In the reaction 2Mg (s) + O2 (g) + 2H2O (l) → 2Mg2 + (aq) + 4OH - (aq), which substance is reduced?

A. Mg

B. O

C. H2O

D. OH

E. None of the Above (This is not a redox reaction.)

F. None of the above

No O atoms are present as reacting substances, only O_2 and H_2O molecules.

O_2 + 2H_2O + 2e^ (-) → 4OH^ (-)

We must use oxidation numbers to decide whether oxygen or water is the substance reduced.

The oxidation number of O changes from 0 in O_2 to - 2 in OH^ (-).

A decrease in oxidation number is reduction, so O_2 is the substance reduced.

The oxidation number of O is - 2 in both H_2O and OH^ (-), so water is neither oxidized nor reduced.