Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium
From Proceedings of the Symposium on Oxygen Electrochemistry and Water Research 2002, 36 (1), 85-94:

$$
\begin{array}{lcc}
\text{Formation}&amp;:&amp;~\ce{2H+ + O2 + 2e- &amp;-&gt;&amp; H2O2} &amp;\quad E^0=\pu{0.440V} \text{ vs. SCE}\\
\text{Reduction}&amp;:&amp; ~\ce{H2O2 + 2H+ + 2e- &amp;-&gt;&amp; 2H2O} &amp;\quad E^0=\pu{1.534V} \text{ vs. SCE}\\
\text{Direct water}&amp;:&amp; ~\ce{2H+ + 1/2O2 + 2e- &amp;-&gt;&amp; H2O}\\
\text{Decomposition}&amp;:&amp;  ~\ce{H2O2 &amp;-&gt;&amp; H2O + 1/2 O2}
\end{array}$$

($\text{SCE}$ stands for Saturated Calomel Electrode)

Another strange suggested reaction involves the anion $\ce{HO2-}$ (Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. Monatshefte für Chemie 1952 83(2) 422–39.):

$$\text{Decomposition 2}:\ce{H2O2 + H2O- -&gt; H2O + O2 + OH-}$$



At the cathode occur at the same time the hydrogen gas evolution:

$$\ce{2H+ + 2e- -&gt; H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$

Meanwhile, at the anode, oxygen gas is evolved:

$$~\ce{2H2O -&gt; 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$