Electrolysis of Brine: Competing Reactions and Product Prediction
A student electrolyses concentrated aqueous sodium chloride using inert graphite electrodes. At the cathode, a gas is evolved and the surrounding solution becomes alkaline. At the anode, a pale yellow-green gas with a pungent smell is produced. The student expected sodium metal at the cathode and oxygen at the anode, based on the electrode potentials of Na⁺/Na (E° = −2.71 V) and O2/H2O (E° = +1.23 V). Standard reduction potentials: Na⁺ + e⁻ → Na, −2.71 V; 2H2O + 2e⁻ → H2 + 2OH⁻, −0.83 V; Cl2 + 2e⁻ → 2Cl⁻, +1.36 V; O2 + 4H⁺ + 4e⁻ → 2H2O, +1.23 V.
1. Explain why hydrogen, not sodium, forms at the cathode, and why chlorine, not oxygen, forms at the anode.
Read the worked answer
Cathode. Reduction occurs at the cathode. The two candidates are Na⁺ + e⁻ → Na (E° = −2.71 V) and 2H2O + 2e⁻ → H2 + 2OH⁻ (E° = −0.83 V). The water reduction has the far less negative (more positive) potential, so it is thermodynamically favoured. Sodium metal would require a much larger applied voltage and, in water, any sodium formed would immediately react with water to give H2 and NaOH. So hydrogen is observed and the solution becomes alkaline from the OH⁻ produced - consistent with the observations.
Anode. Oxidation occurs at the anode; the species that is oxidised more readily (lower oxidation potential, i.e. less positive reduction potential) wins. For chlorine: Cl2 + 2e⁻ → 2Cl⁻, E° = +1.36 V. For oxygen from water: O2 + 4H⁺ + 4e⁻ → 2H2O, E° = +1.23 V. On standard potentials alone oxygen is slightly favoured, but in concentrated brine the chloride concentration is high, so the Cl2/Cl⁻ potential shifts favourably (Nernst effect) and the overpotential for O2 evolution on graphite is large. In practice chlorine is produced. The pale yellow-green pungent gas confirms Cl2.
Conclusion. The student's prediction based on raw standard potentials ignores concentration effects, overpotential, and the reactivity of sodium with water.
2. Write the overall cell equation for the observed electrolysis and explain the role of the applied voltage relative to the thermodynamic minimum.
Read the worked answer
Half-equations. Cathode (reduction): 2H2O + 2e⁻ → H2 + 2OH⁻ Anode (oxidation): 2Cl⁻ → Cl2 + 2e⁻
Overall. 2H2O + 2Cl⁻ → H2 + Cl2 + 2OH⁻, or in molecular form with the spectator Na⁺: 2NaCl + 2H2O → 2NaOH + H2 + Cl2.
This shows why the solution becomes alkaline (NaOH accumulates) and why the gases are H2 and Cl2.
Applied voltage. The thermodynamic minimum is E°cell = E°cathode − E°anode = (−0.83) − (+1.36) = −2.19 V, so an external voltage of at least about 2.19 V is needed; the negative sign indicates the reaction is non-spontaneous and must be driven. In practice a higher voltage is applied to overcome overpotential at the electrodes (especially for gas evolution on graphite) and the resistance of the electrolyte. The student should understand that the applied voltage must exceed the thermodynamic minimum plus overpotentials; simply exceeding 2.19 V is necessary but not sufficient for the observed products, because kinetics and concentration determine which competing reaction dominates.