A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is ___ . [Given : molar mass of aluminium and chlorine are 27 g mol –1 and 35.5 g mol –1 respectively, Faraday constant = 96500 C mol –1 ] (1) 1.660 g (2) 1.007 g (3) 0.336 g (4) 0.441 g

🌟 Key Concepts

1. Faraday’s First Law of Electrolysis

   • Mass of a substance deposited ($m$) at an electrode is directly proportional to the charge ($Q$) passed through the electrolyte.
     $m = ZQ$
     where $Z$ is the electro-chemical equivalent.

2. Relationship between Charge, Current & Time
   $Q = I \times t$
   with $I$ in amperes (coulombs per second) and $t$ in seconds.

3. Moles of Electrons

   • 1 mole of electrons carries a charge of 1 Faraday ($F = 96500\,\text{C mol}^{-1}$).
     $n_{e^-} = \frac{Q}{F}$

4. Stoichiometry of the Electrode Reaction

   • For aluminium:
     $\text{Al}^{3+} + 3e^- \rightarrow \text{Al (s)}$
   • 3 moles of electrons deposit 1 mole of Al.

5. Mass–Mole Connection
   $m = n \times M$
   where $M$ is molar mass.

🔍 Logical Steps a Student Follows

1. Compute total charge: use current and time.
2. Convert charge to moles of electrons: divide by $F$.
3. Relate electrons to moles of aluminium: use the 3-to-1 ratio.
4. Find mass of aluminium: multiply moles of Al by its molar mass.
5. Match your value with the options and pick the closest.

🎈 What’s happening?

Imagine you have a little factory where tiny electric cars (electrons) drive metal atoms out of a liquid and park them on a shiny plate.

• Current (2 A) tells you how many cars leave the garage each second.
• Time (30 min) tells you how long the garage gate stays open.
• Faraday’s constant (96500 C mol⁻¹) tells you how many cars are needed to carry one big box of charge (1 mole of electrons).
• Aluminium needs 3 cars for every one atom to leave the liquid and sit on the plate.

So, you just count the total cars, divide them into boxes, and then see how many aluminium atoms can hitch a ride. Finally, multiply by how heavy each aluminium atom is and—tada!—you know the weight of aluminium left on the plate.

The answer turns out to be about 0.336 g (option 3).

Step-by-Step Calculation

1. Convert time to seconds
   $t = 30\,\text{min} \times 60\,\frac{\text{s}}{\text{min}} = 1800\,\text{s}$

2. Total charge passed
   $Q = I \times t = 2\,\text{A} \times 1800\,\text{s} = 3600\,\text{C}$

3. Moles of electrons delivered
   $n_{e^-} = \frac{Q}{F} = \frac{3600}{96500} \approx 0.0373\,\text{mol e}^-$

4. Moles of aluminium deposited
   The cathode reaction is $\text{Al}^{3+}+3e^- \rightarrow \text{Al}$
   Therefore, 3 mol e⁻ produce 1 mol Al:
   $n_{\text{Al}} = \frac{n_{e^-}}{3} = \frac{0.0373}{3} \approx 0.0124\,\text{mol}$

5. Mass of aluminium formed
   $m_{\text{Al}} = n_{\text{Al}} \times M_{\text{Al}} = 0.0124 \times 27 \approx 0.336\,\text{g}$

Final Answer

$\boxed{0.336\,\text{g}}$ (Option 3)