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Montano1993 [528]
4 years ago
12

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

Chemistry
1 answer:
Bogdan [553]4 years ago
6 0

F. <em>None of the above </em>

<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.

O_2 + 2H_2O + 2e^(-) → 4OH^(-)

We must use <em>oxidation numbers</em> 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 <em>reduction</em>, so O_2 is the substance  reduced.

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

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