`H_(2)O_(2)` can be produced by the Ammonium hydrogen sulphate. Reactions occuring in electrolytic cell.
`NH_(4)HSO_(4) rarr NH_(4)^(+) +H^(+) +SO_(4)^(2-)`
`2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-)` (Anode)
`2H^(+) +2e^(-) rarr H_(2)` (cathode)
`(NH_(4))_(2)S_(2)O_(8) +2H_(2)O rarr 2NH_(4)HSO_(4) +H_(2)O_(2)`
Hydrolysis of Ammonium persulphate
Assume `100%` yield of hydrolysis reaction.
How many moles of electrons are to be passed in order to produce enough `H_(2)O_(2)` which when reacted with excess of `KI` then liberated iodine required 100 ml of centimolar Hypo solution.

Options

A. `10^(-1)`
B. `10^(-2)`
C. `10^(-3)`
D. `5 xx 10^(-4)`

Answer

To solve the problem, we need to determine how many moles of electrons are required to produce enough \( H_2O_2 \) that will react with excess \( KI \) to liberate iodine, which in turn will react with a given volume of centimolar hypo solution.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction producing \( H_2O_2 \) is:
\[ (NH_4)_2S_2O_8 + 2H_2O \rightarrow 2NH_4HSO_4 + H_2O_2 \]
- From the electrolysis reactions, we know:
- At the anode:
\[ 2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^- \]
- At the cathode:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
- Each mole of \( H_2O_2 \) produced requires 2 moles of electrons.

2. **Determine the Amount of \( H_2O_2 \) Required**:
- The \( H_2O_2 \) will react with \( KI \) to liberate iodine (\( I_2 \)):
\[ H_2O_2 + 2KI \rightarrow 2K^+ + 2I^- + 2H^+ \]
- This means 1 mole of \( H_2O_2 \) produces 1 mole of \( I_2 \).

3. **Relating \( I_2 \) to Hypo Solution**:
- The liberated iodine (\( I_2 \)) reacts with hypo solution (\( Na_2S_2O_4 \)):
\[ I_2 + 2Na_2S_2O_4 \rightarrow 2NaI + Na_2S_4O_6 \]
- This indicates that 1 mole of \( I_2 \) reacts with 2 moles of hypo solution.

4. **Calculate Moles of Hypo Solution**:
- Given that we have 100 mL of centimolar hypo solution:
- Centimolar means \( 0.01 \, mol/L \).
- Therefore, the moles of hypo solution in 100 mL (or 0.1 L) is:
\[ \text{Moles of hypo} = 0.01 \, mol/L \times 0.1 \, L = 0.001 \, mol \]

5. **Relate Moles of Hypo to Moles of \( H_2O_2 \)**:
- Since 1 mole of \( I_2 \) requires 2 moles of hypo, we can find the moles of \( I_2 \):
\[ \text{Moles of } I_2 = \frac{0.001 \, mol}{2} = 0.0005 \, mol \]
- Therefore, the moles of \( H_2O_2 \) required (since 1 mole of \( H_2O_2 \) produces 1 mole of \( I_2 \)):
\[ \text{Moles of } H_2O_2 = 0.0005 \, mol \]

6. **Calculate Moles of Electrons**:
- Since 2 moles of electrons are required to produce 1 mole of \( H_2O_2 \):
\[ \text{Moles of electrons} = 2 \times 0.0005 \, mol = 0.001 \, mol \]

### Final Answer:
The number of moles of electrons required is \( 0.001 \, mol \) or \( 10^{-3} \, mol \).