To find the final balanced equation by combining the given half-reactions, let's follow a step-by-step approach:
### Oxidation Half-Reaction:
[tex]\[ Cu \longrightarrow Cu^{2+} + 2e^- \][/tex]
### Reduction Half-Reaction:
[tex]\[ NO_3^- + 2e^- + 2H^+ \longrightarrow NO_2^- + H_2O \][/tex]
### Step-by-Step Combination:
1. Determine Number of Electrons:
The oxidation half-reaction generates [tex]\( 2 \)[/tex] electrons, and the reduction half-reaction consumes [tex]\( 2 \)[/tex] electrons. Since the number of electrons lost in oxidation is equal to the number of electrons gained in reduction, the electrons are already balanced.
2. Combine the Half-Reactions:
Combine the oxidation and reduction half-reactions ensuring that the electrons cancel out.
Oxidation:
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
Reduction:
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
When combined, the electrons cancel out:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
Therefore, the balanced equation is:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
Among the provided choices, this matches the second equation:
[tex]\[ Cu + NO_3^- + 2H^+ \longrightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
Hence, the correct answer is:
[tex]\[ 2 \][/tex]
