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Example Questions

Example Question #681 : Ap Chemistry

What is the cell potential of the following cell:

Zn (s) + 2 H⁺(aq) -> Zn²⁺ (aq) + H₂ (g) E^o = 0.76 V

When the [Zn²⁺] = 1.0 M; P_H2 = 1 atm, and the pH in the cathode is 5.2?

Possible Answers:

1.32 V

1.59 V

0.45 V

1.80 V

0.90 V

Correct answer:

0.45 V

Explanation:

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

$[H^+ ] = 10^{-pH}= 10^{-5.2} = 6.31 x 10^{-6}$

$E_{cell} = 0.76 V - \frac{0.0592}{2} \: log \: \frac{[Zn^{2+} ] P_{H_2} }{[H^+ ]^2 }$

$E_{cell} = 0.76 - \frac{0.0592}{2} \: log\: \frac{[1.0]1}{[6.31 x 10^{-6} ]^2} = 0.45 V$

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Example Question #682 : Ap Chemistry

Calculate the standard cell potential of the following reaction:

Cd(s) + MnO₂ (s) + 4 H⁺(aq) + -> Cd²⁺ (aq) + Mn²⁺ (aq) + 2 H2O (l)

Given:

MnO₂ (s) + 4 H⁺ (aq) + 2e- -> Mn²⁺ (aq) + 2 H₂O (l) E^o = 1.23 V

Cd²⁺ (aq) + 2 e- -> Cd (s) E^o = -0.40 V

Possible Answers:

0.83 V

0.0 V

1.63 V

-0.83 V

-1.63 V

Correct answer:

1.63 V

Explanation:

E^o_cell = E^o_cathode- E^o_anode

E^o_cell = 1.23 – (-0.40) = 1.63 V

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Example Question #683 : Ap Chemistry

Calculate the standard cell potential of the following reaction:

Zn (s) + Cu²⁺(aq) -> Zn²⁺(aq) + Cu (s)

Given:

Zn²⁺(aq)+ 2 e^--> Zn (s) E^o= -0.76 V

Cu²⁺(aq)+ 2 e^--> Cu (s) E^o = 0.34 V

Possible Answers:

1.10 V

-0.42 V

0.0 V

0.42 V

-1.10 V

Correct answer:

1.10 V

Explanation:

E^o_cell = E^o_cathode- E^o_anode

E^o_cell = 0.34 – (-0.76) = 1.10 V

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Example Question #684 : Ap Chemistry

Calculate the standard cell potential of the following reaction:

Zn (s) + 2 Ag¹⁺(aq) -> Zn²⁺(aq) + 2 Ag (s)

Given:

Zn²⁺(aq)+ 2 e^--> Zn (s) E^o= -0.76 V

Ag¹⁺(aq)+ 1 e^--> Ag (s) E^o = 0.80 V

Possible Answers:

-1.56 V

1.56 V

-0.04 V

0.04 V

2.36 V

Correct answer:

1.56 V

Explanation:

E^o_cell = E^o_cathode- E^o_anode

E^o_cell = 0.80 – (-0.76) = 1.56 V

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Example Question #685 : Ap Chemistry

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

5.61 V

1.37 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 #686 : Ap Chemistry

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:

2.41 V

3.71 V

0.95 V

1.89 V

1.76 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 #687 : Ap Chemistry

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:

56.3 g

38.8 g

19.4 g

103 g

12.5 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 #688 : Ap Chemistry

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

Possible Answers:

0.84 g

1.25 g

6.72 g

3.36 g

1.68 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 #689 : Ap Chemistry

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

Possible Answers:

125 amps

50.0 amps

75.3 amps

100. amps

10.5 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 #690 : Ap Chemistry

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:

1.1 g

8.6 g

2.2 g

4.3 g

18.3 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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AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #681 : Ap Chemistry

Example Question #682 : Ap Chemistry

Example Question #683 : Ap Chemistry

Example Question #684 : Ap Chemistry

Example Question #685 : Ap Chemistry

Example Question #686 : Ap Chemistry

Example Question #687 : Ap Chemistry

Example Question #688 : Ap Chemistry

Example Question #689 : Ap Chemistry

Example Question #690 : Ap Chemistry

All AP Chemistry Resources

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