In the reaction,
`4Zn + NO_(3)^(-) + 7H_(2)O rarr 4Zn^(2+) + NH_(4)^(+) + 10OH^(-)`, the substance which gets reduced is `"______________"`.

Options

A. `Zn` is reduced to `Zn^(2+)`
B. `H_(2)O`
C. `NO_(3)^(-)`
D. `NH_(4)^(-)`

Answer

To determine which substance gets reduced in the reaction

\[ 4\text{Zn} + \text{NO}_3^{-} + 7\text{H}_2\text{O} \rightarrow 4\text{Zn}^{2+} + \text{NH}_4^{+} + 10\text{OH}^{-} \]

we need to analyze the oxidation states of the elements involved in the reaction.

### Step 1: Identify the oxidation states of the reactants and products.

1. **Zinc (Zn)**: In its elemental form, the oxidation state of zinc is 0.
2. **Nitrate ion (\( \text{NO}_3^{-} \))**:
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of nitrogen (N) be \( x \).
- The equation for the oxidation state is:
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
- Therefore, the oxidation state of nitrogen in \( \text{NO}_3^{-} \) is +5.
3. **Water (\( \text{H}_2\text{O} \))**:
- Hydrogen (H) has an oxidation state of +1, and oxygen (O) is -2.
4. **Zinc ion (\( \text{Zn}^{2+} \))**: The oxidation state is +2.
5. **Ammonium ion (\( \text{NH}_4^{+} \))**:
- Let the oxidation state of nitrogen be \( y \).
- The equation for the oxidation state is:
\[ y + 4(+1) = +1 \implies y + 4 = +1 \implies y = -3 \]
- Therefore, the oxidation state of nitrogen in \( \text{NH}_4^{+} \) is -3.
6. **Hydroxide ion (\( \text{OH}^{-} \))**:
- Oxygen is -2, and hydrogen is +1.

### Step 2: Determine changes in oxidation states.

- **Zinc (Zn)**: Changes from 0 to +2 (oxidation).
- **Nitrogen in \( \text{NO}_3^{-} \)**: Changes from +5 to -3 (reduction).
- **Nitrogen in \( \text{NH}_4^{+} \)**: Remains at -3.
- **Oxygen and Hydrogen**: Their oxidation states do not change significantly in this context.

### Step 3: Identify the substance that gets reduced.

Reduction is defined as the gain of electrons. In this reaction:
- The nitrogen in \( \text{NO}_3^{-} \) goes from +5 to -3, indicating that it has gained electrons.

### Conclusion:

Thus, the substance that gets reduced in the reaction is:

**"NO3^-"**

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