For the production of X L `H_2` at STP at cathode, cost of electricity is x then cost of production of X L `O_2` at STP at anode from water will be

Options

A. x
B. `x/2`
C. 2x
D. 4x

Answer

To solve the problem, we need to understand the electrolysis of water and the relationship between the volumes of hydrogen and oxygen produced.

### Step-by-Step Solution:

1. **Understanding Electrolysis of Water**:
- When water (H₂O) undergoes electrolysis, it splits into hydrogen (H₂) and oxygen (O₂).
- The balanced chemical equation for the electrolysis of water is:
\[ 2H_2O \rightarrow 2H_2 + O_2 \]
- This indicates that for every 2 moles of water, 2 moles of hydrogen and 1 mole of oxygen are produced.

2. **Volume Ratio**:
- According to Avogadro's law, at STP (Standard Temperature and Pressure), equal volumes of gases contain equal numbers of molecules.
- Therefore, the volume ratio of hydrogen to oxygen produced during electrolysis is:
\[ V_{H_2} : V_{O_2} = 2 : 1 \]
- This means for every 2 liters of hydrogen produced, 1 liter of oxygen is produced.

3. **Given Data**:
- We are given that the production of X liters of hydrogen (H₂) requires a cost of electricity of X.
- From the volume ratio, if X liters of H₂ are produced, then the volume of O₂ produced will be:
\[ V_{O_2} = \frac{X}{2} \text{ liters} \]

4. **Cost Calculation**:
- The cost of producing X liters of H₂ is X.
- Since the volume of O₂ produced is half that of H₂, we can use a unitary method to find the cost of producing O₂.
- If \(\frac{X}{2}\) liters of O₂ costs X, then:
\[ \text{Cost for } X \text{ liters of } O_2 = 2 \times X = 2X \]

5. **Conclusion**:
- Therefore, the cost of production of X liters of O₂ at STP from water will be \(2X\).

### Final Answer:
The cost of production of X liters of O₂ at STP will be \(2X\).

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