In electrolysis of very dilute of `NaOH` using platinum electrodes

Options

A. `H_(2) ` is evolved at cathode and `O_(2)` at anode
B. `NH_(3)` is produced at anode
C. `Cl_(2)` is obtained at cathode
D. `O_(2)` is produced at cathode and `H_(2)` at anode.

Explanation

**Step-by-Step Solution:**

1. **Identify the Electrolyte and Its Components:**
- In this case, we are using very dilute NaOH (sodium hydroxide) as the electrolyte.
- The dissociation of NaOH in water gives us Na⁺ (sodium ions) and OH⁻ (hydroxide ions). Additionally, water can dissociate to provide H⁺ (hydrogen ions).

2. **Determine the Reactions at the Electrodes:**
- In electrolysis, reduction occurs at the cathode and oxidation occurs at the anode.
- At the cathode, we need to determine which ion will be reduced. The relevant ions are H⁺ and Na⁺.
- The standard reduction potential for H⁺ is higher than that for Na⁺, meaning H⁺ will be reduced preferentially.

3. **Write the Cathode Reaction:**
- The reduction of H⁺ at the cathode can be represented as:
\[ 2H^+ + 2e^- \rightarrow H_2(g) \]
- This indicates that hydrogen gas (H₂) is evolved at the cathode.

4. **Identify the Oxidation Reaction at the Anode:**
- At the anode, the only negatively charged species present is OH⁻.
- The oxidation of OH⁻ can be represented as:
\[ 4OH^- \rightarrow 2H_2O + O_2(g) + 4e^- \]
- This indicates that oxygen gas (O₂) is evolved at the anode.

5. **Summarize the Observations:**
- During the electrolysis of very dilute NaOH:
- Hydrogen gas (H₂) is produced at the cathode.
- Oxygen gas (O₂) is produced at the anode.

6. **Conclusion:**
- The correct observations from the electrolysis of very dilute NaOH using platinum electrodes are:
- H₂ is evolved at the cathode.
- O₂ is evolved at the anode.

**Final Answer:**
- At the cathode, hydrogen gas (H₂) is produced, and at the anode, oxygen gas (O₂) is produced.

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