Calculating the standard reduction potential for the oxidation of water
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:

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