The electrolytic decomposition of dilute sulphuric acid with platinum electrode, cathodic reaction is :

Options

A. (a) reduction of `H^+`
B. (b) oxidation of `SO_4^(2-)`
C. (c) reduction `SO_3^(2-)`
D. (d) oxidation of `H_2O`

Answer

To determine the cathodic reaction during the electrolytic decomposition of dilute sulfuric acid (H₂SO₄) using platinum electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- Dilute sulfuric acid (H₂SO₄) in aqueous solution means we have H₂SO₄ dissociated into ions:
- H₂SO₄ → 2H⁺ + SO₄²⁻
- Additionally, water (H₂O) can dissociate into:
- H₂O → H⁺ + OH⁻

2. **Understand the Electrolytic Cell**:
- In an electrolytic cell, there are two electrodes: the anode (positive) and the cathode (negative).
- At the cathode, a reduction reaction occurs, while at the anode, an oxidation reaction occurs.

3. **Determine the Ions at the Cathode**:
- Since the cathode is negatively charged, it attracts positive ions. The relevant positive ions present are H⁺ ions from both the dissociation of sulfuric acid and water.

4. **Identify the Cathodic Reaction**:
- At the cathode, H⁺ ions will gain electrons (reduction) to form hydrogen gas (H₂):
- 2H⁺ + 2e⁻ → H₂
- This indicates that hydrogen ions are reduced to hydrogen gas at the cathode.

5. **Check for Charge and Mass Balance**:
- The reaction is balanced in terms of both charge and mass:
- Left side: 2 positive charges from 2H⁺ and 2 negative charges from 2e⁻.
- Right side: 0 charge from H₂ (neutral molecule).

6. **Conclusion**:
- The cathodic reaction in the electrolytic decomposition of dilute sulfuric acid with platinum electrodes is:
- **2H⁺ + 2e⁻ → H₂**