During electrolysis of `H_2SO_4`(aq) with high charge density, `H_2S_2O_8` formed as by product. In such electrolysis 22.4L `H_2(g)` and 8.4 L `O_2(g)` liberated at 1 atm and 273 K at electrode. The moles of `H_2S_2O_8` formed is :

Options

A. 0.25
B. 0.5
C. 0.75
D. 1

Answer

To find the moles of `H2S2O8` formed during the electrolysis of `H2SO4`, we can follow these steps:

### Step 1: Calculate the moles of `H2` and `O2` produced
Using the ideal gas law, we can calculate the number of moles of hydrogen and oxygen produced from the volumes given.

- Volume of `H2` = 22.4 L
- Volume of `O2` = 8.4 L
- At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L.

**Moles of `H2`:**
\[ \text{Moles of } H2 = \frac{\text{Volume of } H2}{\text{Volume at STP}} = \frac{22.4 \, \text{L}}{22.4 \, \text{L/mol}} = 1 \, \text{mol} \]

**Moles of `O2`:**
\[ \text{Moles of } O2 = \frac{\text{Volume of } O2}{\text{Volume at STP}} = \frac{8.4 \, \text{L}}{22.4 \, \text{L/mol}} = 0.375 \, \text{mol} \]

### Step 2: Write the balanced equation for the electrolysis
The electrolysis of `H2SO4` can be represented as:
\[ 2H2SO4 \rightarrow 2H2 + O2 + H2S2O8 \]

### Step 3: Determine the stoichiometry
From the balanced equation, we can see that:
- 2 moles of `H2` are produced for every mole of `H2S2O8` formed.
- 1 mole of `O2` is produced for every mole of `H2S2O8` formed.

### Step 4: Set up the relationships based on moles produced
Let \( x \) be the moles of `H2S2O8` formed. Then:
- Moles of `H2` produced = \( 2x \)
- Moles of `O2` produced = \( x \)

From the moles calculated:
- For `H2`: \( 2x = 1 \)
- For `O2`: \( x = 0.375 \)

### Step 5: Solve for `x`
From the equation \( 2x = 1 \):
\[ x = \frac{1}{2} = 0.5 \, \text{mol} \]

From the equation \( x = 0.375 \):
This indicates that the limiting reagent is `O2`, and thus we should use this value to find the moles of `H2S2O8`.

### Conclusion
The moles of `H2S2O8` formed is:
\[ \text{Moles of } H2S2O8 = 0.375 \, \text{mol} \]