Question:easy

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

Show 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!
Updated On: Jun 12, 2026
  • $\text{Na}\ (\text{s})$
  • $\text{Cl}_2\ (\text{g})$
  • $\text{O}_2\ (\text{g})$
  • $\text{Na}\ (\text{l})$
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Recall electrode roles.
In electrolysis the anode is positive and is where oxidation (loss of electrons) happens; the cathode is negative and is where reduction happens. The question asks for the anode product.
Step 2: Identify the ions present.
Molten (fused) NaCl fully dissociates into free Na$^+$ cations and Cl$^-$ anions, with no water involved.
Step 3: Decide which ion goes to the anode.
The negative Cl$^-$ ions are attracted to the positive anode.
Step 4: Write the anode reaction.
At the anode, chloride ions are oxidised: $2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$. The product is chlorine gas.
Step 5: Note the cathode for contrast.
Na$^+$ ions go to the cathode and are reduced to molten sodium: $\text{Na}^+ + e^- \rightarrow \text{Na}$. This is the cathode product, not what is asked.
Step 6: State the anode answer.
Since the question wants the anode product, it is chlorine gas, $\text{Cl}_2(g)$ - option (2). Oxygen cannot form here because there is no water.
\[ \boxed{\text{Anode product: Cl}_2(g)\ \text{(option 2)}} \]
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