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.