The gas evolved at the anode when `K_(2)SO_(4)(aq)` is electrolyzed between `Pt` electrode is

Options

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

Explanation

To determine the gas evolved at the anode when K₂SO₄ (aq) is electrolyzed between platinum electrodes, we can follow these steps:

### Step 1: Identify the Electrolysis Process
In the electrolysis of K₂SO₄ (aq), we need to consider the reactions occurring at both the anode and cathode. The anode is where oxidation occurs, and the cathode is where reduction occurs.

### Step 2: Determine the Species Present
In an aqueous solution of K₂SO₄, the dissociated ions are:
- K⁺ (potassium ions)
- SO₄²⁻ (sulfate ions)
- H₂O (water molecules)

### Step 3: Analyze Oxidation Reactions at the Anode
At the anode, we have two potential species that can undergo oxidation:
1. SO₄²⁻ (sulfate ions)
2. H₂O (water)

### Step 4: Compare Oxidation Potentials
To determine which species will be oxidized, we need to compare their oxidation potentials:
- The oxidation potential of SO₄²⁻ is lower than that of H₂O. This means that H₂O has a greater tendency to be oxidized compared to SO₄²⁻.

### Step 5: Write the Oxidation Reaction for Water
Since water is oxidized at the anode, we can write the half-reaction for the oxidation of water:
\[ 2 H₂O \rightarrow O₂ + 4 H⁺ + 4 e⁻ \]
This equation shows that oxygen gas (O₂) is produced along with hydrogen ions (H⁺) and electrons (e⁻).

### Step 6: Conclude the Gas Evolved
From the oxidation of water, we see that the gas evolved at the anode is oxygen (O₂).

### Final Answer
The gas evolved at the anode when K₂SO₄ (aq) is electrolyzed between platinum electrodes is **O₂** (oxygen).

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