Question

During the electrolysis of fused NaCl, the product obtained at anode is

• A. $\text{Na}\ (\text{s})$
• B. $\text{Cl}_2\ (\text{g})$
• C. $\text{O}_2\ (\text{g})$
• D. $\text{Na}\ (\text{l})$

Hint:

Always keep the classic acronym AN OX and RED CAT in mind: Anode = Oxidation; Reduction = Cathode. Since oxidation involves negative ions losing their negative charge, the neutral non-metal gas always escapes at the anode!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the specific chemical product liberated at the positive electrode (anode) during the electrochemical electrolysis of molten (fused) sodium chloride (NaCl).

Step 2: Key Formula or Approach:
During electrolysis:
The anode is the positive electrode where oxidation (loss of electrons) occurs.
The cathode is the negative electrode where reduction (gain of electrons) occurs.
Molten NaCl dissociates entirely into free-moving $\text{Na}^+$ cations and $\text{Cl}^-$ anions.

Step 3: Detailed Explanation:
When an electric current is passed through fused NaCl:
1. The negatively charged chloride ions ($\text{Cl}^-$) migrate toward the positive anode. At the anode surface, they lose electrons and oxidize to form green chlorine gas molecules:
$$\text{Anode reaction (Oxidation):}\quad 2\text{Cl}^- \rightarrow \text{Cl}_2\ (\text{g}) + 2\text{e}^-$$ 2. Simultaneously, the positively charged sodium ions ($\text{Na}^+$) migrate toward the negative cathode where they capture electrons to reduce into neutral metallic sodium:
$$\text{Cathode reaction (Reduction):}\quad \text{Na}^+ + \text{e}^- \rightarrow \text{Na}\ (\text{l/s})$$ Since the question specifically requests the product formed at the anode, it must be chlorine gas ($\text{Cl}_2$). This matches option (B).

Step 4: Final Answer:
The product obtained at the anode during the electrolysis of fused NaCl is $\text{Cl}_2\ (\text{g})$, corresponding to option (B).