During electrolysis of acidified water, `O_(2)` gas is formed at the anode . To produce `O_(2)` gas at the anode at the rate of `0.224mL` per second at `STP`, current passed is

Options

A. `0.224A`
B. `2.24A`
C. `9.65A`
D. `3.86A`

Explanation

To solve the problem of calculating the current required to produce \( O_2 \) gas at the anode during the electrolysis of acidified water, we can follow these steps:

### Step 1: Determine the volume of \( O_2 \) produced
Given that \( O_2 \) gas is produced at a rate of \( 0.224 \, \text{mL/s} \).

### Step 2: Convert the volume of \( O_2 \) to moles
At Standard Temperature and Pressure (STP), 1 mole of gas occupies \( 22400 \, \text{mL} \). Therefore, we can find the number of moles of \( O_2 \) produced per second:
\[ \text{Moles of } O_2 = \frac{0.224 \, \text{mL}}{22400 \, \text{mL/mol}} = 1 \times 10^{-5} \, \text{mol} \]

### Step 3: Calculate the mass of \( O_2 \) produced
Using the molar mass of \( O_2 \) (which is \( 32 \, \text{g/mol} \)):
\[ \text{Mass of } O_2 = \text{Moles} \times \text{Molar Mass} = 1 \times 10^{-5} \, \text{mol} \times 32 \, \text{g/mol} = 3.2 \times 10^{-4} \, \text{g} \]

### Step 4: Use the electrochemical equivalent formula
The formula relating weight (W), electrochemical equivalent (Z), current (I), and time (T) is:
\[ W = Z \cdot I \cdot T \]
Rearranging for current (I):
\[ I = \frac{W}{Z \cdot T} \]

### Step 5: Determine the electrochemical equivalent (Z)
For the electrolysis of water, \( O_2 \) is produced by the reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
This indicates that 4 moles of electrons are required to produce 1 mole of \( O_2 \). The electrochemical equivalent \( Z \) for \( O_2 \) can be calculated as:
\[ Z = \frac{Molar \, Mass}{nF} = \frac{32 \, \text{g/mol}}{4 \cdot 96500 \, \text{C/mol}} = \frac{32}{386000} \, \text{g/C} \]

### Step 6: Calculate the current (I)
Substituting the values into the rearranged formula:
\[ I = \frac{3.2 \times 10^{-4} \, \text{g}}{\left(\frac{32}{386000} \, \text{g/C}\right) \cdot 1 \, \text{s}} = \frac{3.2 \times 10^{-4} \cdot 386000}{32} \]
Calculating this gives:
\[ I \approx 3.86 \, \text{A} \]

### Final Answer
The current passed to produce \( O_2 \) gas at the anode at the rate of \( 0.224 \, \text{mL/s} \) is approximately \( 3.86 \, \text{A} \).

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