To balance the given reaction [tex]\( \text{Zn} + \text{H}^+ \rightarrow \text{Zn}^{2+} + \text{H}_2 \)[/tex], we need to break it down into its oxidation and reduction half-reactions.
Step 1: Identify the Oxidation and Reduction Half-Reactions
1. Oxidation Half-Reaction (Zinc is oxidized):
[tex]\[
\text{Zn} \rightarrow \text{Zn}^{2+}
\][/tex]
Here, zinc (Zn) is losing electrons to form zinc ion ([tex]\(\text{Zn}^{2+}\)[/tex]). To balance the charge, we need to add electrons to the right side of the equation:
[tex]\[
\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-
\][/tex]
This shows that zinc is oxidized, releasing 2 electrons.
2. Reduction Half-Reaction (Hydrogen is reduced):
[tex]\[
\text{H}^+ \rightarrow \text{H}_2
\][/tex]
Hydrogen ions ([tex]\(\text{H}^+\)[/tex]) gain electrons to form hydrogen gas ([tex]\(\text{H}_2\)[/tex]). Each hydrogen molecule ([tex]\(\text{H}_2\)[/tex]) requires 2 hydrogen ions ([tex]\(\text{H}^+\)[/tex]) and 2 electrons to balance the reaction:
[tex]\[
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
\][/tex]
This shows that hydrogen ions are reduced by gaining 2 electrons.
Step 2: Balance the Electrons Between Half-Reactions
Each half-reaction produces and consumes the same number of electrons (2 in this case), so the electrons are already balanced.
Step 3: Write the Complete Balanced Half-Reactions
- Oxidation Half-Reaction:
[tex]\[
\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-
\][/tex]
- Reduction Half-Reaction:
[tex]\[
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
\][/tex]
Summary:
- Number of electrons produced in the oxidation half-reaction: [tex]\(2\)[/tex]
- Number of electrons used in the reduction half-reaction: [tex]\(2\)[/tex]
So, the balanced half-reactions are:
[tex]\[
2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2
\][/tex]
[tex]\[
\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-
\][/tex]
