The volume of gases evolved at `STP` by passing `0.1A` of current for `965sec`, through an aqueous solution of potassium acetate
(a)`22.4mL`
(b)`11.2mL`
(c)`89.6mL`
(d)`44.8mL`

Options

A. `22.4mL`
B. `11.2mL`
C. `89.6mL`
D. `44.8mL`

Explanation

To solve the problem of determining the volume of gases evolved at STP by passing 0.1 A of current for 965 seconds through an aqueous solution of potassium acetate, we can follow these steps:

### Step 1: Calculate the total charge (in coulombs) passed through the solution.
Using the formula:
\[ \text{Charge (Q)} = \text{Current (I)} \times \text{Time (t)} \]
Where:
- Current (I) = 0.1 A
- Time (t) = 965 s

\[ Q = 0.1 \, \text{A} \times 965 \, \text{s} = 96.5 \, \text{C} \]

### Step 2: Determine the number of moles of electrons transferred.
Using Faraday's constant (approximately 96500 C/mol), we can find the number of moles of electrons (n) using the formula:
\[ n = \frac{Q}{F} \]
Where:
- \( F \) = 96500 C/mol

\[ n = \frac{96.5 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.001 \, \text{mol of electrons} \]

### Step 3: Identify the reactions occurring at the electrodes.
At the anode (oxidation), acetate ions (CH₃COO⁻) lose electrons to form gaseous products (like methane and carbon dioxide). The reaction can be simplified to show that 2 moles of acetate produce 3 moles of gas (2 CO₂ and 1 CH₄).

At the cathode (reduction), water is reduced to produce hydrogen gas:
\[ 2H₂O + 2e⁻ \rightarrow H₂ + 2OH⁻ \]
This means that 2 moles of electrons produce 1 mole of hydrogen gas.

### Step 4: Calculate the total moles of gas produced.
From the reactions:
- 2 moles of acetate produce 3 moles of gas at the anode.
- 2 moles of electrons produce 1 mole of hydrogen gas at the cathode.

Thus, for every 2 moles of electrons (which corresponds to 0.002 moles of electrons), we get:
- 3 moles of gas from the anode
- 1 mole of gas from the cathode

This totals to 4 moles of gas for 2 moles of electrons.

### Step 5: Calculate the volume of gas produced at STP.
Using the molar volume of a gas at STP (22.4 L/mol), we can calculate the volume of gas produced:
\[ \text{Volume} = \text{Moles of gas} \times 22.4 \, \text{L/mol} \]
For 0.002 moles of electrons, we have:
\[ \text{Moles of gas produced} = 0.002 \, \text{mol} \times 4 = 0.008 \, \text{mol} \]
Thus:
\[ \text{Volume} = 0.008 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.1792 \, \text{L} = 179.2 \, \text{mL} \]

### Step 6: Adjust for the actual charge passed.
Since we calculated for 0.002 moles of electrons, we need to adjust our calculations based on the charge passed:
\[ \text{Total Volume} = \frac{96.5 \, \text{C}}{96500 \, \text{C/mol}} \times 4 \times 22.4 \, \text{L/mol} \]
Calculating this gives us:
\[ \text{Total Volume} = 44.8 \, \text{mL} \]

### Final Answer
The volume of gases evolved at STP is **44.8 mL** (Option d).

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