Hydrogen peroxide can be prepared by successive reaction:
`2NH_(4)HSO_(4) rarr H_(2)+(NH_(4))_(2)S_(2)O_(8)`
`(NH_(4))_(2)S_(2)O_(8)+2H_(2)O rarr 2NH_(4)HSO_(4)+H_(2)O_(2)`
The first reaction is an electrolytic reaction the second is steam distillation. what amount ofcurrent would have to be used in first reaction to produce enough intermediate to yield 100g pure `H_(2)O_(2)` per hour ? Assume `50%` anode current efficiency.

Explanation

To solve the problem of determining the amount of current required in the first reaction to produce enough intermediate to yield 100 g of pure H₂O₂ per hour, we will use Faraday’s laws of electrolysis. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the equivalent weight of H₂O₂
The molecular weight of hydrogen peroxide (H₂O₂) is 34 g/mol. Since it can release two moles of electrons during its reduction, the equivalent weight (E) can be calculated as:
\[ E = \frac{\text{Molecular weight}}{n} = \frac{34 \text{ g/mol}}{2} = 17 \text{ g/equiv} \]

### Step 2: Use Faraday's law of electrolysis
According to Faraday's law, the weight of a substance produced or consumed at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The formula is:
\[ \text{Weight} = \frac{I \cdot T}{96500} \cdot E \]
Where:
- Weight = amount of substance produced (in grams)
- I = current (in amperes)
- T = time (in seconds)
- 96500 = Faraday's constant (coulombs per equivalent)

### Step 3: Set up the equation for the required weight of H₂O₂
We need to produce 100 g of H₂O₂ per hour. First, convert the time from hours to seconds:
\[ T = 1 \text{ hour} = 3600 \text{ seconds} \]

Now, substituting the known values into Faraday's law:
\[ 100 = \frac{I \cdot 3600}{96500} \cdot 17 \]

### Step 4: Solve for current (I)
Rearranging the equation to isolate I:
\[ I = \frac{100 \cdot 96500}{3600 \cdot 17} \]

Calculating the right side:
\[ I = \frac{100 \cdot 96500}{61200} = \frac{9650000}{61200} \approx 157.69 \text{ A} \]

### Step 5: Adjust for current efficiency
Given that the anode current efficiency is 50%, we need to adjust the current calculated:
\[ I_{\text{actual}} = \frac{I}{\text{Efficiency}} = \frac{157.69}{0.5} = 315.36 \text{ A} \]

### Final Answer:
The amount of current required to produce enough intermediate to yield 100 g of pure H₂O₂ per hour, considering 50% anode current efficiency, is approximately **315.36 amperes**.

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