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AP Chemistry : Electrolysis and Faraday's Law

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Example Questions

Example Question #1 : Electrolysis And Faraday's Law

How many grams of Cr can be obtained by the electrolysis of a Cr(NO₃)₃ if 10 amps are passed through the cell for 6 hours?

Possible Answers:

103 g

19.4 g

38.8 g

12.5 g

56.3 g

Correct answer:

38.8 g

Explanation:

$Cr^{3+} + 3 e- \rightarrow Cr(s)$

$mass = \frac{10 A \, x\, 6 hr \, x \, \frac{3600 s}{1 hr}}{ \frac{96485 C}{mol e-} } \, x \, \frac{1 mol \, Cr}{3 mol e-} \, x \, \frac{51.996 g \, Cr}{1 mol \, Cr}= 38.8 g$

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Example Question #2 : Electrolysis And Faraday's Law

How much Al would be precipitated if an AlCl₃ solution was electrolyzed for 1.00 hours with 5.00 amps?

Possible Answers:

1.68 g

3.36 g

0.84 g

1.25 g

6.72 g

Correct answer:

1.68 g

Explanation:

$Al^{3+} + 3 e- \rightarrow Al(s)$

$mass = \frac{5 A x 1 hr \, x \, \frac{3600 s}{1 hr} }{\frac{96485 C}{mol e-}} \, x \, \frac{1 mol \, Al }{3 \, mol \, e-}\, x\, \frac{26.982 g Al}{1 mol Al} = 1.68 g$

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Example Question #3 : Electrolysis And Faraday's Law

What current must be past through a solution of AlCl₃ for 1.19 hours to produce 40g of solid Al?

Possible Answers:

125 amps

75.3 amps

10.5 amps

50.0 amps

100. amps

Correct answer:

100. amps

Explanation:

$Al^{3+} + 3 e- \rightarrow Al(s)$

$40g \, Al \: x \, \frac{1\, mol \, Al}{26.982g}\, x\, \frac{3 \, mol \, e-}{1 \, mol \, Al} \, x \, \frac{96485 C}{1 \, mol \, e^-} = 429108 \, C$

$Current \, x \, time\, (sec) = Coulombs$

$Current = \frac{429108 C}{1.19hr \, x \, \frac{3600 s}{1hr}} = 100 A$

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Example Question #1 : Electrolysis And Faraday's Law

In the electrolysis of CaCl₂ a 1.15 amp current is passed through the cell for 5.0 hours. How much Ca is produced?

Possible Answers:

2.2 g

4.3 g

1.1 g

18.3 g

8.6 g

Correct answer:

4.3 g

Explanation:

$Ca^{2+} + 2 e- \rightarrow Ca(s)$

$mass = \frac{1.15 A \, x \, 5 \, hr \, x\, \frac{ 3600 \, s}{1 \, hr} }{\frac{96485\, C}{mol \, e-}}\, x \, \frac{1 mol Ca }{2\, mol \, e-}\, x \, \frac{40.078 g \, Al}{1 mol \, Al}= 4.3 g$

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Example Question #4 : Electrolysis And Faraday's Law

How long would it take to electroplate 28.3g of silver at a constant current of 2.0 amps from a solution of silver nitrate?

Possible Answers:

53 min

844 min

211 min

105 min

422 min

Correct answer:

211 min

Explanation:

$Ag^+ + 1 e- \rightarrow Ag(s)$

$28.3g \: Ag \: x \: \frac{1 mol Ag}{107.87g} \: x \: \frac{1 mol e-}{1 mol Ag^+ } \: x \: \frac{96485 C}{1 mol e^-} = 25313 C$

$Current \, x \, time (sec) = Coulombs$

$time = \frac{25313 C}{2 A}\: x \: \frac{1 min}{60 sec} = 211 min$

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AP Chemistry : Electrolysis and Faraday's Law

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Example Question #1 : Electrolysis And Faraday's Law

Example Question #2 : Electrolysis And Faraday's Law

Example Question #3 : Electrolysis And Faraday's Law

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