On electrolysis of a solution of dilute `H_(2)SO_(4)` between platinum electrodes , the gas evolved at the anode is
(a)`SO_(2)`
(b)`SO_(3)`
(c)`O_(2)`
(d)`H_(2)`

Options

A. `SO_(2)`
B. `SO_(3)`
C. `O_(2)`
D. `H_(2)`

Explanation

To determine the gas evolved at the anode during the electrolysis of a dilute solution of sulfuric acid (H₂SO₄) between platinum electrodes, we can follow these steps:

### Step 1: Understand the Electrolysis Process
During electrolysis, an electric current is passed through the solution, causing chemical reactions at the electrodes. The solution of dilute sulfuric acid dissociates into ions: H⁺ (hydrogen ions) and SO₄²⁻ (sulfate ions), along with water which dissociates into H⁺ and OH⁻ (hydroxide ions).

### Step 2: Identify the Ions Present
In a dilute solution of sulfuric acid, the ions present are:
- H⁺ (from sulfuric acid and water)
- SO₄²⁻ (sulfate ions)
- OH⁻ (from water)

### Step 3: Determine the Reactions at the Electrodes
At the anode, oxidation occurs, which means that negatively charged ions will migrate towards the anode and lose electrons. The two main candidates for oxidation are:
- OH⁻ (hydroxide ions)
- SO₄²⁻ (sulfate ions)

### Step 4: Apply the Concept of Preferential Discharge
According to the preferential discharge theory, the ion that is more easily oxidized will react first. The oxidation potential of hydroxide ions (OH⁻) is higher than that of sulfate ions (SO₄²⁻). Therefore, OH⁻ ions will be discharged at the anode.

### Step 5: Write the Half-Reaction for the Anode
The oxidation of hydroxide ions can be represented by the following half-reaction:
\[ 4OH⁻ \rightarrow 2H₂O + O₂ + 4e⁻ \]
This shows that oxygen gas (O₂) is produced at the anode.

### Step 6: Conclusion
Thus, the gas evolved at the anode during the electrolysis of dilute sulfuric acid is oxygen (O₂).

### Final Answer
The gas evolved at the anode is (c) O₂.

---