An aqueous solution of `NaCl` on electrolysis gives `H_(2)(g), Cl_(2)(g),` and `NaOH` accroding to the reaction `:`
`2Cl^(c-)(aq)+2H_(2)Orarr2overset(c-)(O)H(aq)+H_(2)(g)+Cl_(2)(g)`
A direct current of `25A` with a current efficiency of `62%` is passed through `20L` of `NaCl` solution `(20%` by weight`)`. Write down the reactions taking place at the anode and cathode. How long will it take to produce `1 kg ` of `Cl_(2)`? `(` Assume no loss due to evaporation . `)`

Explanation

To solve the problem step by step, we will break it down into two parts: identifying the reactions at the anode and cathode, and calculating the time required to produce 1 kg of Cl₂.

### Step 1: Identify the Reactions at the Anode and Cathode

1. **At the Anode (Oxidation Reaction)**:
- Chloride ions (Cl⁻) are oxidized to chlorine gas (Cl₂).
- The half-reaction is:
\[ 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \]

2. **At the Cathode (Reduction Reaction)**:
- Water is reduced to produce hydrogen gas (H₂) and hydroxide ions (OH⁻).
- The half-reaction is:
\[ 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \]

### Step 2: Calculate the Time to Produce 1 kg of Cl₂

1. **Determine the Molar Mass of Cl₂**:
- The molar mass of Cl₂ is \( 2 \times 35.5 \, \text{g/mol} = 71 \, \text{g/mol} \).

2. **Calculate the Number of Moles of Cl₂**:
- For 1 kg (1000 g) of Cl₂:
\[ \text{Moles of Cl}_2 = \frac{1000 \, \text{g}}{71 \, \text{g/mol}} \approx 14.08 \, \text{mol} \]

3. **Calculate the Total Charge Required**:
- Each mole of Cl₂ requires 2 moles of electrons (from the half-reaction).
- Therefore, the total moles of electrons needed:
\[ \text{Moles of electrons} = 2 \times 14.08 \, \text{mol} \approx 28.16 \, \text{mol} \]
- The total charge (Q) required can be calculated using Faraday's constant (F = 96500 C/mol):
\[ Q = \text{Moles of electrons} \times F = 28.16 \, \text{mol} \times 96500 \, \text{C/mol} \approx 2720000 \, \text{C} \]

4. **Calculate the Effective Current**:
- Given the current efficiency is 62%, the effective current (I_eff) is:
\[ I_{\text{eff}} = 25 \, \text{A} \times 0.62 = 15.5 \, \text{A} \]

5. **Calculate the Time (T)**:
- Using the formula \( Q = I \times T \):
\[ T = \frac{Q}{I_{\text{eff}}} = \frac{2720000 \, \text{C}}{15.5 \, \text{A}} \approx 175484.84 \, \text{s} \]

6. **Convert Time to Hours**:
- To convert seconds to hours:
\[ T \approx \frac{175484.84 \, \text{s}}{3600 \, \text{s/h}} \approx 48.7 \, \text{hours} \]

### Final Answers:
- **Reactions**:
- Anode: \( 2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- \)
- Cathode: \( 2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \)
- **Time to produce 1 kg of Cl₂**: Approximately 48.7 hours.