Equivalent weight of oxidising agent will be -
`2H_(2) + O_(2) rarr 2H_(2)O`

Options

A. 1
B. 32
C. 2
D. 8

Answer

To find the equivalent weight of the oxidizing agent in the reaction \(2H_2 + O_2 \rightarrow 2H_2O\), we can follow these steps:

### Step 1: Identify the Oxidizing Agent
In the given reaction, we need to identify which species is being reduced. The oxidation states of the elements involved are:
- Hydrogen (H) in \(H_2\) is in the 0 oxidation state.
- Oxygen (O) in \(O_2\) is in the 0 oxidation state.
- In water (\(H_2O\)), hydrogen is in the +1 oxidation state and oxygen is in the -2 oxidation state.

### Step 2: Determine Changes in Oxidation States
- Hydrogen (H) goes from 0 in \(H_2\) to +1 in \(H_2O\), indicating it is being oxidized.
- Oxygen (O) goes from 0 in \(O_2\) to -2 in \(H_2O\), indicating it is being reduced.

Since oxygen is being reduced, it is the oxidizing agent.

### Step 3: Calculate the Change in Oxidation Number
The change in oxidation number for oxygen is:
- From 0 to -2, which is a change of 2 units per oxygen atom.
- Since there are 2 oxygen atoms in \(O_2\), the total change is \(2 \times 2 = 4\).

### Step 4: Determine the n-factor
The n-factor for the oxidizing agent (O2) is the total number of electrons gained during the reduction process. Here, oxygen gains 4 electrons in total.

### Step 5: Calculate the Molecular Weight of O2
The molecular weight of \(O_2\) is calculated as follows:
- Atomic weight of oxygen (O) = 16 g/mol.
- Therefore, molecular weight of \(O_2\) = \(16 \times 2 = 32\) g/mol.

### Step 6: Calculate the Equivalent Weight
The equivalent weight (EW) of an oxidizing agent can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent Weight of } O_2 = \frac{32}{4} = 8 \text{ g/equiv} \]

### Final Answer
The equivalent weight of the oxidizing agent \(O_2\) is **8 g/equiv**.

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