In a half reaction, nitrate is reduced by `6e^(-)` reduction to x as follows
`7H^(+)+NO_(3)^(-)+6e^(-)rarr 2H_(2)O+x`. The 'x' in the reaction is

Options

A. `NO`
B. `NH_(2)NH_(2)`
C. `NH_(3)`
D. `NH_(2)OH`

Answer

To solve the given half-reaction and determine the value of 'x', we will follow these steps:

### Step 1: Write the half-reaction
The half-reaction provided is:
\[ 7H^+ + NO_3^- + 6e^- \rightarrow 2H_2O + x \]

### Step 2: Identify the components
- On the left side, we have:
- 7 protons (H⁺)
- 1 nitrate ion (NO₃⁻)
- 6 electrons (e⁻)

- On the right side, we have:
- 2 water molecules (2H₂O)
- The unknown 'x'

### Step 3: Analyze the options for 'x'
We need to check the possible options for 'x' to see which one balances the equation.

#### Option 1: NO
- Nitrogen: 1 on both sides (balanced)
- Hydrogen: 7 on the left, 4 on the right (unbalanced)
- Conclusion: Incorrect.

#### Option 2: NH₂NH₂ (Hydrazine)
- Nitrogen: 2 on the right (unbalanced)
- Conclusion: Incorrect.

#### Option 3: NH₃ (Ammonia)
- Nitrogen: 1 on both sides (balanced)
- Hydrogen: 7 on the left, 4 (from 2H₂O) + 3 (from NH₃) = 7 on the right (balanced)
- Oxygen: 3 on the left (from NO₃⁻), 2 (from 2H₂O) on the right (unbalanced)
- Conclusion: Incorrect.

#### Option 4: NH₂OH (Hydroxylamine)
- Nitrogen: 1 on both sides (balanced)
- Hydrogen: 7 on the left, 4 (from 2H₂O) + 3 (from NH₂OH) = 7 on the right (balanced)
- Oxygen: 3 on the left (from NO₃⁻), 2 (from 2H₂O) + 1 (from NH₂OH) = 3 on the right (balanced)
- Conclusion: Correct.

### Step 4: Final Answer
The value of 'x' in the half-reaction is:
\[ x = NH_2OH \]