Using electrolytic method, the cost of production of 5L of oxygen at STP, is Rs X, the cost of production of same volume of hydrogen at STP, will be

Options

A. `2C`
B. `(X)/(2)`
C. `8X`
D. `(X)/(8)`

Answer

To solve the problem of determining the cost of producing hydrogen when the cost of producing 5 liters of oxygen at STP is Rs. X, we can follow these steps:

### Step 1: Understand the Electrolysis Process
The electrolysis of water produces oxygen and hydrogen. The half-reactions involved are:
- **Oxidation half-reaction**: \( 2H_2O \rightarrow O_2 + 4e^- + 4H^+ \)
- **Reduction half-reaction**: \( 2H^+ + 2e^- \rightarrow H_2 \)

### Step 2: Calculate Moles of Oxygen Produced
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, to find the number of moles of oxygen produced from 5 liters:
\[ \text{Moles of } O_2 = \frac{5 \text{ L}}{22.4 \text{ L/mol}} \approx 0.223 \text{ moles} \]

### Step 3: Calculate Electricity Required for Oxygen Production
From the oxidation half-reaction, we know that to produce 1 mole of \( O_2 \), 4 Faraday's of electricity are required. Therefore, for \( n \) moles of \( O_2 \):
\[ \text{Electricity required for } n \text{ moles of } O_2 = 4n \text{ Faraday's} \]
For \( 0.223 \) moles of \( O_2 \):
\[ \text{Electricity required} = 4 \times 0.223 \approx 0.892 \text{ Faraday's} \]

### Step 4: Calculate Moles of Hydrogen Produced
From the reduction half-reaction, we know that to produce 1 mole of \( H_2 \), 2 Faraday's of electricity are required. Thus, for \( n \) moles of \( H_2 \):
\[ \text{Electricity required for } n \text{ moles of } H_2 = 2n \text{ Faraday's} \]
Since we are producing the same volume (and thus the same number of moles) of hydrogen, we will also have \( 0.223 \) moles of \( H_2 \):
\[ \text{Electricity required} = 2 \times 0.223 \approx 0.446 \text{ Faraday's} \]

### Step 5: Relate Cost of Production
The cost of producing oxygen is given as Rs. X for 5 liters (or \( 0.223 \) moles). The cost per mole of \( O_2 \) is:
\[ \text{Cost per mole of } O_2 = \frac{X}{0.223} \]
For hydrogen, since it requires less electricity, we can calculate the cost:
\[ \text{Cost for } H_2 = \text{Cost per mole of } O_2 \times \text{Electricity required for } H_2 \]
This gives us:
\[ \text{Cost for } H_2 = \frac{X}{0.223} \times 0.446 \]
Simplifying this, we find:
\[ \text{Cost for } H_2 = \frac{X \times 0.446}{0.223} = 2X \]

### Step 6: Conclusion
Thus, the cost of production of the same volume of hydrogen at STP is Rs. \( \frac{X}{2} \).

### Final Answer
The cost of production of the same volume of hydrogen at STP will be Rs. \( \frac{X}{2} \).

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