The number of electrons involved in the reduction of nitrate `(NO_(3)^(ө))` to hydrazine `(N_(2)H_(4))` is

Options

A. `8`
B. `7`
C. `3`
D. `5`

Explanation

To determine the number of electrons involved in the reduction of nitrate (NO₃⁻) to hydrazine (N₂H₄), we can follow these steps:

### Step 1: Write the unbalanced reaction
The first step is to write the unbalanced equation for the reduction of nitrate to hydrazine:
\[ 2 \text{NO}_3^- \rightarrow \text{N}_2\text{H}_4 \]

### Step 2: Balance the nitrogen atoms
In the product (N₂H₄), there are 2 nitrogen atoms. Therefore, we need 2 nitrate ions to balance the nitrogen:
\[ 2 \text{NO}_3^- \rightarrow \text{N}_2\text{H}_4 \]

### Step 3: Balance the oxygen atoms
On the left side, we have 6 oxygen atoms (from 2 NO₃⁻). To balance the oxygen, we need to add 6 water molecules (H₂O) to the right side:
\[ 2 \text{NO}_3^- \rightarrow \text{N}_2\text{H}_4 + 6 \text{H}_2\text{O} \]

### Step 4: Balance the hydrogen atoms
Now, we have 12 hydrogen atoms from 6 water molecules on the right side. To balance the hydrogen, we need to add 16 hydrogen ions (H⁺) to the left side:
\[ 2 \text{NO}_3^- + 16 \text{H}^+ \rightarrow \text{N}_2\text{H}_4 + 6 \text{H}_2\text{O} \]

### Step 5: Balance the charge
Now, we need to balance the charge. The left side has a total charge of +14 (from 16 H⁺ and 2 NO₃⁻ which is -2), and the right side is neutral. To balance the charge, we need to add electrons (e⁻):
\[ 2 \text{NO}_3^- + 16 \text{H}^+ + 14 \text{e}^- \rightarrow \text{N}_2\text{H}_4 + 6 \text{H}_2\text{O} \]

### Step 6: Calculate the number of electrons for one nitrate ion
Since the balanced equation shows that 2 nitrate ions are reduced, the number of electrons involved in the reduction of one nitrate ion is:
\[ \text{Electrons for one NO}_3^- = \frac{14 \text{ e}^-}{2} = 7 \text{ e}^- \]

### Final Answer
The number of electrons involved in the reduction of nitrate (NO₃⁻) to hydrazine (N₂H₄) is **7 electrons**.

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