Answer :
To find the final, balanced equation formed by combining the two given half-reactions, we need to follow a series of steps to ensure both mass and charge are balanced. Let's look at the given half-reactions again:
1.
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
2.
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
Step-by-Step Solution:
### Step 1: Identify the Half-Reactions
The two half-reactions are:
- Oxidation half-reaction (copper losing electrons):
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
- Reduction half-reaction (nitrate gaining electrons):
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
### Step 2: Balance the Electrons
Each half-reaction involves the transfer of 2 electrons. Since the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction, we can combine them directly.
### Step 3: Add the Half-Reactions Together
Combine the two half-reactions, ensuring that the electrons cancel out:
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
Adding these together:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
### Step 4: Verify the Balance
- Atoms:
- Copper (Cu): 1 atom on each side.
- Nitrate (NO3-): 1 nitrogen atom on each side.
- Hydrogen (H+): 2 atoms on the left, and in the product [tex]\( H_2O \)[/tex] (2 H atoms).
- Oxygen (O): 3 atoms from [tex]\( NO_3^- \)[/tex] on the left, with 1 oxygen in [tex]\( NO_2^- \)[/tex] and another 1 in [tex]\( H_2O \)[/tex] on the right.
- Charge:
- Left side: No net charge because [tex]\( Cu \)[/tex] is neutral, [tex]\( NO_3^- \)[/tex] contributes -1, and [tex]\( 2H^+ \)[/tex] contributes +2, making the total charge +1.
- Right side: [tex]\( Cu^{2+} \)[/tex] contributes +2, [tex]\( NO_2^- \)[/tex] contributes -1, making the total charge also +1.
Both mass and charge are balanced.
### Final Balanced Equation:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
Therefore, the final balanced equation you proposed is:
[tex]\[ Cu + NO_3^- + 2H^+ \longrightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
1.
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
2.
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
Step-by-Step Solution:
### Step 1: Identify the Half-Reactions
The two half-reactions are:
- Oxidation half-reaction (copper losing electrons):
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
- Reduction half-reaction (nitrate gaining electrons):
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
### Step 2: Balance the Electrons
Each half-reaction involves the transfer of 2 electrons. Since the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction, we can combine them directly.
### Step 3: Add the Half-Reactions Together
Combine the two half-reactions, ensuring that the electrons cancel out:
[tex]\[ Cu \rightarrow Cu^{2+} + 2e^- \][/tex]
[tex]\[ NO_3^- + 2e^- + 2H^+ \rightarrow NO_2^- + H_2O \][/tex]
Adding these together:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
### Step 4: Verify the Balance
- Atoms:
- Copper (Cu): 1 atom on each side.
- Nitrate (NO3-): 1 nitrogen atom on each side.
- Hydrogen (H+): 2 atoms on the left, and in the product [tex]\( H_2O \)[/tex] (2 H atoms).
- Oxygen (O): 3 atoms from [tex]\( NO_3^- \)[/tex] on the left, with 1 oxygen in [tex]\( NO_2^- \)[/tex] and another 1 in [tex]\( H_2O \)[/tex] on the right.
- Charge:
- Left side: No net charge because [tex]\( Cu \)[/tex] is neutral, [tex]\( NO_3^- \)[/tex] contributes -1, and [tex]\( 2H^+ \)[/tex] contributes +2, making the total charge +1.
- Right side: [tex]\( Cu^{2+} \)[/tex] contributes +2, [tex]\( NO_2^- \)[/tex] contributes -1, making the total charge also +1.
Both mass and charge are balanced.
### Final Balanced Equation:
[tex]\[ Cu + NO_3^- + 2H^+ \rightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
Therefore, the final balanced equation you proposed is:
[tex]\[ Cu + NO_3^- + 2H^+ \longrightarrow Cu^{2+} + NO_2^- + H_2O \][/tex]
