Electrolysis products in concentrated NaCl(aq)

Steps:

• At cathode (reduction): H₂O + 2e⁻ → H₂ + 2OH⁻ occurs over Na⁺ + e⁻ → Na, as water reduces easier.
• At anode (oxidation): 2Cl⁻ → Cl₂ + 2e⁻ dominates in concentrated solution over 2H₂O → O₂ + 4H⁺ + 4e⁻, due to higher Cl⁻ concentration.
• Inert electrodes ensure no reaction interference.
• Overall: H₂ at cathode, Cl₂ at anode.

Why A is correct:

• In concentrated NaCl(aq), cathode favors water reduction (E° = -0.83 V) over Na⁺ (E° = -2.71 V); anode favors Cl⁻ oxidation (E° = -1.36 V) over water (E° = -1.23 V) per standard electrode potentials.

Why the others are wrong:

• B: Oxygen forms at anode in dilute NaCl, not concentrated.
• C: Sodium metal cannot form from aqueous Na⁺ due to water's priority in reduction.
• D: Neither sodium nor oxygen forms in this setup.

Final answer: A