What will be the volume of `O_(2)` Liberated at NTP by passing 5 A current For 193 sec. through acidified water.

Options

A. 56 mL
B. 112 mL
C. 158 mL
D. 965 mL

Answer

To solve the problem of finding the volume of \( O_2 \) liberated at NTP by passing a 5 A current for 193 seconds through acidified water, we can follow these steps:

### Step 1: Understand the electrolysis of water
During the electrolysis of water, oxygen is produced at the anode. The balanced half-reaction for the oxidation of water to produce oxygen is:
\[ 2 H_2O \rightarrow O_2 + 4 H^+ + 4 e^- \]
From this equation, we can see that 1 mole of \( O_2 \) is produced for every 4 moles of electrons.

### Step 2: Calculate the total charge (Q) passed through the solution
Using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)

Substituting the values:
\[ Q = 5 \, \text{A} \times 193 \, \text{s} = 965 \, \text{C} \]

### Step 3: Calculate the number of moles of electrons (n)
Using Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \), we can find the number of moles of electrons:
\[ n = \frac{Q}{F} = \frac{965 \, \text{C}}{96500 \, \text{C/mol}} = 0.01 \, \text{mol} \]

### Step 4: Relate moles of electrons to moles of \( O_2 \)
From the balanced equation, we know that 4 moles of electrons produce 1 mole of \( O_2 \). Therefore, the number of moles of \( O_2 \) produced can be calculated as:
\[ \text{Moles of } O_2 = \frac{n}{4} = \frac{0.01}{4} = 0.0025 \, \text{mol} \]

### Step 5: Calculate the volume of \( O_2 \) at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Thus, the volume of \( O_2 \) produced can be calculated as:
\[ \text{Volume} = \text{Moles of } O_2 \times 22.4 \, \text{L/mol} = 0.0025 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.056 \, \text{L} \]

### Final Answer
The volume of \( O_2 \) liberated at NTP is:
\[ 0.056 \, \text{L} \text{ or } 56 \, \text{mL} \]