# Question #0d806

May 17, 2017

$M g$ is being oxidized,
$H$ is being reduced,
($S {O}_{4}^{2 -}$ is merely a spectator ion)

#### Explanation:

Oxidation is the increase in oxidation state due to the loss of electrons; reduction is the lowering in oxidation state due to the gain of electrons.

$M g \left(s\right)$ starts as a pure element, and all pure substances in a chemical reaction have an oxidation state of $0$; after the reaction, it is ionized (it loses two electrons) to form a compound with the sulfate ion, and then has an oxidation state of $2 +$. Since its oxidation state is increased from $0$ to $2 +$, $M g$ is being oxidized.

$H$ starts off as the cation (with a charge and oxidation state of $1 +$) in the sulfuric acid component, and as the reaction proceeds, it becomes pure ${H}_{2} \left(g\right)$, so it then has an oxidation state of $0$. Since its oxidation state is lowered from $1 +$ to $0$, $H$ is being reduced.

Lastly, although not part of the question, the sulfate ion $S {O}_{4}^{2} -$ plays no direct role in the net ionic equation, and is considered a spectator ion. All oxidation-reduction reactions have at least one spectator ion.