Answer :
To write a balanced half-reaction for the reduction of the nitrate ion (NO3^-) to nitrogen dioxide (NO2) in a basic aqueous solution, we will follow several systematic steps. Here is the detailed, step-by-step solution:
### Step 1: Identify the Reactants and Products
We begin by identifying the reactant and product in the reduction half-reaction:
- Reactant: Nitrate ion (NO3^-)
- Product: Nitrogen dioxide (NO2)
### Step 2: Write the Unbalanced Half-Reaction
First, we write the unbalanced equation with just the reactants and products:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 \][/tex]
### Step 3: Balance All Atoms Other Than Hydrogen and Oxygen
In this case, we have one nitrogen atom on both sides of the equation, so the nitrogen atoms are already balanced:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 \][/tex]
### Step 4: Balance Oxygen Atoms by Adding Water Molecules
Next, we balance the oxygen atoms. The nitrate ion (NO3^-) has 3 oxygen atoms, and nitrogen dioxide (NO2) has 2 oxygen atoms. To balance the oxygen atoms, we need to add one water molecule (H2O) to the product side:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \][/tex]
### Step 5: Balance Hydrogen Atoms by Adding Hydroxide Ions (OH^-)
Since we are in a basic solution, we balance the hydrogen atoms by adding hydroxide ions (OH^-) on the reactant side. Each water molecule (H2O) has 2 hydrogen atoms, so we add two hydroxide ions (2 OH^-) to the reactant side:
[tex]\[ \text{NO}_3^- + 2 \text{OH}^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \][/tex]
### Step 6: Balance the Charge by Adding Electrons (e^-)
Now, we balance the charges by adding electrons. The left side has a total charge of -3 (nitrate ion: -1, two hydroxides: -2). The right side has a charge of 0 (nitrogen dioxide is neutral, and water is neutral). To balance the charges, we add two electrons (2e^-) to the left side:
[tex]\[ \text{NO}_3^- + 2 \text{H}_2\text{O} + 2e^- \rightarrow \text{NO}_2 + 2 \text{OH}^- \][/tex]
### Step 7: Verify the Balancing
Finally, we verify that both the atoms and the charges are balanced on both sides of the equation.
- Hydrogen atoms: 2 (from 2 OH^-) on the left = 2 (from H2O) on the right
- Oxygen atoms: 3 (from NO3^-) + 2 (from 2 OH^-) on the left = 2 (from NO2) + 1 (from H2O) on the right
- Charge: -1 (from NO3^-) + -2 (from 2 OH^-) + 2e^- = -3 on the left, and -2 (from 2 OH^-) on the right
Thus, the balanced half-reaction for the reduction of nitrate ion to nitrogen dioxide in a basic aqueous solution is:
[tex]\[ \text{NO}_3^- (\text{aq}) + \text{H}_2\text{O} (\text{l}) + 2e^- \rightarrow \text{NO}_2 (\text{g}) + 2\text{OH}^- (\text{aq}) \][/tex]
### Step 1: Identify the Reactants and Products
We begin by identifying the reactant and product in the reduction half-reaction:
- Reactant: Nitrate ion (NO3^-)
- Product: Nitrogen dioxide (NO2)
### Step 2: Write the Unbalanced Half-Reaction
First, we write the unbalanced equation with just the reactants and products:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 \][/tex]
### Step 3: Balance All Atoms Other Than Hydrogen and Oxygen
In this case, we have one nitrogen atom on both sides of the equation, so the nitrogen atoms are already balanced:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 \][/tex]
### Step 4: Balance Oxygen Atoms by Adding Water Molecules
Next, we balance the oxygen atoms. The nitrate ion (NO3^-) has 3 oxygen atoms, and nitrogen dioxide (NO2) has 2 oxygen atoms. To balance the oxygen atoms, we need to add one water molecule (H2O) to the product side:
[tex]\[ \text{NO}_3^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \][/tex]
### Step 5: Balance Hydrogen Atoms by Adding Hydroxide Ions (OH^-)
Since we are in a basic solution, we balance the hydrogen atoms by adding hydroxide ions (OH^-) on the reactant side. Each water molecule (H2O) has 2 hydrogen atoms, so we add two hydroxide ions (2 OH^-) to the reactant side:
[tex]\[ \text{NO}_3^- + 2 \text{OH}^- \rightarrow \text{NO}_2 + \text{H}_2\text{O} \][/tex]
### Step 6: Balance the Charge by Adding Electrons (e^-)
Now, we balance the charges by adding electrons. The left side has a total charge of -3 (nitrate ion: -1, two hydroxides: -2). The right side has a charge of 0 (nitrogen dioxide is neutral, and water is neutral). To balance the charges, we add two electrons (2e^-) to the left side:
[tex]\[ \text{NO}_3^- + 2 \text{H}_2\text{O} + 2e^- \rightarrow \text{NO}_2 + 2 \text{OH}^- \][/tex]
### Step 7: Verify the Balancing
Finally, we verify that both the atoms and the charges are balanced on both sides of the equation.
- Hydrogen atoms: 2 (from 2 OH^-) on the left = 2 (from H2O) on the right
- Oxygen atoms: 3 (from NO3^-) + 2 (from 2 OH^-) on the left = 2 (from NO2) + 1 (from H2O) on the right
- Charge: -1 (from NO3^-) + -2 (from 2 OH^-) + 2e^- = -3 on the left, and -2 (from 2 OH^-) on the right
Thus, the balanced half-reaction for the reduction of nitrate ion to nitrogen dioxide in a basic aqueous solution is:
[tex]\[ \text{NO}_3^- (\text{aq}) + \text{H}_2\text{O} (\text{l}) + 2e^- \rightarrow \text{NO}_2 (\text{g}) + 2\text{OH}^- (\text{aq}) \][/tex]
