1000mL 1M `CuSO_4`(aq) is electrolysed by 9.65 A current for 100 sec using Pt-electrode which is /are correct statements?

Options

A. Blue colour intensity decreases during electrolysis
B. Blue colour intensity remains constant if Cu-electrode used.
C. pH of solution is 8 after electrolysis.
D. 28mL of `CH_4` at 1 atm and 273 K required for its combustion by `O_2`, liberated during electrolysis.

Answer

To solve the problem, we will analyze each statement provided in the question regarding the electrolysis of 1000 mL of 1M CuSO₄ solution using a current of 9.65 A for 100 seconds. We will evaluate the correctness of each statement step by step.

### Step-by-Step Solution:

1. **Understanding Electrolysis of CuSO₄**:
- When CuSO₄ is electrolyzed, copper ions (Cu²⁺) are reduced at the cathode to form solid copper (Cu), which has a brownish-red color. The blue color of the CuSO₄ solution is due to the presence of Cu²⁺ ions.

2. **Evaluating Statement 1**:
- **Statement**: "Blue color intensity decreases during electrolysis."
- **Analysis**: As Cu²⁺ ions are reduced to solid copper, the concentration of Cu²⁺ ions in the solution decreases, leading to a decrease in the blue color intensity of the solution.
- **Conclusion**: This statement is **correct**.

3. **Evaluating Statement 2**:
- **Statement**: "Blue color intensity remains constant if Cu electrode used."
- **Analysis**: If a copper electrode is used, the copper from the anode dissolves back into the solution as Cu²⁺ ions, maintaining the blue color intensity. Therefore, the concentration of Cu²⁺ ions does not change significantly.
- **Conclusion**: This statement is **correct**.

4. **Evaluating Statement 3**:
- **Statement**: "pH of solution is 8 after electrolysis."
- **Analysis**: The electrolysis of CuSO₄ produces H⁺ ions at the anode, leading to an acidic solution. The pH of an acidic solution is typically less than 7, not 8.
- **Conclusion**: This statement is **incorrect**.

5. **Evaluating Statement 4**:
- **Statement**: "28 mL of CH₄ at 1 atm and 273 K required for its combustion by O₂ liberated during electrolysis."
- **Calculation**:
- First, calculate the total charge passed during electrolysis:
\[ Q = I \times t = 9.65 \, \text{A} \times 100 \, \text{s} = 965 \, \text{C} \]
- Using Faraday’s law, the amount of substance produced can be calculated:
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{965}{96500} = 0.01 \, \text{mol} \]
- For the formation of O₂, the reaction at the anode is:
\[ 2 \, \text{H}_2O \rightarrow O_2 + 4 \, \text{H}^+ + 4 \, e^- \]
This means 4 moles of electrons produce 1 mole of O₂. Thus, 0.01 mol of electrons will produce:
\[ \text{Moles of O₂} = \frac{0.01}{4} = 0.0025 \, \text{mol} \]
- Converting moles of O₂ to volume at STP (22.4 L/mol):
\[ \text{Volume of O₂} = 0.0025 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.056 \, \text{L} = 56 \, \text{mL} \]
- The balanced combustion reaction for CH₄ is:
\[ CH₄ + 2 O₂ \rightarrow CO₂ + 2 H₂O \]
From the stoichiometry, 1 mole of CH₄ reacts with 2 moles of O₂. Therefore, 56 mL of O₂ will react with:
\[ \text{Volume of CH₄} = \frac{56 \, \text{mL}}{2} = 28 \, \text{mL} \]
- **Conclusion**: This statement is **correct**.

### Final Conclusion:
The correct statements are:
1. Blue color intensity decreases during electrolysis. (Correct)
2. Blue color intensity remains constant if Cu electrode used. (Correct)
3. pH of solution is 8 after electrolysis. (Incorrect)
4. 28 mL of CH₄ at 1 atm and 273 K required for its combustion by O₂ liberated during electrolysis. (Correct)

### Summary of Correct Statements:
- The correct statements are 1, 2, and 4.