Question:
How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten \(\mathrm{CaCl}_2\)?
How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten \(\mathrm{CaCl}_2\)?
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\(t\) (in seconds) = \(\frac{w \times 96500}{E \times I}\). Convert to hours by dividing by 3600.
Updated On: Apr 20, 2026
- 27 h
- 8.3 h
- 11 h
- 16 h
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Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Use Faraday's first law: \(w = ZIt = \frac{E}{96500} \times I \times t\).
Step 2: Detailed Explanation:
Equivalent weight of Ca = Atomic weight / valency = 40 / 2 = 20.
\(w = 60\) g, \(I = 5\) A.
\(60 = \frac{20}{96500} \times 5 \times t\)
\(t = \frac{60 \times 96500}{20 \times 5} = \frac{60 \times 96500}{100} = 60 \times 965 = 57900\) seconds.
Convert to hours: \(57900 / 3600 = 16.08 \approx 16\) hours.
Step 3: Final Answer:
16 h
Use Faraday's first law: \(w = ZIt = \frac{E}{96500} \times I \times t\).
Step 2: Detailed Explanation:
Equivalent weight of Ca = Atomic weight / valency = 40 / 2 = 20.
\(w = 60\) g, \(I = 5\) A.
\(60 = \frac{20}{96500} \times 5 \times t\)
\(t = \frac{60 \times 96500}{20 \times 5} = \frac{60 \times 96500}{100} = 60 \times 965 = 57900\) seconds.
Convert to hours: \(57900 / 3600 = 16.08 \approx 16\) hours.
Step 3: Final Answer:
16 h
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