Passage of 96500 coulomb of electricity liberates …… litre of `O_2` at NTP during electrolysis-

Options

A. 5.6
B. 6.5
C. 22.2
D. 11.2

Answer

To solve the question regarding the liberation of oxygen gas during electrolysis when 96500 coulombs of electricity is passed, we can follow these steps:

### Step 1: Understand the Charge and Faraday's Law
The passage of 96500 coulombs of electricity corresponds to 1 Faraday. According to Faraday's laws of electrolysis, 1 Faraday of charge liberates 1 gram equivalent of a substance.

### Step 2: Determine the Reaction Involved
In the electrolysis of water (H₂O), the reaction can be represented as:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]
From this reaction, we can see that 4 moles of electrons are required to produce 1 mole of O₂.

### Step 3: Calculate the Equivalent Weight of Oxygen
The equivalent weight of oxygen can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
where \( n \) is the number of electrons involved in the reaction. For O₂, the molar mass is 32 g/mol, and since 4 electrons are involved, we have:
\[ \text{Equivalent Weight of O}_2 = \frac{32 \text{ g/mol}}{4} = 8 \text{ g} \]

### Step 4: Relate Charge to Weight Liberated
Using Faraday's second law of electrolysis:
\[ W = \frac{E \times Q}{96500} \]
where \( W \) is the weight of the substance liberated, \( E \) is the equivalent weight, and \( Q \) is the charge (96500 C in this case). Thus:
\[ W = \frac{8 \text{ g} \times 96500 \text{ C}}{96500} = 8 \text{ g} \]

### Step 5: Calculate Moles of O₂
Now, we can calculate the number of moles of O₂ liberated:
\[ \text{Number of moles} = \frac{\text{Weight}}{\text{Molar Mass}} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 6: Convert Moles to Volume at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of O₂ liberated can be calculated as:
\[ \text{Volume} = \text{Number of moles} \times 22.4 \text{ L/mol} = 0.25 \text{ moles} \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

### Final Answer
Thus, the volume of O₂ liberated during the electrolysis when 96500 coulombs of electricity is passed is **5.6 liters**.

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