What is the thermodynamic minimum cell potential for water electrolysis at standard conditions (25°C)?

• A. 0.00 V
• B. 1.23 V
• C. -1.23 V
• D. 0.59 V

Answer: (B)

Driving water splitting thermodynamically requires supplying the exact amount of free energy to overcome the formation of H2 and O2. At standard conditions, the overall reaction 2 H2O → 2 H2 + O2 has a standard Gibbs free energy change of about +474 kJ per mole of O2 produced (i.e., per 4 electrons transferred). With four electrons involved, the minimum external voltage is E_min = ΔG°/(nF) = 474,000 J/mol ÷ (4 × 96485 C/mol) ≈ 1.23 V. So, the thermodynamic minimum potential you must apply to drive the reaction is about 1.23 V. In practice, extra voltage is needed due to overpotentials and ohmic losses.