To form a balanced equation from the given half-reactions, we need to combine them in such a way that the electrons are balanced and all elements on both sides of the equation are accounted for.
The given half-reactions are:
1. [tex]\( \text{Oxidation reaction:} \)[/tex]
[tex]\[ \text{Cu} \longrightarrow \text{Cu}^{2+} + 2e^{-} \][/tex]
2. [tex]\( \text{Reduction reaction:} \)[/tex]
[tex]\[ \text{NO}_3^{-} + 2e^{-} + 2H^{+} \longrightarrow \text{NO}_2^{-} + \text{H}_2\text{O} \][/tex]
To balance these half-reactions, the electrons lost in the oxidation reaction must equal the electrons gained in the reduction reaction. In both half-reactions, 2 electrons are involved, so they already match.
Now, we can add the two half-reactions together:
[tex]\[ \text{Cu} \longrightarrow \text{Cu}^{2+} + 2e^{-} \][/tex]
[tex]\[ \text{NO}_3^{-} + 2e^{-} + 2H^{+} \longrightarrow \text{NO}_2^{-} + \text{H}_2\text{O} \][/tex]
When we add them up, the 2 electrons on the right side of the oxidation reaction and the 2 electrons on the left side of the reduction reaction cancel each other out. So we can write:
[tex]\[ \text{Cu} + \text{NO}_3^{-} + 2H^{+} \longrightarrow \text{Cu}^{2+} + \text{NO}_2^{-} + \text{H}_2\text{O} \][/tex]
This equation is now balanced with respect to all elements and charge. Consequently, the final balanced equation is:
[tex]\[ \text{Cu} + \text{NO}_3^{-} + 2H^{+} \longrightarrow \text{Cu}^{2+} + \text{NO}_2^{-} + \text{H}_2\text{O} \][/tex]
