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 find the volume of \( O_2 \) liberated at Normal Temperature and Pressure (NTP) by passing a current through acidified water, we can follow these steps:

### Step 1: Determine the number of electrons required for the reaction
The electrolysis of water can be represented by the following half-reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
From this equation, we see that to produce 1 mole of \( O_2 \), 4 moles of electrons are required.

### Step 2: Calculate the total charge passed through the solution
The total charge \( Q \) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current (in amperes),
- \( t \) is the time (in seconds).

Given:
- \( I = 5 \, A \)
- \( t = 193 \, s \)

Calculating \( Q \):
\[ Q = 5 \, A \times 193 \, s = 965 \, C \]

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

### Step 4: Calculate the moles of \( O_2 \) produced
Since 4 moles of electrons produce 1 mole of \( O_2 \), we can find the moles of \( O_2 \) produced:
\[ \text{Moles of } O_2 = \frac{n}{4} = \frac{0.01 \, mol}{4} = 0.0025 \, mol \]

### Step 5: Calculate the volume of \( O_2 \) at NTP
At NTP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume \( V \) of \( O_2 \) can be calculated as:
\[ V = \text{Moles of } O_2 \times 22.4 \, L/mol \]
Substituting the values:
\[ V = 0.0025 \, mol \times 22.4 \, L/mol = 0.056 \, L \]

### Step 6: Convert volume to milliliters
To convert liters to milliliters, we multiply by 1000:
\[ V = 0.056 \, L \times 1000 = 56 \, mL \]

### Final Answer
The volume of \( O_2 \) liberated at NTP is **56 mL**.

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