What is the overall reaction for the electrolysis of water with inert electrodes at standard conditions, neglecting overpotentials?

• A. 2 H2O(l) → 2 H2(g) + O2(g)
• B. H2O(l) → H2(g) + 1/2 O2(g)
• C. H2O(l) + electricity → H2(g) + O2(g) under standard conditions
• D. 2 H2(g) + O2(g) → 2 H2O(l)

Answer: (C)

Splitting water with electricity is driven by energy input, not spontaneous chemistry. When you use inert electrodes, the overall process is water being decomposed into hydrogen and oxygen, with electrical energy supplying the drive.

This can be written as water plus electricity yielding hydrogen and oxygen under standard conditions. Framing it this way emphasizes that energy is the reactant driving the reaction, which is the hallmark of electrolysis. It’s the same overall transformation you’d write as 2 H2O → 2 H2 + O2 if you show the exact stoichiometry, but including electricity highlights the external energy requirement as part of the process.

The other expressions either omit the energy input or present the data in a less direct way: a purely chemical equation without noting energy, a per-molecule form, or the reverse reaction forming water. The key idea is that electricity drives the decomposition of water into H2 and O2.