Question #0d806

1 Answer
May 17, 2017

Answer:

#Mg# is being oxidized,
#H# is being reduced,
(#SO_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.

#Mg(s)# 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+#, #Mg# 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(g)#, 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 #SO_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.