Answered

What is the final, balanced equation that is formed by combining these two half-reactions?

[tex]\[
\begin{array}{l}
Cu \longrightarrow Cu^{2+} + 2e^{-} \\
NO_3^{-} + 2e^{-} + 2H^{+} \longrightarrow NO_2^{-} + H_2O
\end{array}
\][/tex]

A. [tex]\[ Cu^{2+} + NO_3^{-} + 4e^{-} + 2H^{+} \longrightarrow Cu + NO_2^{-} + H_2O \][/tex]

B. [tex]\[ Cu + NO_3^{-} + 2H^{+} \longrightarrow Cu^{2+} + NO_2^{-} + H_2O \][/tex]

C. [tex]\[ 2Cu + NO_3^{-} + 2H^{+} \longrightarrow 2Cu^{2+} + NO_2^{-} + H_2O \][/tex]

D. [tex]\[ Cu + NO_3^{-} + 2H^{+} \longrightarrow Cu^{2+} + NO_2^{-} + 2H_2O \][/tex]



Answer :

GinnyAnswer
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]