First, let's write down the given half-reactions for magnesium and hydrogen:
1. [tex]\( \text{Mg} \longrightarrow \text{Mg}^{2+}(\text{aq}) + 2e^- \)[/tex]
2. [tex]\( 2\text{H}^+(\text{aq}) + 2e^- \longrightarrow \text{H}_2(\text{g}) \)[/tex]
Next, we need to combine these half-reactions to form the overall balanced equation. Notice that in the first half-reaction, magnesium ([tex]\(\text{Mg}\)[/tex]) is oxidized to [tex]\(\text{Mg}^{2+}(\text{aq})\)[/tex] with a release of 2 electrons. In the second half-reaction, two protons ([tex]\(2\text{H}^+\)[/tex]) gain 2 electrons to form hydrogen gas ([tex]\(\text{H}_2(\text{g})\)[/tex]).
Since the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction, they can be combined directly:
[tex]\[
\text{Mg}(s) + 2\text{H}^+(\text{aq}) \longrightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})
\][/tex]
This is the overall balanced equation for the redox reaction.
Therefore, the correct answer is:
[tex]\[
\text{Mg}(s) + 2\text{H}^+(\text{aq}) \longrightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})
\][/tex]
