`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.
What is the current efficiency when 100 Amp current is passed for 965 sec, in order to produce 17 gm of `H_(2)O_(2)`.

Options

A. `20%`
B. `40%`
C. `80%`
D. `100%`

Answer

To solve the problem, we need to calculate the current efficiency when producing hydrogen peroxide (H₂O₂) using the provided reactions and data. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reactions
We have the following reactions occurring in the electrolytic cell:
1. **Dissociation of Ammonium Hydrogen Sulfate:**
\[ NH_4HSO_4 \rightarrow NH_4^+ + H^+ + SO_4^{2-} \]
2. **Oxidation at the Anode:**
\[ 2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^- \]
3. **Reduction at the Cathode:**
\[ 2H^+ + 2e^- \rightarrow H_2 \]
4. **Hydrolysis of Ammonium Persulfate:**
\[ (NH_4)_2S_2O_8 + 2H_2O \rightarrow 2NH_4HSO_4 + H_2O_2 \]

### Step 2: Calculate the Equivalent Mass of H₂O₂
The equivalent mass of H₂O₂ can be calculated using its molar mass. The molar mass of H₂O₂ is approximately 34 g/mol. Since H₂O₂ has a basicity of 2 (it can donate 2 electrons), the equivalent mass is:
\[ \text{Equivalent mass of H}_2O_2 = \frac{34 \, \text{g/mol}}{2} = 17 \, \text{g/equiv} \]

### Step 3: Calculate the Theoretical Current (I)
Using Faraday's law of electrolysis, we can relate the deposited mass of the substance to the current and time:
\[ \frac{\text{Deposited mass}}{\text{Equivalent mass}} = \frac{I \cdot t}{F} \]
Where:
- Deposited mass = 17 g (mass of H₂O₂ produced)
- Equivalent mass = 17 g/equiv (calculated above)
- \(t\) = 965 seconds (time the current is passed)
- \(F\) = 96500 C/mol (Faraday's constant)

Substituting the values into the equation:
\[ \frac{17 \, \text{g}}{17 \, \text{g/equiv}} = \frac{I \cdot 965 \, \text{s}}{96500 \, \text{C/mol}} \]
This simplifies to:
\[ 1 = \frac{I \cdot 965}{96500} \]
Rearranging gives:
\[ I = \frac{96500}{965} = 100 \, \text{A} \]

### Step 4: Calculate Current Efficiency
Current efficiency is defined as:
\[ \text{Current Efficiency} = \left(\frac{\text{Theoretical Current}}{\text{Practical Current}}\right) \times 100 \]
From our calculations:
- Theoretical Current (I) = 100 A
- Practical Current = 100 A (given in the problem)

Substituting these values:
\[ \text{Current Efficiency} = \left(\frac{100}{100}\right) \times 100 = 100\% \]

### Final Answer
The current efficiency when 100 Amp current is passed for 965 seconds to produce 17 g of H₂O₂ is **100%**.