The product of electrolysis of an aqueous solution of `K_(2)SO_(4)` using inert electrodes, at anode and cathode respectively are

Options

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

Answer

To solve the question regarding the products of electrolysis of an aqueous solution of potassium sulfate (K₂SO₄) using inert electrodes, we will follow these steps:

### Step 1: Understand the Electrolysis Process
When K₂SO₄ is dissolved in water, it dissociates into its ions:
- K₂SO₄ → 2 K⁺ + SO₄²⁻

Additionally, water also dissociates into:
- H₂O → H⁺ + OH⁻

Thus, the ions present in the solution are:
- Cations: K⁺, H⁺
- Anions: SO₄²⁻, OH⁻

### Step 2: Identify the Reactions at the Electrodes
Electrolysis involves reactions at both the anode (positive electrode) and cathode (negative electrode).

#### At the Anode:
The anions present are SO₄²⁻ and OH⁻. We need to determine which ion will be discharged first based on their oxidation potentials. The hydroxide ion (OH⁻) has a higher tendency to discharge compared to the sulfate ion (SO₄²⁻).

The reaction at the anode can be represented as:
\[ 4 OH⁻ \rightarrow 2 H₂O + O₂ + 4 e⁻ \]
This indicates that oxygen gas (O₂) is evolved at the anode.

#### At the Cathode:
The cations present are K⁺ and H⁺. The hydrogen ion (H⁺) has a higher reduction potential than potassium ion (K⁺), meaning H⁺ will be discharged first.

The reaction at the cathode can be represented as:
\[ 2 H⁺ + 2 e⁻ \rightarrow H₂ \]
This indicates that hydrogen gas (H₂) is evolved at the cathode.

### Step 3: Summarize the Products
From the reactions at the anode and cathode, we conclude:
- At the anode: Oxygen gas (O₂) is produced.
- At the cathode: Hydrogen gas (H₂) is produced.

### Final Answer
The products of electrolysis of an aqueous solution of K₂SO₄ using inert electrodes are:
- At the anode: Oxygen (O₂)
- At the cathode: Hydrogen (H₂)