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 solve the problem of determining the moles of `H2S2O8` formed during the electrolysis of `H2SO4`, we will follow these steps:

### Step 1: Calculate moles of `H2` and `O2` liberated
Using the ideal gas law, we can calculate the moles of hydrogen and oxygen gas liberated during electrolysis.

- **Volume of `H2` liberated** = 22.4 L
- **Volume of `O2` liberated** = 8.4 L
- **Standard molar volume at STP** = 22.4 L/mol

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

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

### Step 2: Determine the number of equivalents at the anode and cathode
Next, we will find the equivalents of `H2` and `O2` produced, which will help us relate it to the formation of `H2S2O8`.

- **For `H2`**: The reaction at the cathode is:
\[ 2H^+ + 2e^- \rightarrow H2 \]
This means 1 mole of `H2` corresponds to 2 equivalents.

\[ \text{Equivalents of } H2 = 1 \, \text{mol} \times 2 = 2 \, \text{equivalents} \]

- **For `O2`**: The reaction at the anode is:
\[ 2OH^- \rightarrow O2 + 4H^+ + 4e^- \]
This means 0.375 moles of `O2` corresponds to 4 equivalents (since 2 moles of `OH^-` produce 1 mole of `O2`).

\[ \text{Equivalents of } O2 = 0.375 \, \text{mol} \times 4 = 1.5 \, \text{equivalents} \]

### Step 3: Set up the relationship between the equivalents
At the anode, the formation of `H2S2O8` can be represented as:
\[ 2HSO4^- \rightarrow H2S2O8 + 2e^- \]
This means that for every mole of `H2S2O8` formed, 2 equivalents of `HSO4^-` are consumed.

Let \( x \) be the moles of `H2S2O8` formed. The total equivalents at the anode can be expressed as:
\[ \text{Equivalents from } H2 + \text{Equivalents from } O2 = 2 + 1.5 = 3.5 \]
And this must equal the equivalents from `H2S2O8`:
\[ 2x = 3.5 \]

### Step 4: Solve for \( x \)
\[ x = \frac{3.5}{2} = 1.75 \, \text{moles of } H2S2O8 \]

### Final Answer
The moles of `H2S2O8` formed is **1.75 moles**.

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