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

Options

A. `0.224A`
B. `2.24 A`
C. `9.64 A`
D. `3.86 A`

Explanation

To solve the problem of determining the current required to produce \( O_2 \) gas at the anode during the electrolysis of acidified water at a rate of \( 0.224 \, \text{ml/s} \) at STP, we can follow these steps:

### Step 1: Write the Electrolysis Reaction
The electrolysis of water can be represented by the following half-reaction at the anode:
\[ 2 \, \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \, \text{H}^+ + 4 \, e^- \]
This indicates that 2 moles of water produce 1 mole of oxygen gas and requires 4 moles of electrons.

### Step 2: Calculate the Moles of \( O_2 \) Produced
At STP, 1 mole of any gas occupies \( 22.4 \, \text{L} \) (or \( 22400 \, \text{ml} \)). We need to find out how many moles of \( O_2 \) are produced per second when \( 0.224 \, \text{ml} \) of \( O_2 \) is generated.
\[ \text{Moles of } O_2 = \frac{0.224 \, \text{ml}}{22400 \, \text{ml/mol}} = 1 \times 10^{-5} \, \text{mol/s} \]

### Step 3: Calculate the Moles of Electrons Required
From the balanced equation, we know that to produce 1 mole of \( O_2 \), 4 moles of electrons are required. Therefore, for \( 1 \times 10^{-5} \, \text{mol/s} \) of \( O_2 \):
\[ \text{Moles of electrons} = 4 \times (1 \times 10^{-5}) = 4 \times 10^{-5} \, \text{mol/s} \]

### Step 4: Calculate the Charge Required
The charge of one mole of electrons is given by Faraday's constant, \( F = 96500 \, \text{C/mol} \). Thus, the total charge \( Q \) required per second is:
\[ Q = \text{Moles of electrons} \times F = 4 \times 10^{-5} \, \text{mol/s} \times 96500 \, \text{C/mol} = 3.86 \, \text{C/s} \]

### Step 5: Calculate the Current
Current \( I \) is defined as the charge per unit time:
\[ I = \frac{Q}{t} \]
Since we are calculating the current for one second (\( t = 1 \, \text{s} \)):
\[ I = \frac{3.86 \, \text{C}}{1 \, \text{s}} = 3.86 \, \text{A} \]

### Conclusion
The current required to produce \( O_2 \) gas at the anode at the rate of \( 0.224 \, \text{ml/s} \) at STP is:
\[ \boxed{3.86 \, \text{A}} \]