Electrolysis of 50% `H_(2)SO_(4)` gives

Options

A. `H_(2)O`
B. `D_(2)O`
C. `H_(2)O_(2)`
D. `H_(2)`

Answer

To solve the question regarding the electrolysis of 50% H₂SO₄, we will break down the process step by step.

### Step 1: Understanding Electrolysis
Electrolysis is a chemical process that uses an electric current to drive a non-spontaneous reaction. In this case, we are looking at the electrolysis of sulfuric acid (H₂SO₄) dissolved in water.

### Step 2: Dissociation of Sulfuric Acid
When sulfuric acid (H₂SO₄) is dissolved in water, it dissociates into ions:
\[ H₂SO₄ \rightarrow 2H^+ + SO₄^{2-} \]
In a 50% solution, it primarily dissociates into hydrogen ions (H⁺) and bisulfate ions (HSO₄⁻).

### Step 3: Reactions at the Electrodes
During electrolysis, reactions occur at both the cathode and the anode.

- **At the Cathode**:
The hydrogen ions (H⁺) gain electrons (reduction) to form hydrogen gas (H₂):
\[ 2H^+ + 2e^- \rightarrow H₂(g) \]
Thus, hydrogen gas is produced at the cathode.

- **At the Anode**:
The bisulfate ions (HSO₄⁻) undergo oxidation:
\[ 2HSO₄^- \rightarrow H₂S₂O₈ + 2e^- \]
This reaction produces peroxodisulfuric acid (H₂S₂O₈), which can further react with water to produce hydrogen peroxide (H₂O₂) and sulfuric acid (H₂SO₄).

### Step 4: Final Products
From the electrolysis of 50% H₂SO₄, we obtain:
- Hydrogen gas (H₂) at the cathode.
- Peroxodisulfuric acid (H₂S₂O₈) at the anode, which can yield hydrogen peroxide (H₂O₂) upon further reaction with water.

### Conclusion
The primary product of the electrolysis of 50% H₂SO₄ is hydrogen peroxide (H₂O₂), along with hydrogen gas (H₂) at the cathode.

### Final Answer
The correct answer is **Hydrogen Peroxide (H₂O₂)**.

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