# Calculating Electrode Potential for Electrolytic Cell at Non-Standard States

Recently, I came across an interesting question that left me rather baffled. I was asked to calculate the minimum voltage required for the electrolysis of 1.0 M of NaCl in neutral solution using the following equation:

$$\ce{2H2O + 2Cl- (1M) -> H2 (1 atm) + Cl2 (1 atm) + 2OH-} (\pu{1*10^{-7}~M})$$

I was left in a dilemma, if I were to use the Nernst Equation, I would get:

$$E = E_0 - \frac{RT}{nF \ln Q}$$

$$E = -1.36-(+0.83) - 0.012839\times\ln \pu{(1*10^{-7})^2}$$

$$E= \pu{-2.19 V} - (-0.4135) = \pu{-1.78 V}$$

So $+1.78$ V is needed to drive the reaction, theoretically and assuming ideal conditions.

However, I went online to search a bit, and came across a line written in Wikipedia that

... the standard potential of the water electrolysis cell is −1.23 V at 25 °C at pH 0 ([H+] = 1.0 M). At 25 °C with pH 7 ($\ce{[H+]} = \pu{1.0×10^−7 M}$), the potential is unchanged based on the Nernst equation.

Does that mean that regardless of the pH, electrode potential for water is still the same, or is there some rule or standard definitions that I'm missing out?

In this case, does this scenario apply only to water, what about other cases where other reagents are used at non-standard states, will my original method (using the Nernst Equation) work?