Assume that during electrolysis of `AgNO_(3)` only `H_(2)O` is electrolyesd and `O_(2)` is formed at the anode as :
`2H_(2)Orarr 4H^(+)+O_(2)+4e^(-)`
`O_(2)` formed at NTP due to passage of 2 amperes of current for 965 sec is :

Options

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

Answer

To solve the problem, we need to calculate the volume of oxygen (O₂) formed during the electrolysis of water at the anode when a current of 2 amperes is passed for 965 seconds. Here are the steps to arrive at the solution:

### Step 1: Calculate the total charge (Q) passed during electrolysis
The total charge can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) = current in amperes (2 A)
- \( t \) = time in seconds (965 s)

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

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's constant, which states that 1 mole of electrons corresponds to approximately 96500 coulombs, we can find the number of moles of electrons:
\[ n = \frac{Q}{F} \]
where:
- \( F \) = Faraday's constant (96500 C/mol)

Substituting the values:
\[ n = \frac{1930 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.0200 \, \text{mol} \]

### Step 3: Determine the moles of O₂ produced
From the electrolysis reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
It can be seen that 4 moles of electrons produce 1 mole of O₂. Thus, the number of moles of O₂ produced is:
\[ \text{moles of } O_2 = \frac{n}{4} = \frac{0.0200 \, \text{mol}}{4} = 0.0050 \, \text{mol} \]

### Step 4: Calculate the volume of O₂ at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of O₂ produced can be calculated as:
\[ \text{Volume} = \text{moles of } O_2 \times 22.4 \, \text{L/mol} \]
Substituting the values:
\[ \text{Volume} = 0.0050 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.112 \, \text{L} \]

### Final Answer
The volume of O₂ formed at NTP is **0.112 liters**.

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