Oxidation of water — what is the half reaction for water?

During the electrolysis of a solution of copper sulfate, copper is reduced to form a solid on the inert electrode while water is oxidised at the anode. What is the half equation for water?

\begin{align} \text{Is it}&& \ce{4 OH−(aq) &-> O2(g) + 2 H2O(l) + 4 e−}\\ \text{or}&& \ce{2H2O(l) &-> 4H+(aq) + O2(g) + 4e–}? \end{align}

What is the difference? And what is the standard reduction potential in an electrolytic cell?

Both your reactions are equivalent. Here’s why. We start from $(1)$.

$$\ce{4 OH- -> O2 + 2 H2O + 4 e-}\tag{1}$$

(I have omitted the phase descriptors for clarity.) In any chemical reaction, we can add or subtract spectators as we see fit; this is akin to mathematical equations where we can add or substract expressions as we see fit as long as we do the same to both sides. For example, I want to add four protons to $(1)$ to give $(2)$.

$$\ce{4 OH- + 4 H+ -> O2 + 2 H2O + 4 e- + 4 H+}\tag{2}$$

If this were now a mathematical equation, I could combine (or divide) parts as I see fit. In a chemical reaction, we need to be a bit more careful. But as long as our focus is not on proton transfers, we can combine anything basic enough with acidic protons. Here, you can combine the four hydroxides and four protons on the reactant side of $(2)$ to give $(3)$.

$$\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\tag{3}$$

Now, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.

$$\ce{2 H2O -> 4 H+ + O2 + 4 e-}\tag{4}$$

It so turns out that $(4)$ is your second suggested reaction. It is the same equation as $(1)$ as I have just shown; the difference is merely in the type of water molecules present.

Now which one should you use? Notice that one reaction consumes hydroxide while the other produces protons (formally! The consumption of hydroxide is equivalent to the production of protons). Thus, if your reaction medium is basic, you should use $(1)$ for your redox reaction. If, however, your reaction mixture is acidic, $(4)$ is more appropriate. In short, choose the one with the same charged particle as is used in the other half-reaction to make your life simpler.

You are looking for a reaction where water is oxidized, i.e. loses electrons. In the chemical equation, the electrons must appear on the opposite side as $\ce{H2O}$, since electrons are products of an oxidation half-reaction.

On the other hand, any time we write $\ce{X + e-}$ we are depicting a reduction of $\ce{X}$. Which of your proposed reactions shows water being oxidized (and not reduced)?

• Strictly speaking, both of them. It is just a matter of pH. Apr 11, 2017 at 12:25
• You're arguing that although the first reaction produces two water molecules, it consumes (oxidizes) four water molecules, which are represented here by hydroxide intermediate derived from the autoprotolysis of water? Apr 11, 2017 at 15:49
• Well, yeah, you may put it like that. Apr 11, 2017 at 15:50
• What is hydroxide, if it is not a form of water minus a proton?
– Jan
Sep 13, 2017 at 9:15