The product formed when concentrated sodium chloride solution is electrolysed using carbon electrodes is

Analyzing the question: We are electrolysing concentrated sodium chloride solution (brine) with inert carbon electrodes. We need the product formed.

Concept: In the electrolysis of brine, the solution contains four ions:

• From NaCl: Na⁺ and Cl⁻
• From water: H⁺ and OH⁻

Selective discharge decides which ions are discharged:

• At the cathode (–): H⁺ and Na⁺ compete. H⁺ is lower in the electrochemical series, so it is discharged: 2H⁺ + 2e⁻ → H₂(g)
• At the anode (+): Cl⁻ and OH⁻ compete. Because the solution is concentrated, Cl⁻ is preferentially discharged: 2Cl⁻ → Cl₂(g) + 2e⁻

What remains in solution: Na⁺ and OH⁻ ions, which means sodium hydroxide (NaOH) is formed.

Overall: 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂

Checking the options:

• A. Chlorine water: this would need Cl₂ dissolved in water. The Cl₂ escapes as a gas, so it is not the product left in the cell.
• B. Hydrochloric acid: H⁺ and Cl⁻ are both discharged, so HCl is not formed.
• C. Sodium hydroxide: Na⁺ and OH⁻ remain in solution. ✔
• D. Sodium oxochlorate(I) (NaOCl): this forms only if Cl₂ is allowed to mix with cold NaOH, such as in a cell without separation, and it is not the direct product.

Answer: The products are hydrogen at the cathode, chlorine at the anode, and sodium hydroxide in solution.

Tip: Remember "concentrated brine gives H₂, Cl₂, and NaOH." If the solution is dilute, OH⁻ is discharged instead and O₂ is produced at the anode. The concentration of the solution tells you which ion wins at the anode.