When the redox reaction
`Cu_(2)O+NO_(3)^(-)toLCu^(++)+NO` is balanced by ion elecrtron method acidic medium.
What is correct co-efficient of `Cu^(++)`

Answer

To balance the redox reaction \( \text{Cu}_2\text{O} + \text{NO}_3^- \rightarrow \text{Cu}^{2+} + \text{NO} \) using the ion-electron method in acidic medium, we will follow these steps:

### Step 1: Identify the oxidation and reduction half-reactions
- **Oxidation half-reaction**: \( \text{Cu}_2\text{O} \) is oxidized to \( \text{Cu}^{2+} \).
- **Reduction half-reaction**: \( \text{NO}_3^- \) is reduced to \( \text{NO} \).

### Step 2: Write the oxidation half-reaction
The oxidation half-reaction involves converting \( \text{Cu}_2\text{O} \) to \( \text{Cu}^{2+} \):
\[ \text{Cu}_2\text{O} \rightarrow 2 \text{Cu}^{2+} + \text{H}_2\text{O} \]
Now, we need to balance the oxygen and hydrogen:
- There is 1 oxygen in \( \text{Cu}_2\text{O} \) and 1 in \( \text{H}_2\text{O} \), so they are balanced.
- We have 2 copper on the left and 2 copper on the right.

### Step 3: Balance the hydrogen ions
To balance the hydrogen, we add \( 2 \text{H}^+ \):
\[ \text{Cu}_2\text{O} + 2 \text{H}^+ \rightarrow 2 \text{Cu}^{2+} + \text{H}_2\text{O} \]

### Step 4: Balance the charges
On the left, we have a total charge of \( +2 \) from \( 2 \text{H}^+ \) and on the right, we have \( 2 \times (+2) = +4 \) from \( 2 \text{Cu}^{2+} \). To balance the charges, we add 2 electrons to the right:
\[ \text{Cu}_2\text{O} + 2 \text{H}^+ \rightarrow 2 \text{Cu}^{2+} + \text{H}_2\text{O} + 2 e^- \]

### Step 5: Write the reduction half-reaction
The reduction half-reaction involves converting \( \text{NO}_3^- \) to \( \text{NO} \):
\[ \text{NO}_3^- \rightarrow \text{NO} + 2 \text{H}_2\text{O} \]
Now, we need to balance the hydrogen and oxygen:
- Add \( 4 \text{H}^+ \) to the left:
\[ \text{NO}_3^- + 4 \text{H}^+ \rightarrow \text{NO} + 2 \text{H}_2\text{O} \]

### Step 6: Balance the charges
The left side has a total charge of \( +4 - 1 = +3 \) and the right side has \( +0 \). To balance the charges, we add 3 electrons to the left:
\[ \text{NO}_3^- + 4 \text{H}^+ + 3 e^- \rightarrow \text{NO} + 2 \text{H}_2\text{O} \]

### Step 7: Equalize the number of electrons
The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 3 electrons. To equalize, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
- Oxidation:
\[ 3 \text{Cu}_2\text{O} + 6 \text{H}^+ \rightarrow 6 \text{Cu}^{2+} + 3 \text{H}_2\text{O} + 6 e^- \]
- Reduction:
\[ 2 \text{NO}_3^- + 8 \text{H}^+ + 6 e^- \rightarrow 2 \text{NO} + 4 \text{H}_2\text{O} \]

### Step 8: Combine the half-reactions
Now, we can add the two half-reactions:
\[ 3 \text{Cu}_2\text{O} + 2 \text{NO}_3^- + 14 \text{H}^+ \rightarrow 6 \text{Cu}^{2+} + 2 \text{NO} + 7 \text{H}_2\text{O} \]

### Conclusion
The coefficient of \( \text{Cu}^{2+} \) in the balanced equation is **6**.

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