For the redox reaction,`Zn + NO_(3)^(-) to Zn^(2+) + NH_(4)^(+)` In basic medium, coefficients of Zn, `NO_(3)^(-)` and `OH^(-)` in the balanced equation are respectively.

Options

A. 7,4,1
B. 1,4,10
C. 4,1,10
D. 4,1,7

Explanation

To balance the redox reaction \( \text{Zn} + \text{NO}_3^{-} \rightarrow \text{Zn}^{2+} + \text{NH}_4^{+} \) in a basic medium, we can follow these steps:

### Step 1: Assign Oxidation States
- **Zinc (Zn)**: In elemental form, Zn has an oxidation state of 0. In \( \text{Zn}^{2+} \), the oxidation state is +2.
- **Nitrogen in \( \text{NO}_3^{-} \)**: The oxidation state of nitrogen (N) can be calculated as follows:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
- **Nitrogen in \( \text{NH}_4^{+} \)**: The oxidation state of nitrogen can be calculated as follows:
\[ x + 4(+1) = +1 \implies x + 4 = +1 \implies x = -3 \]

### Step 2: Identify Oxidation and Reduction
- **Oxidation**: Zn goes from 0 to +2 (oxidation).
- **Reduction**: N in \( \text{NO}_3^{-} \) goes from +5 to -3 (reduction).

### Step 3: Calculate Changes in Oxidation States
- **Change for Zn**: \( 0 \rightarrow +2 \) (change of +2)
- **Change for N**: \( +5 \rightarrow -3 \) (change of -8)

### Step 4: Balance the Changes
To balance the total changes in oxidation states:
- For Zn, we have a change of +2.
- For N, we have a change of -8.

To balance these, we need to multiply the Zn change by 4:
\[ 4 \times (+2) = +8 \]
Thus, we will have 4 Zn atoms reacting.

### Step 5: Write the Half-Reactions
1. **Oxidation Half-Reaction**:
\[ 4 \text{Zn} \rightarrow 4 \text{Zn}^{2+} + 8e^{-} \]
2. **Reduction Half-Reaction**:
\[ \text{NO}_3^{-} + 8e^{-} \rightarrow \text{NH}_4^{+} \]

### Step 6: Combine the Half-Reactions
Combining the half-reactions gives:
\[ 4 \text{Zn} + \text{NO}_3^{-} \rightarrow 4 \text{Zn}^{2+} + \text{NH}_4^{+} \]

### Step 7: Balance the Charge with OH⁻
In a basic medium, we need to balance the charges:
- Left side: \( 4 \text{Zn} \) (neutral) + \( \text{NO}_3^{-} \) (charge -1) = total charge -1
- Right side: \( 4 \text{Zn}^{2+} \) (charge +8) + \( \text{NH}_4^{+} \) (charge +1) = total charge +9

To balance the charges, we need to add 10 OH⁻ to the left side:
\[ 4 \text{Zn} + \text{NO}_3^{-} + 10 \text{OH}^- \rightarrow 4 \text{Zn}^{2+} + \text{NH}_4^{+} + 10 \text{H}_2\text{O} \]

### Step 8: Balance the Water
Now, we need to balance the water molecules:
- On the left side, we have 10 OH⁻ which corresponds to 10 H atoms.
- On the right side, we have 4 NH₄⁺ which corresponds to 14 H atoms.
Thus, we need to add 7 H₂O to the left side:
\[ 4 \text{Zn} + \text{NO}_3^{-} + 10 \text{OH}^- \rightarrow 4 \text{Zn}^{2+} + \text{NH}_4^{+} + 7 \text{H}_2\text{O} \]

### Final Balanced Equation
The final balanced equation is:
\[ 4 \text{Zn} + \text{NO}_3^{-} + 10 \text{OH}^- \rightarrow 4 \text{Zn}^{2+} + \text{NH}_4^{+} + 7 \text{H}_2\text{O} \]

### Coefficients
- Coefficient of Zn: 4
- Coefficient of \( \text{NO}_3^{-} \): 1
- Coefficient of \( \text{OH}^- \): 10