An aqueous solution of `Na_2SO_4` is electrolysed using Pt electrodes. The products at the cathode and anode are respectively:

Options

A. `H_2,SO_2`
B. `O_2,NaOH`
C. `H_2,O_2`
D. `O_2,SO_2`

Answer

To solve the question regarding the electrolysis of an aqueous solution of sodium sulfate (Na₂SO₄) using platinum electrodes, we will follow these steps:

### Step 1: Identify the Electrolysis Process
In electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. We need to determine the reactions that take place at both electrodes.

### Step 2: Analyze the Cathode Reaction
At the cathode, reduction takes place. The aqueous solution contains Na⁺ ions, SO₄²⁻ ions, and water (H₂O). The possible reductions are:
- Reduction of water:
\[ 2H_2O + 2e^- \rightarrow H_2 + 2OH^- \]
- Reduction of sodium ions:
\[ Na^+ + e^- \rightarrow Na \]

To determine which reaction occurs, we compare their standard reduction potentials. The reduction potential for water is approximately -0.83 V, while for sodium it is -2.71 V. Since the reduction potential for water is less negative, water will be reduced at the cathode.

**Result at Cathode:** Hydrogen gas (H₂) is produced.

### Step 3: Analyze the Anode Reaction
At the anode, oxidation takes place. The possible oxidations are:
- Oxidation of water:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
- Oxidation of sulfate ions (SO₄²⁻): However, sulfate is already in a high oxidation state (+6), so it cannot be oxidized further.

Thus, water will undergo oxidation at the anode.

**Result at Anode:** Oxygen gas (O₂) is produced.

### Step 4: Summarize the Products
From the analysis:
- At the cathode, the product is hydrogen gas (H₂).
- At the anode, the product is oxygen gas (O₂).

### Final Answer
The products at the cathode and anode are respectively:
- Cathode: H₂
- Anode: O₂