Question

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

Answer 1

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.