During an electrolysis of conc. `H_(2)SO_(4)`, peroxydisulphuric acid `(H_(2)S_(2)O_(8))` and `O_(2)` form in an equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be

Options

A. Thrice that of `O_(2)`
B. Twice that of `O_(2)`
C. Equal to that of `O_(2)`
D. Half of the of `O_(2)`

Answer

To solve the problem, we need to analyze the electrolysis of concentrated sulfuric acid (H₂SO₄) and understand the reactions occurring at the anode and cathode.

### Step-by-Step Solution:

1. **Identify the Products of Electrolysis**:
During the electrolysis of concentrated H₂SO₄, two main products are formed: peroxydisulfuric acid (H₂S₂O₈) and oxygen gas (O₂). According to the problem, these are formed in equimolar amounts.

2. **Determine the Reactions at the Electrodes**:
- At the **anode**, the oxidation reaction occurs, leading to the formation of O₂. The half-reaction can be represented as:
\[ 2H₂O \rightarrow O₂ + 4H^+ + 4e^- \]
- At the **cathode**, the reduction reaction occurs, where H⁺ ions are reduced to form hydrogen gas (H₂):
\[ 2H^+ + 2e^- \rightarrow H₂ \]

3. **Calculate the Electrons Required for Each Reaction**:
- For the formation of 1 mole of O₂, 4 moles of electrons are required.
- For the formation of 1 mole of H₂S₂O₈, 2 moles of electrons are required.

4. **Total Electron Requirement**:
Since H₂S₂O₈ and O₂ are formed in equimolar amounts, let's denote the amount formed as \( x \) moles for each.
- For \( x \) moles of O₂, the total electrons required will be \( 4x \).
- For \( x \) moles of H₂S₂O₈, the total electrons required will be \( 2x \).
- Therefore, the total electrons required for the formation of both products will be:
\[ \text{Total electrons} = 4x + 2x = 6x \]

5. **Relate Electrons to Hydrogen Production**:
Each mole of H₂ produced requires 2 moles of electrons:
\[ 2e^- \rightarrow H₂ \]
Therefore, the moles of H₂ produced from \( 6x \) moles of electrons will be:
\[ \text{Moles of H₂} = \frac{6x}{2} = 3x \]

6. **Conclusion**:
Since \( x \) is the amount of O₂ and H₂S₂O₈ formed, and they are in equimolar amounts, the moles of H₂ produced will be \( 3 \) times the moles of O₂ (or H₂S₂O₈) formed. Thus, if 1 mole of O₂ is formed, then:
\[ \text{Moles of H₂} = 3 \times 1 = 3 \text{ moles} \]

### Final Answer:
The moles of H₂ that will be formed simultaneously are **3 moles**.

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