A solution of sodium sulphate was electrolyzed using some inert electrode. The product at the electrodes are

Options

A. `O_(2),H_(2)`
B. `O_(2),Na`
C. `O_(2),SO_(2)`
D. `O_(2),S_(2)O_(8)^(2-)`

Explanation

To determine the products formed at the electrodes during the electrolysis of a sodium sulfate (Na2SO4) solution using inert electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ions in the Solution:**
Sodium sulfate dissociates in water to form sodium ions (Na⁺) and sulfate ions (SO₄²⁻). The complete ionic dissociation is:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-} \]

2. **Determine the Reactions at the Electrodes:**
In electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. The relevant half-reactions are:
- At the **anode**: Oxidation (loss of electrons)
- At the **cathode**: Reduction (gain of electrons)

3. **Analyze the Anode Reaction:**
The sulfate ion (SO₄²⁻) cannot be oxidized because its standard reduction potential (SRP) is less favorable than that of water. Therefore, water is oxidized at the anode:
\[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \]
This reaction produces oxygen gas (O₂) at the anode.

4. **Analyze the Cathode Reaction:**
At the cathode, the sodium ion (Na⁺) cannot be reduced because its SRP is less favorable than that of hydrogen ions (H⁺). Therefore, water is reduced:
\[ 2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \]
This reaction produces hydrogen gas (H₂) at the cathode.

5. **Combine the Reactions:**
To find the overall reaction, we can add the half-reactions together. The half-reaction for the anode produces 4 electrons, while the cathode consumes 2 electrons. We need to multiply the cathode reaction by 2 to balance the electrons:
\[ 2(2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-) \]
This gives:
\[ 4\text{H}_2\text{O} + 4e^- \rightarrow 2\text{H}_2 + 4\text{OH}^- \]

6. **Final Overall Reaction:**
Combining the balanced half-reactions gives:
\[ 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad \text{(anode)} \]
\[ 4\text{H}_2\text{O} + 4e^- \rightarrow 2\text{H}_2 + 4\text{OH}^- \quad \text{(cathode)} \]
The overall balanced reaction is:
\[ 2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2 \]

### Products at the Electrodes:
- **At the Anode:** Oxygen gas (O₂)
- **At the Cathode:** Hydrogen gas (H₂)