Hydrogen Produced**:
- The half-reaction for the reduction of H⁺ to H₂ is:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]
- This means that 2 moles of electrons produce 1 mole of H₂.
- Therefore, the moles of H₂ produced from 0.5001 moles of electrons is:
\[ \text{Moles of H}_2 = \frac{0.5001 \text{ mol e}^-}{2} = 0.25005 \text{ mol H}_2 \]

6. **Calculate the Volume of Hydrogen at STP**:
- At Standard Temperature and Pressure (STP), 1 mole of gas occupies 22.4 liters.
- Thus, the volume of H₂ produced is:
\[ \text{Volume of H}_2 = 0.25005 \text{ mol H}_2 \times 22.4 \text{ L/mol} = 5.6001 \text{ L} \]

### Final Answer:
The volume of hydrogen gas (H₂) produced at STP is approximately **5.6 liters**.

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