Substances that are oxidized and reduced in the following reaction are respectively.
`N_(2)H_(4)(l)+2H_(2)O_((l))rarrN_(2(g))+4H_(2)O_((l))`

Options

A. `N_(2)H_(4), H_(2)O`
B. `N_(2)H_(4), H_(2)O_(2)`
C. `N_(2), H_(2)O_(2)`
D. `H_(2)O, N_(2)`

Answer

To determine which substances are oxidized and reduced in the reaction:

\[ \text{N}_2\text{H}_4(l) + 2\text{H}_2\text{O}_2(l) \rightarrow \text{N}_2(g) + 4\text{H}_2\text{O}(l) \]

we will follow these steps:

### Step 1: Assign Oxidation States
- For **N2H4** (hydrazine):
- Let the oxidation state of nitrogen be \( x \).
- The equation for the oxidation state of nitrogen in N2H4 is:
\[ 2x + 4(+1) = 0 \quad \text{(since there are 4 hydrogen atoms)} \]
\[ 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \]
- Thus, the oxidation state of nitrogen in N2H4 is **-2**.

- For **H2O2** (hydrogen peroxide):
- The oxidation state of hydrogen is +1.
- Let the oxidation state of oxygen be \( y \). The equation is:
\[ 2(+1) + 2y = 0 \quad \text{(since there are 2 oxygen atoms)} \]
\[ 2 + 2y = 0 \implies 2y = -2 \implies y = -1 \]
- Thus, the oxidation state of oxygen in H2O2 is **-1**.

### Step 2: Determine Oxidation States in Products
- For **N2** (nitrogen gas):
- The oxidation state of nitrogen in its elemental form is **0**.

- For **H2O** (water):
- The oxidation state of hydrogen is +1 and for oxygen is -2.

### Step 3: Identify Changes in Oxidation States
- **N2H4**: Nitrogen changes from -2 to 0 (oxidation).
- **H2O2**: Oxygen changes from -1 to -2 (reduction).

### Step 4: Conclusion
- The substance that is oxidized is **N2H4** (hydrazine).
- The substance that is reduced is **H2O2** (hydrogen peroxide).

### Final Answer:
- Substances that are oxidized and reduced in the reaction are respectively:
- **Oxidized**: N2H4
- **Reduced**: H2O2

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