In the resction `2NH_(4)^(+)+6NO_(3)^(-)(aq)+4H^(+)(aq)rarr 6NO_(2)(g)+N_(2)(g)+6H_(2)O` the reducing agent is

Options

A. `NH_(4)^(+)`
B. `NO_(3)^(-)`
C. `H^(+)(aq)`
D. `NO_(2)`

Explanation

To determine the reducing agent in the reaction

\[ 2NH_4^+ + 6NO_3^- + 4H^+ \rightarrow 6NO_2 + N_2 + 6H_2O, \]

we will follow these steps:

### Step 1: Identify the oxidation states of nitrogen in the reactants.
- In \( NH_4^+ \):
- Let the oxidation state of nitrogen be \( x \).
- The four hydrogens contribute \( +4 \) (each hydrogen is \( +1 \)).
- The total charge of the ion is \( +1 \).
- Therefore, we have:
\[ x + 4 = +1 \implies x = -3. \]
- So, the oxidation state of nitrogen in \( NH_4^+ \) is \( -3 \).

- In \( NO_3^- \):
- Let the oxidation state of nitrogen be \( y \).
- The three oxygens contribute \( -6 \) (each oxygen is \( -2 \)).
- The total charge of the ion is \( -1 \).
- Therefore, we have:
\[ y - 6 = -1 \implies y = +5. \]
- So, the oxidation state of nitrogen in \( NO_3^- \) is \( +5 \).

### Step 2: Identify the oxidation states of nitrogen in the products.
- In \( NO_2 \):
- Let the oxidation state of nitrogen be \( z \).
- The two oxygens contribute \( -4 \).
- Therefore, we have:
\[ z - 4 = 0 \implies z = +4. \]
- So, the oxidation state of nitrogen in \( NO_2 \) is \( +4 \).

- In \( N_2 \):
- In elemental form, nitrogen has an oxidation state of \( 0 \).

### Step 3: Determine the changes in oxidation states.
- In \( NH_4^+ \), nitrogen changes from \( -3 \) to \( 0 \) (oxidation).
- In \( NO_3^- \), nitrogen changes from \( +5 \) to \( +4 \) (reduction).

### Step 4: Identify the reducing agent.
- The reducing agent is the species that gets oxidized while reducing another species. Here, \( NH_4^+ \) is oxidized from \( -3 \) to \( 0 \) and reduces \( NO_3^- \) from \( +5 \) to \( +4 \).

### Conclusion:
Thus, the reducing agent in the reaction is \( NH_4^+ \).