During the electrolysis of `AgNO_(3)`, the volume of `O_(2)` formed at `STP` due to passage of `2A` of current for `965 s` is

Options

A. `0.112L`
B. `0.224L`
C. `11.2L`
D. `22.4L`

Explanation

To solve the problem of finding the volume of \( O_2 \) formed during the electrolysis of \( AgNO_3 \) with a current of \( 2A \) for \( 965s \), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Total Charge (Q) Passed:**
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where \( I \) is the current in amperes and \( t \) is the time in seconds.

Given:
- \( I = 2A \)
- \( t = 965s \)

Substituting the values:
\[ Q = 2 \, \text{A} \times 965 \, \text{s} = 1930 \, \text{C} \]

2. **Convert Charge to Faraday:**
We know that 1 Faraday (F) is equivalent to \( 96500 \, C \). To convert the total charge into Faraday, we use the formula:
\[ \text{Faraday} = \frac{Q}{96500} \]
Substituting the value of \( Q \):
\[ \text{Faraday} = \frac{1930}{96500} \approx 0.0200 \, F \]

3. **Determine Moles of Electrons:**
Since 1 Faraday corresponds to 1 mole of electrons, the number of moles of electrons (n) can be calculated as:
\[ n = \text{Faraday} = 0.0200 \, \text{mol} \]

4. **Use the Electrolysis Reaction:**
The electrolysis of water can be represented as:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
From the balanced equation, we see that 4 moles of electrons produce 1 mole of \( O_2 \). Therefore, the moles of \( O_2 \) produced can be calculated as:
\[ \text{Moles of } O_2 = \frac{n}{4} = \frac{0.0200}{4} = 0.00500 \, \text{mol} \]

5. **Calculate the Volume of \( O_2 \) at STP:**
At STP (Standard Temperature and Pressure), 1 mole of gas occupies \( 22.4 \, L \). Therefore, the volume of \( O_2 \) produced can be calculated as:
\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times 22.4 \, L/mol \]
Substituting the values:
\[ \text{Volume of } O_2 = 0.00500 \, \text{mol} \times 22.4 \, L/mol = 0.112 \, L \]

### Final Answer:
The volume of \( O_2 \) formed at STP is \( 0.112 \, L \).

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