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)
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)
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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!
Updated On: Jun 12, 2026
- 3.0 g
- 9.0 g
- 13.6 g
- 0.9 g
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The problem asks for the mass of Aluminum metal (Al) deposited at the negative cathode during the electrolysis of molten $\text{AlCl}_3$ when a constant electrical current of $1\ \text{A}$ passes through the cell for a duration of $9650\ \text{seconds}$.
Step 2: Key Formula or Approach:
According to Faraday's First Law of Electrolysis, the mass ($W$) of a substance deposited is calculated by:
$$W = \frac{I \times t \times M}{n \times F}$$ Where:
$I =$ electrical current in amperes ($1\ \text{A}$)
$t =$ time duration in seconds ($9650\ \text{s}$)
$M =$ molar atomic mass of the element ($27\ \text{g\ mol}^{-1}$)
$n =$ number of moles of electrons exchanged per mole of metal (valency factor)
$F =$ Faraday's constant ($\approx 96500\ \text{C\ mol}^{-1}$)
Step 3: Detailed Explanation:
First, identify the reduction half-reaction taking place at the cathode for the $\text{Al}^{3+}$ ion:
$$\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}\ (\text{s})$$ This shows that $n = 3$ moles of electrons are required to deposit 1 mole of Aluminum metal.
Now, substitute the parameters into Faraday's equation:
$$W = \frac{1\ \text{A} \times 9650\ \text{s} \times 27\ \text{g/mol}}{3 \times 96500\ \text{C/mol}}$$ Simplify the fraction by canceling common multiples between 9650 and 96500:
$$\frac{9650}{96500} = \frac{1}{10}$$ Substitute this back into the equation:
$$W = \frac{1}{10} \times \frac{27}{3}$$ $$W = \frac{1}{10} \times 9 = 0.9\ \text{g}$$ This calculated mass corresponds exactly to option (D).
Step 4: Final Answer:
The mass of Aluminum deposited at the cathode is 0.9 g, which corresponds to option (D).
The problem asks for the mass of Aluminum metal (Al) deposited at the negative cathode during the electrolysis of molten $\text{AlCl}_3$ when a constant electrical current of $1\ \text{A}$ passes through the cell for a duration of $9650\ \text{seconds}$.
Step 2: Key Formula or Approach:
According to Faraday's First Law of Electrolysis, the mass ($W$) of a substance deposited is calculated by:
$$W = \frac{I \times t \times M}{n \times F}$$ Where:
$I =$ electrical current in amperes ($1\ \text{A}$)
$t =$ time duration in seconds ($9650\ \text{s}$)
$M =$ molar atomic mass of the element ($27\ \text{g\ mol}^{-1}$)
$n =$ number of moles of electrons exchanged per mole of metal (valency factor)
$F =$ Faraday's constant ($\approx 96500\ \text{C\ mol}^{-1}$)
Step 3: Detailed Explanation:
First, identify the reduction half-reaction taking place at the cathode for the $\text{Al}^{3+}$ ion:
$$\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}\ (\text{s})$$ This shows that $n = 3$ moles of electrons are required to deposit 1 mole of Aluminum metal.
Now, substitute the parameters into Faraday's equation:
$$W = \frac{1\ \text{A} \times 9650\ \text{s} \times 27\ \text{g/mol}}{3 \times 96500\ \text{C/mol}}$$ Simplify the fraction by canceling common multiples between 9650 and 96500:
$$\frac{9650}{96500} = \frac{1}{10}$$ Substitute this back into the equation:
$$W = \frac{1}{10} \times \frac{27}{3}$$ $$W = \frac{1}{10} \times 9 = 0.9\ \text{g}$$ This calculated mass corresponds exactly to option (D).
Step 4: Final Answer:
The mass of Aluminum deposited at the cathode is 0.9 g, which corresponds to option (D).
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