`4.5 g` of aluminium (at mass `27 u`) is deposited at cathode from `Al^(3+)` solution by a certain quantity of electric charge. The volume of hydrogen gas produced at `STP` from `H^(+)` ions in solution by the same quantity of electric charge will be:

Options

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

Explanation

To solve the problem step by step, we will follow the principles of electrochemistry and Faraday's laws of electrolysis.

### Step 1: Calculate the number of equivalents of Aluminium deposited
Given:
- Mass of Aluminium (Al) = 4.5 g
- Atomic mass of Aluminium = 27 u
- Valency of Aluminium (from Al³⁺) = 3

First, we calculate the equivalent weight of Aluminium:
\[ \text{Equivalent weight of Al} = \frac{\text{Atomic weight}}{\text{Valency}} = \frac{27}{3} = 9 \text{ g/equiv} \]

Now, we can find the number of equivalents of Aluminium deposited:
\[ \text{Number of equivalents of Al} = \frac{\text{Mass of Al}}{\text{Equivalent weight of Al}} = \frac{4.5 \text{ g}}{9 \text{ g/equiv}} = 0.5 \text{ equiv} \]

### Step 2: Relate the equivalents of Aluminium to the equivalents of Hydrogen
According to Faraday's laws of electrolysis, the same amount of electric charge will deposit the same number of equivalents of different substances. Therefore, the equivalents of Hydrogen (H₂) produced will also be 0.5 equiv.

### Step 3: Calculate the volume of Hydrogen gas produced at STP
For Hydrogen gas (H₂), the valency factor is 2 because 2 H⁺ ions are required to produce 1 molecule of H₂.

Using the equivalent volume of Hydrogen gas at STP:
\[ \text{Volume of 1 equivalent of H₂} = 22.4 \text{ L} \]
Since 1 equivalent of H₂ corresponds to 1 mole of H₂, and we have 0.5 equivalents:
\[ \text{Volume of H₂ produced} = \text{Number of equivalents} \times \text{Volume of 1 equivalent} = 0.5 \times 22.4 \text{ L} = 11.2 \text{ L} \]

### Final Answer
The volume of hydrogen gas produced at STP from H⁺ ions by the same quantity of electric charge is **11.2 L**.

---