Calculating the standard reduction potential for the oxidation of water

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*.




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:


$$\begin{align}\Delta G\_1+\Delta G\_2&=\Delta G\_3\\
z\_1E\_1+z\_2E\_2&=z\_3E\_3\\
2\times0.695\ \mathrm V+2\times1.763\ \mathrm V&=4\times1.229\ \mathrm V
\end{align}$$


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