Question

\(96.5\) amperes current is passed through the molten \(AlCl_3\) for \(100\) seconds. The mass of aluminium deposited at the cathode is
Atomic weight of \(Al=27\,u\)

• A. \(0.90\,g\)
• B. \(0.45\,g\)
• C. \(1.35\,g\)
• D. \(1.8\,g\)

Hint:

For electrolysis, \[ m=\frac{EIt}{F} \] where \[ E=\frac{\text{atomic mass}}{\text{valency}} \] For aluminium deposition, \[ Al^{3+}+3e^-\rightarrow Al \] so valency \(=3\).

Answer 1

The Correct Option is A

Step 1: Write the electrode reaction for aluminium deposition.
During electrolysis of molten AlCl$_3$, aluminium ions are reduced at the cathode: \[ Al^{3+} + 3e^- \rightarrow Al \] So 3 moles of electrons (3 Faradays) are required to deposit 1 mole (27 g) of aluminium.
Step 2: Recall Faraday's first law of electrolysis.
The mass of substance deposited is given by: \[ m = \frac{M \times I \times t}{n \times F} \] where $M$ = molar mass, $I$ = current (A), $t$ = time (s), $n$ = number of electrons, $F$ = Faraday constant $\approx 96500$ C/mol.
Step 3: Identify the given values.
$I = 96.5$ A, $t = 100$ s, $M = 27$ g/mol (aluminium), $n = 3$, $F = 96500$ C/mol.
Step 4: Calculate the charge passed.
\[ Q = I \times t = 96.5 \times 100 = 9650 \text{ C} \]
Step 5: Calculate the moles of electrons passed.
\[ \text{moles of electrons} = \frac{Q}{F} = \frac{9650}{96500} = 0.1 \text{ mol} \]
Step 6: Calculate the mass of aluminium deposited.
Since 3 mol electrons deposit 1 mol (27 g) of Al: \[ \text{moles of Al} = \frac{0.1}{3} \approx 0.0333 \text{ mol} \] \[ m = 0.0333 \times 27 = 0.90 \text{ g} \]
Step 7: Match with options.
$0.90$ g matches option 1 exactly.
\[ \boxed{m = 0.90 \text{ g}} \]