How much chlorine will be liberated on passing one ampere current for 30 minutes through NaCl solution ?

Options

A. 0.66 mole
B. 0.33 mole
C. 0.66 g
D. 0.33 g

Answer

To determine how much chlorine will be liberated when passing a current of 1 ampere for 30 minutes through a NaCl solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
- When NaCl is dissolved in water, it dissociates into Na⁺ and Cl⁻ ions. At the anode, Cl⁻ ions are oxidized to form chlorine gas (Cl₂).
- The half-reaction at the anode can be represented as:
\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \text{e}^- \]
- This indicates that 2 moles of electrons are required to produce 1 mole of Cl₂.

2. **Calculate the Total Charge (Q)**:
- The formula for charge is given by:
\[ Q = I \times t \]
- Here, \( I = 1 \, \text{A} \) and \( t = 30 \, \text{minutes} = 30 \times 60 = 1800 \, \text{seconds} \).
- Therefore, the total charge is:
\[ Q = 1 \, \text{A} \times 1800 \, \text{s} = 1800 \, \text{C} \]

3. **Relate Charge to Moles of Electrons**:
- According to Faraday's law, 1 Faraday (F) corresponds to the charge of approximately 96500 coulombs and is equivalent to 1 mole of electrons.
- The number of moles of electrons (n) can be calculated as:
\[ n = \frac{Q}{F} = \frac{1800 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.01863 \, \text{mol} \]

4. **Calculate Moles of Cl₂ Produced**:
- Since 2 moles of electrons are required to produce 1 mole of Cl₂, the moles of Cl₂ produced can be calculated as:
\[ \text{Moles of } \text{Cl}_2 = \frac{n}{2} = \frac{0.01863}{2} \approx 0.009315 \, \text{mol} \]

5. **Convert Moles of Cl₂ to Mass**:
- The molar mass of Cl₂ is approximately 71 g/mol.
- Therefore, the mass of Cl₂ produced is:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.009315 \, \text{mol} \times 71 \, \text{g/mol} \approx 0.66 \, \text{g} \]

### Final Answer:
The amount of chlorine liberated is approximately **0.66 grams**.

---