Question

What is the weight of Al deposited at cathode when 1 ampere current is passed through molten $\text{AlCl}_3$ for 9650 seconds? (At mass of Al = 27)

• A. 3.0 g
• B. 9.0 g
• C. 13.6 g
• D. 0.9 g

Hint:

Notice how clean the numbers are! $9650$ seconds is exactly $0.1$ of a Faraday ($96500\ \text{C}$). Since 1 full Faraday deposits 1 equivalent weight of Al ($\frac{27}{3} = 9\ \text{g}$), then $0.1$ Faraday will deposit exactly $0.1 \times 9\ \text{g} = 0.9\ \text{g}$ instantly!

Answer 1

The Correct Option is D

Step 1: Write the cathode reaction.
At the cathode, $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$, so 3 moles of electrons deposit one mole of aluminium.
Step 2: Find the charge passed.
Charge $Q = I \times t = 1 \times 9650 = 9650$ C.
Step 3: Convert charge to moles of electrons.
Moles of electrons $= \frac{Q}{F} = \frac{9650}{96500} = 0.1$ mol.
Step 4: Convert electrons to moles of Al.
Since 3 electrons give one Al atom, moles of Al $= \frac{0.1}{3}$ mol.
Step 5: Convert to mass.
Mass $= \text{moles} \times M = \frac{0.1}{3} \times 27$.
Step 6: Do the arithmetic.
$\frac{0.1}{3} \times 27 = 0.1 \times 9 = 0.9$ g.
Step 7: Report the answer.
The aluminium deposited is 0.9 g, which is option (4).
\[ \boxed{W = \dfrac{0.1}{3} \times 27 = 0.9\ \text{g}} \]