In this region, part of the *OH will possibly also be reduced to H 2 O, leaving a free Ru site behind. With more negative potentials, the *OH coverage decreases, decreasing the H 2 O 2 yield as well, and the 4-electron process via direct contact of O 2 with the Ru surface becomes more competitive and dominant. To explain the rapidly increasing tendency for the 4-electron process, the reactivity of the bare Ru sites for the latter pathway must be significantly higher than that of the *OH-covered sites, which are responsible for the H 2 O 2 formation.