Question:
The number of Faradays involved in the conversion of 0.25 mol of $Al^{3+}$ to Al is x and number of Faradays involved in the conversion of 1100 mL of 0.5 M $Cu^{2+}$ to Cu is y. The values of x and y respectively are:
The number of Faradays involved in the conversion of 0.25 mol of $Al^{3+}$ to Al is x and number of Faradays involved in the conversion of 1100 mL of 0.5 M $Cu^{2+}$ to Cu is y. The values of x and y respectively are:
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Faradays = Moles of electrons transferred!
Updated On: Jun 6, 2026
- 0.75, 1.1
- 0.25, 2.2
- 0.50, 3.3
- 1.00, 2.2
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Faraday's law: Moles of electrons required $= n \times \text{moles}$.
Step 2: Meaning
$Al^{3+} + 3e^- \rightarrow Al$; $Cu^{2+} + 2e^- \rightarrow Cu$.
Step 3: Analysis
$x = 3 \times 0.25 = 0.75$ F. Moles of $Cu^{2+} = 1.1 \text{ L} \times 0.5 \text{ M} = 0.55$ mol. $y = 2 \times 0.55 = 1.1$ F.
Step 4: Conclusion
x = 0.75, y = 1.1.
Final Answer: (A)
Faraday's law: Moles of electrons required $= n \times \text{moles}$.
Step 2: Meaning
$Al^{3+} + 3e^- \rightarrow Al$; $Cu^{2+} + 2e^- \rightarrow Cu$.
Step 3: Analysis
$x = 3 \times 0.25 = 0.75$ F. Moles of $Cu^{2+} = 1.1 \text{ L} \times 0.5 \text{ M} = 0.55$ mol. $y = 2 \times 0.55 = 1.1$ F.
Step 4: Conclusion
x = 0.75, y = 1.1.
Final Answer: (A)
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