Question

How many amphere of current is needed to electrolysis one litre water

Answer 1

Splitting water takes 4 electrons per molecule of O2 and requires a potential of 1.23 V. This is a well-studied problem of water electrolysis. I’ll give a simple answer to your question but keep in mind actual industrial electrolysis makes use of catalysts (like platinum) and special chemistries (like sodium mixtures) and also applies an overvoltage to drive the reaction quickly.

One liter of water contains 1000 grams of water or about 55.56 mols of water. Each molecule of O2 produced requires 4 electrons in the reaction. The 55.56 mols of water can convert to 27.78 mols of O2. Thus, 111.1 mols of electrons are required to complete the reaction entirely. This is about 10.7 MC (mega-Coulombs) of charge that needs to be provided.

If you drove the reaction with 1 ampere and at 1.5 V (a 0.27 V overpotential) then it would take about 125 days to fully electrolize a liter of water. It would consume 1.5 W of power totalling to 16.1 MJ of energy. A higher current would make this go quicker but the total energy consumed would remain the same, unless you change the overpotential as well.

Keep in mind, you can’t specify the current at a given electrode voltage. The current is determined based on the electrode surface area and electrode separation, as well as the convective transport of the aqueous reactants. What you would do is apply a voltage and the current is whatever it happens to be. This is the case for electrochemistry reactions in general.

Also, regardless of current and voltage, 10.7 MC of charge needs to be provided always for 1 L of water. Higher currents deliver it faster but require a higher overpotential which makes the process less efficient.