Question

A current of $0.5\text{ amp}$ is passed through molten $AlCl_3$ for $96.5\text{ seconds}$. The volume of $Cl_2$ gas liberated at STP at anode (in ml) is ($Cl = 35.5\text{ u}$) ($1F = 96500 \text{ C mol}^{-1}$)

• A. $11.2$
• B. $22.4$
• C. $5.6$
• D. $33.6$

Hint:

Always double-check the units! The question asks for the volume in ml, but standard molar volume is often remembered in Litres ($22.4\text{ L}$). Multiplying by 1000 early can prevent decimal errors.

Answer 1

The Correct Option is C

This problem involves Faraday's First Law of Electrolysis and the concept of molar volume at STP. Step 1: Calculate the total charge ($Q$) $$Q = I \times t$$ $$Q = 0.5\text{ A} \times 96.5\text{ s} = 48.25\text{ Coulombs}$$

Step 2: Identify the electrode reaction for $Cl_2$ At the anode, chloride ions are oxidized: $$2Cl^- \to Cl_2(g) + 2e^-$$ This equation shows that

2 moles of electrons (2 Faraday) are required to produce

1 mole of $Cl_2$ gas.

Step 3: Calculate the moles of $Cl_2$ produced $$\text{Moles of } Cl_2 = \frac{Q}{n \times F}$$ $$\text{Moles of } Cl_2 = \frac{48.25}{2 \times 96500} = \frac{48.25}{193000} = 0.00025\text{ moles}$$

Step 4: Convert moles to volume at STP Molar volume at STP is $22.4\text{ L} = 22400\text{ ml}$. $$\text{Volume} = \text{moles} \times 22400$$ $$\text{Volume} = 0.00025 \times 22400$$ $$\text{Volume} = \frac{1}{4000} \times 22400 = \frac{22.4}{4} = 5.6\text{ ml}$$ Thus, the volume of chlorine gas liberated is $5.6\text{ ml}$.