Assume that during the electrolysis of `AgNO_(3)`, only `H_(2)O` is electrolyzed and `O_(2)` is formed as
`2H_(2)O rarr 4H^(o+)+O_(2)+4e^(-)`
`O_(2)` formed at `NTP` due to passage of 2 amperes of current for 965 second is

Options

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

Explanation

To solve the problem of calculating the volume of oxygen gas formed during the electrolysis of water in the presence of silver nitrate, we can follow these steps:

### Step 1: Write the Electrolysis Reaction
The electrolysis of water can be represented by the following reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
This indicates that 2 moles of water produce 1 mole of oxygen gas and 4 moles of electrons.

### Step 2: Calculate the Total Charge (Q)
We can calculate the total charge (Q) using the formula:
\[ Q = I \times t \]
Where:
- \( I = 2 \, \text{amperes} \)
- \( t = 965 \, \text{seconds} \)

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

### Step 3: Convert Charge to Faraday
The charge in Faraday (F) can be calculated using the relation:
\[ \text{Charge in Faraday} = \frac{Q}{96500 \, \text{C/mol}} \]
Substituting the value of Q:
\[ \text{Charge in Faraday} = \frac{1930}{96500} \approx 0.0200 \, \text{mol} \]

### Step 4: Calculate Moles of Oxygen Produced
From the electrolysis reaction, we know that 4 Faraday produces 1 mole of oxygen. Therefore, the moles of oxygen produced can be calculated as:
\[ \text{Moles of } O_2 = \frac{\text{Charge in Faraday}}{4} \]
Substituting the value:
\[ \text{Moles of } O_2 = \frac{0.0200}{4} = 0.0050 \, \text{mol} \]

### Step 5: Calculate the Volume of Oxygen at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of oxygen produced can be calculated as:
\[ \text{Volume of } O_2 = \text{Moles of } O_2 \times 22.4 \, \text{L/mol} \]
Substituting the value:
\[ \text{Volume of } O_2 = 0.0050 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.112 \, \text{L} \]

### Final Answer
The volume of oxygen gas formed at NTP is:
\[ \boxed{0.112 \, \text{L}} \]

---