Question:
What is the quantity of current required to deposit one mole of metallic magnesium from fused magnesium chloride? (1F = 96500 C)
What is the quantity of current required to deposit one mole of metallic magnesium from fused magnesium chloride? (1F = 96500 C)
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Always check valency: number of electrons transferred determines total charge required.
Updated On: Apr 24, 2026
- \(1.93 \times 10^4 \, C\)
- \(1.93 \times 10^3 \, C\)
- \(9.65 \times 10^3 \, C\)
- \(9.65 \times 10^4 \, C\)
- \(1.93 \times 10^5 \, C\)
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The Correct Option is
Solution and Explanation
Concept:
Faraday’s law:
\[
\text{Charge} = n \times F
\]
where \(n\) = number of electrons transferred.
Step 1: Write electrode reaction.
\[ Mg^{2+} + 2e^- \rightarrow Mg \]
Step 2: Determine electrons required.
1 mole Mg requires 2 moles of electrons: \[ n = 2 \]
Step 3: Calculate charge.
\[ Q = 2 \times 96500 = 1.93 \times 10^5 \, C \]
Step 1: Write electrode reaction.
\[ Mg^{2+} + 2e^- \rightarrow Mg \]
Step 2: Determine electrons required.
1 mole Mg requires 2 moles of electrons: \[ n = 2 \]
Step 3: Calculate charge.
\[ Q = 2 \times 96500 = 1.93 \times 10^5 \, C \]
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