The reaction `H_(2)S+H_(2)O_(2)toS+2H_(2)O`manifests

Options

A. acidic nature of `H_(2)O_(2)`
B. alkaline nature of `H_(2)O_(2)`
C. oxidising action of `H_(2)O_(2)`
D. reducing action of `H_(2)O_(2)`

Answer

To analyze the reaction \( H_2S + H_2O_2 \rightarrow S + 2H_2O \) and determine what it manifests, we can follow these steps:

### Step 1: Write the balanced equation
The given reaction is already balanced:
\[ H_2S + H_2O_2 \rightarrow S + 2H_2O \]

### Step 2: Determine the oxidation states of each element
- In \( H_2S \):
- Hydrogen (H) has an oxidation state of +1.
- Let the oxidation state of sulfur (S) be \( x \). The equation becomes:
\[ 2(+1) + x = 0 \implies x = -2 \]
- Therefore, the oxidation state of sulfur in \( H_2S \) is -2.

- In \( H_2O_2 \):
- Hydrogen (H) has an oxidation state of +1.
- Let the oxidation state of oxygen (O) be \( y \). The equation becomes:
\[ 2(+1) + 2y = 0 \implies 2y = -2 \implies y = -1 \]
- Therefore, the oxidation state of oxygen in \( H_2O_2 \) is -1.

- In the products:
- In elemental sulfur (S), the oxidation state is 0.
- In water (H2O):
- Hydrogen (H) is +1, and oxygen (O) is -2.

### Step 3: Analyze the changes in oxidation states
- For sulfur:
- It changes from -2 in \( H_2S \) to 0 in elemental sulfur (S). This is an increase in oxidation state (oxidation).

- For oxygen:
- It changes from -1 in \( H_2O_2 \) to -2 in water (H2O). This is a decrease in oxidation state (reduction).

### Step 4: Identify the type of reaction
- Since \( H_2S \) is oxidized (its oxidation state increases) and \( H_2O_2 \) is reduced (its oxidation state decreases), \( H_2O_2 \) acts as an oxidizing agent in this reaction.

### Conclusion
Thus, the reaction \( H_2S + H_2O_2 \rightarrow S + 2H_2O \) manifests the oxidizing action of \( H_2O_2 \).

### Final Answer
The reaction manifests the **oxidizing action of \( H_2O_2 \)**.

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