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AP Chemistry : Thermodynamics

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Example Questions

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Example Question #1 : Cell Potential Under Non Standard Conditions

Calculate the standard cell potential of the following reaction:

3 F₂ (g) + 2 Au (s) -> 6 F^- (aq) + 2 Au³⁺

Given:

F₂ (g) + 2 e^- -> 2 F^- (aq) E^o = 2.87 V

Au³⁺(aq)+ 3 e^--> Au (s) E^o= 1.50 V

Possible Answers:

-5.61 V

-1.37 V

1.37 V

5.61 V

4.37 V

Correct answer:

1.37 V

Explanation:

E^o_cell = E^o_cathode- E^o_anode

E^o_cell = 2.87 – (1.50) = 1.37 V

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Example Question #2 : Cell Potential Under Non Standard Conditions

Determine the E_cell for the following reaction at 25 C:

Zn (s) + 2 VO₂⁺ (aq) + 4 H⁺ -> 2 VO²⁺ (aq) + Zn²⁺(aq) + 2 H2O (l)

Given that:

VO₂⁺_(aq)+ 2 H⁺ (aq) + e^- -> VO²⁺(aq) + H2O (l) E^o = 1.00 V

Zn²⁺ (aq) + 2 e^--> Zn (s) E^o = -0.76 V

And

[ VO₂⁺] = 2.0 M; [H⁺] = 0.50 M; [VO²⁺] = 1.0 x 10^-2M; [Zn²⁺] = 1.0 x 10^-1M

Possible Answers:

3.71 V

2.41 V

1.76 V

1.89 V

0.95 V

Correct answer:

1.89 V

Explanation:

$E_{cell} = E^0_{cell} - \frac {0.0592}{n} \: log \: Q$

$E^o_{cell} = 1.00 - (-0.76) = 1.76 V$

$E_{cell} = 1.76 V - \frac{ 0.0592}{2} \: log \: \frac{[VO^{2+} ]^2 [Zn^{2+} ]}{[VO_2^+ ]^2 [H^+ ]^4 )}$

$= 1.76 V - \frac{0.0592}{2}\: log \: \frac{[1.0 x 10^{-2} ]^2 [1.0 x 10^{-1} ]}{[2.0]^2 [0.5]^4} = 1.89$

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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:

12.5 g

38.8 g

19.4 g

103 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 #1 : 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

6.72 g

0.84 g

1.25 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 #1 : 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:

100. amps

50.0 amps

75.3 amps

10.5 amps

125 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:

18.3 g

1.1 g

4.3 g

2.2 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 #1 : 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:

844 min

422 min

105 min

211 min

53 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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Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

Consider the following electrolytic cell:

$Ni + Fe^{2+} \rightarrow Ni^{2+} + Fe$ $E^{\circ} = -0.21V$

What happens at the anode in the electrolytic cell?

Possible Answers:

Nickel is oxidized

Nickel is reduced

Iron is reduced

Iron is oxidized

Correct answer:

Nickel is oxidized

Explanation:

It does not matter if the cell is galvanic or electrolytic; oxidation will always take place at the anode. This means that the nickel loses two electrons and is oxidized at the anode to generate nickel ions.

$Ni + Fe^{2+} \rightarrow Ni^{2+} + Fe$

Nickel ions and iron are products, and are neither oxidized nor reduced during the reaction. Iron ions are reduced at the cathode to generate the iron product.

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Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

Consider the following reaction in a galvanic cell:

$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$

Which of the following takes place at the anode?

Possible Answers:

Copper is reduced

Copper is oxidized

Zinc is oxidized

Zinc is reduced

No reaction takes place at the anode

Correct answer:

Zinc is oxidized

Explanation:

Oxidation takes place at the anode and reduction takes place at the cathode. You can remember this with the pneumonic "An Ox, Red Cat."

In the equation, zinc loses electrons. It goes from a neutral, elemental charge to a charge of $\small +2$ . Since electrons are negative, a loss of electrons will cause an increase in charge. Because zinc loses electrons, it is oxidized. This will take place at the anode.

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Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

Toward which pole do electrons travel in a galvanic cell?

Possible Answers:

More information is needed

Towards the anode

Towards the negative pole

Towards the cathode

Electrons do not travel; only the protons travel

Correct answer:

Towards the cathode

Explanation:

Reduction always occurs at the cathode, and oxidation always occurs at the anode. Since reduction is the addition of electrons, the electrons must flow toward the site of reduction.

In a galvanic cell the positive charge is on the cathode, while the negative charge is on the anode. Since a galvanic cell has a positive potential and is spontaneous, electrons freely flow down their potential gradient. The electrons, which are negatively charged, are traveling towards the cathode, which is positive charged, since opposites attract.

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AP Chemistry : Thermodynamics

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Cell Potential Under Non Standard Conditions

Example Question #2 : Cell Potential Under Non Standard Conditions

Example Question #1 : Electrolysis And Faraday's Law

Example Question #1 : Electrolysis And Faraday's Law

Example Question #1 : Electrolysis And Faraday's Law

Example Question #1 : Electrolysis And Faraday's Law

Example Question #1 : Electrolysis And Faraday's Law

Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

Example Question #1 : Galvanic (Voltai) And Elecrolytic Cells

All AP Chemistry Resources

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