# Energy needed for the electrolysis of water

I am exploring a problem of my own and am confused by the contradictory info that I am finding.

Problem Statement:

What is the minimum energy required to obtain 32 g of diatomic oxygen (one mole) and 4 g of hydrogen (two mole) via the electrolysis of water in the presence of some electrolyte from an external voltage source of 12 V?

I often see sources quote that the minimum energy is the standard enthalpy of the reaction

2H20 (l) --> 2H2 (g) + 02 (g), $$\Delta$$H = 483.6 kJ

Other sources cite that the minimum energy required for an electrolysis reaction is given by

Energy = Work = nFE_ext,

where n is the number of electron moles forced into the system by the external potential, F is the number of Faradays, and E_ext is the external applied voltage.

The oxygen is produced at the anode and is given by the equation

2H20 (l) --> O2 (g) + 4H$$^{+}$$ (aq) + 4e$$^{-}$$, E_red = +1.23 V

The hydrogen is produced at the cathode and is given by the equation

2H20 + 2e$$^{-}$$ --> H2 (g) + 2OH$$^{-}$$, E_red = -0.83 V

Thoughts and Questions

For oxygen production electrons are being forced out of the system, and for hydrogen production electrons are being forced into the system. How do I obtain the values for the variable 'n'? If I assume for oxygen that n = -4 and for hydrogen n = 2, it seems evident that I will obtain a negative value for work (or applied energy), which does not make sense. Perhaps, I am simplifying the scenario too much because of the other oxidation-reduction reactions that take place (e.g. OH$$^{-}$$).

What role does the standard enthalpy value play in this electrolysis reaction? Does the voltage have to supply this energy value too, or can be it be independent of the voltage source and come from the environmental surroundings (which would tend to cool the surroundings).

Any advice would be greatly appreciated.

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