Balance the following redox reactions.
Coppoer reacts with nitric acid, a brown, gas is formed and solution turns blue.
`Cu+NO_(3)^(ө)rarrNO_(2)+Cu^(2+)`

Explanation

To balance the redox reaction of copper reacting with nitric acid, we will follow these steps:

### Step 1: Write the skeletal equation
The skeletal equation provided is:
\[ \text{Cu} + \text{NO}_3^- \rightarrow \text{NO}_2 + \text{Cu}^{2+} \]

### Step 2: Identify oxidation and reduction half-reactions
1. **Oxidation half-reaction**: Copper (Cu) is oxidized to copper ions (Cu²⁺).
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

2. **Reduction half-reaction**: Nitrate ion (NO₃⁻) is reduced to nitrogen dioxide (NO₂).
\[ \text{NO}_3^- + e^- \rightarrow \text{NO}_2 \]

### Step 3: Balance the half-reactions
1. **Oxidation half-reaction**:
- The oxidation half-reaction is already balanced in terms of atoms and charge:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

2. **Reduction half-reaction**:
- The reduction half-reaction needs to balance oxygen and hydrogen.
- We have 3 oxygen on the left and 2 on the right, so we add 1 water molecule to the right:
\[ \text{NO}_3^- + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \]
- Now, we have 2 hydrogen on the right, so we add 2 H⁺ ions to the left:
\[ 2\text{H}^+ + \text{NO}_3^- + e^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \]

### Step 4: Equalize the number of electrons transferred
- The oxidation half-reaction releases 2 electrons, while the reduction half-reaction consumes only 1 electron. To equalize, we multiply the reduction half-reaction by 2:
\[ 2\text{H}^+ + 2\text{NO}_3^- + 2e^- \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O} \]

### Step 5: Combine the half-reactions
Now we can add the two half-reactions together:
1. From the oxidation half-reaction:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

2. From the modified reduction half-reaction:
\[ 2\text{H}^+ + 2\text{NO}_3^- + 2e^- \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O} \]

Combining these gives:
\[ \text{Cu} + 2\text{H}^+ + 2\text{NO}_3^- \rightarrow \text{Cu}^{2+} + 2\text{NO}_2 + 2\text{H}_2\text{O} \]

### Step 6: Final balanced equation
The final balanced redox reaction is:
\[ \text{Cu} + 4\text{H}^+ + 2\text{NO}_3^- \rightarrow \text{Cu}^{2+} + 2\text{NO}_2 + 2\text{H}_2\text{O} \]