MP Board · Class 12 · Chemistry · ElectrochemistryThe quantity of charge required to obtain 1 mole of Al from $\text{Al}3^+$ molten salt is:
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Total Attempts: 6
Accuracy Rate: 83.3%
Step-by-Step Solution
The correct option is $3F$. The reduction reaction is $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$. This shows that 3 moles of electrons (i.e., $3$ Faradays of charge) are required to reduce 1 mole of $\text{Al}^{3+}$ ions to 1 mole of Al metal.
Detailed Options Breakdown
Option : $1F$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrochemistry.
Option 1: $2F$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrochemistry.
Option 2: $3F$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 3: $6F$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrochemistry.
💡 Study Guide: This question tests core syllabus concepts from Electrochemistry. For formulas, key summaries, and mock exam reference guides, read the full Electrochemistry Revision Notes.