Consider the half-reactions below for a chemical reaction:

[tex]\[
\begin{array}{l}
Mg \longrightarrow Mg^{2+}(aq) + 2e^{-} \\
2H^{+}(aq) + 2e^{-} \longrightarrow H_2(g)
\end{array}
\][/tex]

What is the overall equation for this chemical reaction?

A. [tex]\(Mg(s) + 2H^{+}(aq) \longrightarrow Mg^{2+}(aq) + H_2(g)\)[/tex]

B. [tex]\(Mg(s) + 2H^{+}(aq) \longrightarrow Mg^{2+}(aq) + 2e^{-}\)[/tex]

C. [tex]\(Mg^{2+}(aq) + H_2(g) \longrightarrow Mg(s) + 2H^{+}(aq)\)[/tex]

D. [tex]\(Mg(s) + 2H \longrightarrow Mg(aq) + H_2(g)\)[/tex]



Answer :

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]