Calculating the standard reduction potential for the oxidation of water

Question

I was working with Latimer and Frost diagrams for oxygen when I came across what seems to me a contradiction.

From the Latimer diagram for oxygen below, we know the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$ and the standard reduction potential for the reduction of $\ce{H2O2}$ to $\ce{H2O}$, as represented in the following chemical equations:

$$\ce{O2} \overset{+0.70}{\longrightarrow} \ce{H2O2} \overset{+1.76}{\longrightarrow} \ce{H2O}$$

$$\begin{align} \ce{2H+ (aq) + 2e- + H2O2 (aq) &-> 2H2O (l)} & E^\circ &= \pu{+1.76 V} \\ \ce{O2 (g) + 2H+ (aq) + 2e- &-> H2O2 (aq)} & E^\circ &= \pu{+0.70 V} \\ \end{align}$$

If we add these two reduction chemical equations, we can find the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$, which would be $\pu{+2.46 V}$ according to the aformentioned reduction potentials and chemical equations.

The reduction of $\ce{O2}$ to $\ce{H2O}$ is the reverse reaction of the oxidation of water:

$$\ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}\qquad E^\circ = \pu{-1.23 V}$$

Why does the standard reduction potential for the oxidation of water calculated from the Latimer diagram differ from the value reported in the table of standard reduction potentials? More specifically, why is the value calculated from the Latimer diagram double of that reported in the table? I checked my work multiple times and do not understand why this occurs, but I might have still made a silly mistake.

References
Shriver, D. F., Weller, M. T., Overton, T., Rourke, J., & Armstrong, F. A. (2014). Inorganic chemistry 6th Edition.

Answer 1

According to Hess’ law, we expect the Gibbs energy change $\Delta G$ of the two partial reactions to add up to the total Gibbs energy change:

$$\Delta G_1+\Delta G_2=\Delta G_3$$

For an electrochemical reaction, the change in Gibbs energy corresponds to an electrochemical potential $E$ according to

$$\Delta G = -zFE$$

where
$F$ is the Faraday constant and
$z$ is the number of electrons transferred.

Since for the two partial reactions the number of electrons is $z_1=z_2=2$, but for the total reaction $z_3=4$, we expect:

$$\begin{align}\Delta G_1+\Delta G_2&=\Delta G_3\\ z_1E_1+z_2E_2&=z_3E_3\\ 2\times0.695\ \mathrm V+2\times1.763\ \mathrm V&=4\times1.229\ \mathrm V \end{align}$$