Question

A current of 15.0 amperes is passed through a solution of CrCl2, for 45 minutes. The volume of Cl₂ , (in I) obtained at the anode at 1 atm and 273 K is around (IF=96500 Cmol-1, At. wt. of Cl=35.5, R=0.082 L-atmK^-1 mol^-1)

• A. 4.7
• B. 3.7
• C. 2.7
• D. 5.7

Answer 1

The Correct Option is A

Step 1: Understand the Given Information:
The problem states that a current of 15.0 amperes is passed through a solution of CrCl2 for 45 minutes. The volume of Cl2 obtained at the anode at 1 atm and 273 K is to be calculated. The constants provided are:
$1F = 96500 \, \text{C mol}^{-1}$, atomic weight of Cl = 35.5, and \(R = 0.082 \, \text{L.atm K}^{-1} \text{mol}^{-1}\).

Step 2: Use Faraday’s Law of Electrolysis:
We can apply Faraday’s law to calculate the volume of Cl2 gas produced. Faraday’s law states that the amount of substance liberated or deposited at an electrode is directly proportional to the quantity of charge passed:

$ \text{Volume of Cl}_2 = \frac{I \cdot t}{n \cdot F} \cdot \frac{RT}{P} $
where: 
- $I = 15.0$ A (current) 
- $t = 45 \, \text{minutes} = 45 \times 60 \, \text{seconds}$ (time) 
- $n = 2$ (the number of moles of electrons involved in the half-reaction of Cl2)
- $R = 0.082$ L·atm·mol^-1K^-1 (gas constant)
- $T = 273$ K (temperature)
- $P = 1$ atm (pressure)

Step 3: Calculate the total charge passed:
The total charge passed is given by:

$Q = I \cdot t = 15.0 \, \text{A} \times 45 \times 60 \, \text{seconds} = 40500 \, \text{C}$

Step 4: Apply Faraday's Law to Calculate Volume:
The volume of Cl2 produced is calculated as:

$ \text{Volume of Cl}_2 = \frac{40500 \cdot 0.082 \cdot 273}{2 \cdot 96500 \cdot 1} = 4.7 \, \text{L}$

Final Answer:
The volume of Cl2 produced is 4.7 L.