ap 2005 chemistry form b free-response questions = 173 coulombs h2 × 1 mol 1 mol h ( )2 ... 2006...

2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)

Copyright © 2005 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

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Answer EITHER Question 2 below OR Question 3 printed on pages 8 and 9. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent.

2. Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of 0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the electrolysis began. The temperature was 298 K and the atmospheric pressure was 1.00 atm.

(a) Write the balanced equation for the half reaction that took place at the anode.

(b) Calculate the amount of electric charge, in coulombs, that passed through the solution.

(c) Why is the volume of O2(g) collected different from the volume of H2(g) collected, as shown in the diagram?

(d) Calculate the number of moles of H2(g) produced during the electrolysis.

(e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H2(g) produced during the electrolysis.

(f) After the hydrolysis reaction was over, the vertical position of the tube containing the collected H2(g) was adjusted until the water levels inside and outside the tube were the same, as shown in the diagram below. The volume of gas in the tube was measured under these conditions of 298 K and 1.00 atm, and its volume was greater than the volume calculated in part (e). Explain.

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AP® CHEMISTRY 2005 SCORING GUIDELINES (Form B)

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).

Question 2

Water was electrolyzed, as shown in the diagram above, for 5.61 minutes using a constant current of 0.513 ampere. A small amount of nonreactive electrolyte was added to the container before the electrolysis began. The temperature was 298 K and the atmospheric pressure was 1.00 atm.

(a) Write the balanced equation for the half reaction that took place at the anode.

2 H2O(l) → O2(g) + 4 H+(aq) + 4 e−

One point is earned for the correct half reaction.

(b) Calculate the amount of electric charge, in coulombs, that passed through the solution.

0.513 amp = 0.513 coulsec

electric charge = ( )coul0.513 sec × (5.61 min) × 60 sec1 min

⎛ ⎞⎜ ⎟⎝ ⎠

= 173 coulombs

One point is earned for the setup.

One point is earned for the

answer.

(c) Why is the volume of O2(g) collected different from the volume of H2(g) collected, as shown in the diagram?

When water decomposes according to the balanced chemical equation 2 H2O(l) → O2(g) + 2 H2(g) , twice as many moles of hydrogen are produced than moles of oxygen.

One point is earned for the correct explanation based on the

stoichiometry of the decomposition reaction.

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Question 2 (continued) (d) Calculate the number of moles of H2(g) produced during the electrolysis.

The half-reaction that takes place at the cathode is: 2 H2O(l) + 2 e− → H2(g) + OH−(aq)

H2mol = 173 coulombs × 21 mol H ( )1 mol 96,500 coulomb 2 mol

gee

−

−

⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

H2mol = 8.96 × 10− 4 mol

One point is earned for the number of coulombs.

One point is earned for recognizing the 1 : 2 stoichiometry.

(e) Calculate the volume, in liters, at 298 K and 1.00 atm of dry H2(g) produced during the electrolysis.

H2H2

n RTV

P=

4

H2

L atm(8.96 10 mol) 0.0821 (298 K)mol K

1 atmV

− ⎛ ⎞× × ×⎜ ⎟⎝ ⎠= = 0.0219 L

One point is earned for the substitution into the gas law

equation.

One point is earned for the correct answer.

(f) After the hydrolysis reaction was over, the vertical position of the tube containing the collected H2(g) was

adjusted until the water levels inside and outside the tube were the same, as shown in the diagram below. The volume of gas in the tube was measured under these conditions of 298 K and 1.00 atm, and its volume was greater than the volume calculated in part (e). Explain.

Because the electrolysis of water occurs in water, there is some water vaporin the tube of H2(g) that was collected. The volume calculated in part (e) was the volume of only the H2(g) in the tube at the given temperature and pressure. The presence of another gas (water vapor) results in a greater volume at the given temperature and pressure.

One point is earned for recognizing that there is some water vapor in the sample of

hydrogen gas.

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AP® CHEMISTRY 2005 SCORING COMMENTARY (Form B)

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

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Question 2 Sample: 2A Score: 9 This response earned all 9 points: 1 point for part (a), 2 points for part (b), 1 point for part (c), 2 points for part (d), 2 points for part (e), and 1 point for part (f). Sample: 2B Score: 7 The point was not earned in part (a) because the equation is incorrect. The point was not earned in part (f) because the response does not indicate that the partial pressure of water vapor is involved. Sample: 2C Score: 4 The point was not earned in part (a) because the equation is missing H+ ’s and e− ’s to make it balance. Only 1 out of 2 points was earned in part (d) because the response does not recognize that two moles of electrons are transferred for each mole of H2 involved. Neither point was earned in part (e) because the response assumes that the conditions are standard temperature and pressure. The point was not earned in part (f) because the response does not indicate that the partial pressure of water vapor is involved.

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2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)

© 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents).

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Answer EITHER Question 2 OR Question 3 below. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent. 2. Answer the following questions about voltaic cells.

(a) A voltaic cell is set up using Al /Al3+ as one half-cell and Sn /Sn2+ as the other half-cell. The half-cells contain equal volumes of solutions and are at standard conditions.

(i) Write the balanced net-ionic equation for the spontaneous cell reaction.

(ii) Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction.

(iii) Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include units with your answer.

(iv) If the cell operates until [Al3+] is 1.08 M in the Al /Al3+ half-cell, what is [Sn2+] in the Sn /Sn2+ half-cell?

(b) In another voltaic cell with Al /Al3+ and Sn /Sn2+ half-cells, [Sn2+] is 0.010 M and [Al3+] is 1.00 M. Calculate the value, in volts, of the cell potential, Ecell , at 25°C.

3. Answer the following questions about the thermodynamics of the reactions represented below.

Reaction X: 12

I2(s) + 12

Cl2(g) →← ICl(g) fHD = 18 kJ mol–1, 298SD = 78 J K–1 mol–1

Reaction Y: 12

I2(s) + 12

Br2(l) →← IBr(g) fHD = 41 kJ mol–1, 298SD = 124 J K–1 mol–1

(a) Is reaction X , represented above, spontaneous under standard conditions? Justify your answer with a calculation.

(b) Calculate the value of the equilibrium constant, Keq , for reaction X at 25°C.

(c) What effect will an increase in temperature have on the equilibrium constant for reaction X ? Explain your answer.

(d) Explain why the standard entropy change is greater for reaction Y than for reaction X .

(e) Above what temperature will the value of the equilibrium constant for reaction Y be greater than 1.0 ? Justify your answer with calculations.

(f) For the vaporization of solid iodine, I2(s) → I2(g) , the value of 298HD is 62 kJ mol–1. Using this

information, calculate the value of 298HD for the reaction represented below.

I2(g) + Cl2(g) →← 2 ICl(g)

S T O P If you finish before time is called, you may check your work on this part only.

Do not turn to the other part of the test until you are told to do so.

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Question 2

2. Answer the following questions about voltaic cells.

(a) A voltaic cell is set up using Al /Al3+ as one half-cell and Sn /Sn2+ as the other half-cell. The half-cells contain equal volumes of solutions and are at standard conditions.

(i) Write the balanced net-ionic equation for the spontaneous cell reaction.

3 Sn2+ + 2 Al → 3 Sn + 2 Al 3+ One point is earned for the correct direction.

One point is earned for the balanced net-ionic equation.

(ii) Determine the value, in volts, of the standard potential, E°, for the spontaneous cell reaction.

E° = − 0.14 V − (−1.66 V) = 1.52 V (or, 1.52 J C−1)


(Potential must be positive.)

(iii) Calculate the value of the standard free-energy change, ∆G°, for the spontaneous cell reaction. Include units with your answer.

∆G° = − n F E° = − 6 mol

1 mole−

× 96,500 C

1 mol e− × (1.52 J C−1)

= − 8.80 × 105 J mol−1 (or – 880 kJ mol−1)

One point is earned for indicating the correct mol e− to mol reaction ratio.

One point is earned for the correct answer with correct units.

(iv) If the cell operates until [Al3+] is 1.08 M in the Al /Al3+ half-cell, what is [Sn2+] in the Sn /Sn2+ half-cell?

change in [Sn2+] = 3+0.08 mol Al

1 L ×

2+

3+

3 mol Sn

2 mol Al =

2+0.12 mol Sn1 L

[Sn2+] = 1.00 mol L−1 – 0.12 mol L−1 = 0.88 mol L−1


(b) In another voltaic cell with Al /Al3+ and Sn /Sn2+ half-cells, [Sn2+] is 0.010 M and [Al3+] is 1.00 M. Calculate the value, in volts, of the cell potential, Ecell , at 25°C.

Ecell = 1.52 V −

0.0592

6 log

2

3(1.00)

(0.010)

= 1.52 V − 0.0592 V = 1.46 V

Answers must be consistent with part (a)(i). One point is earned for the proper exponents.

One point is earned for the correct substitution of concentrations.


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AP® CHEMISTRY 2006 SCORING COMMENTARY (Form B)


Question 2 Sample: 2A Score: 8 This excellent response earned 8 out of 9 possible points: 2 points for part (a)(i), 1 point for part (a)(ii), 2 points for part (a)(iii), and 3 points for part (b). The point was not earned for part (a)(iv). Sample: 2B Score: 6 In this good response, 1 point was earned in part (a)(iii) for the correct number of moles of electrons, but the second point was not earned because the units are incorrect and the number of significant figures in the answer is different by more than one from the proper number. The point was not earned in part (a)(iv) because the concentrations of Al3+ and Sn2+ are not equal. In part (b) one point was earned for substitution of the correct concentrations, but the point for correct exponents was not earned. The calculation with incorrect exponents is done correctly, so a second point was earned in part (b). Sample: 2C Score: 4 The point was not earned in part (a)(ii) because of a math error in the calculation. This incorrect value was used correctly in part (a)(iii), so part (a)(iii) earned full credit. The point was not earned in part (a)(iv). Part (b) is not attempted.

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2002 AP® CHEMISTRY FREE-RESPONSE QUESTIONS

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

GO ON TO THE NEXT PAGE. 7

Answer EITHER Question 2 below OR Question 3 printed on page 8. Only one of these two questions will be graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score weighting for the question you choose is 20 percent. 2. Answer parts (a) through (e) below, which relate to reactions involving silver ion, Ag+. The reaction between silver ion and solid zinc is represented by the following equation. 2 Ag+(aq) + Zn(s) � Zn2+(aq) + 2 Ag(s)

(a) A 1.50 g sample of Zn is combined with 250. mL of 0.110 M AgNO3 at 25�C.

(i) Identify the limiting reactant. Show calculations to support your answer.

(ii) On the basis of the limiting reactant that you identified in part (i), determine the value of [Zn2+] after the reaction is complete. Assume that volume change is negligible.

(b) Determine the value of the standard potential, E � � for a galvanic cell based on the reaction between AgNO3(aq) and solid Zn at 25�C.

Another galvanic cell is based on the reaction between Ag+(aq) and Cu(s), represented by the equation below.

At 25°C, the standard potential, E � , for the cell is 0.46 V. 2 Ag+(aq) + Cu(s) � Cu2+(aq) + 2 Ag(s)

(c) Determine the value of the standard free-energy change, ∆G�, for the reaction between Ag+(aq) and Cu(s) at 25�C.

(d) The cell is constructed so that [Cu2+] is 0.045 M and [Ag+] is 0.010 M . Calculate the value of the potential, E� for the cell.

(e) Under the conditions specified in part (d), is the reaction in the cell spontaneous? Justify your answer.

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AP® CHEMISTRY 2002 SCORING COMMENTARY

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.

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Question 2

Sample 2A (Score 10) This response earned a total of 10 points — 2 points for part (a)(i), 2 points for part (a)(ii), 1 point for part (b), 2 points for part (c), 2 points for part (d), and 1 point for part (e).

In this excellent response, there is clear understanding of the concepts of limiting reactants, electrochemistry, and thermochemistry as posed in this question. The answers were well constructed: there is a definite logical sequence to answering each question, and the final numerical answers are each circled. Sample 2B (Score 8) This response earned a total of 8 points — 2 points for part (a)(i), 2 points for part (a)(ii), 0 points for part (b), 1 point for part (c), 2 points for part (d), and 1 point for part (e).

In part (b), the voltage of the Ag half-cell is multiplied by 2, which is a common error. In part (c), the units are not given and there are too many significant digits. Sample 2C (Score 6) This response earned a total of 6 points — 1 point for part (a)(i), 2 points for part (a)(ii), 1 point for part (b), 0 points for part (c), 1 point for part (d), and 1 point for part (e).

This response shows some common errors. In part (a)(i), the stoichiometry is forgotten, which leads incorrectly to a determination that Zn is the limiting reactant. The 1 to 2 ratio is not used, but 1 point is earned for correctly continuing on this wrong path and determining that Zn must be consumed first. In part (a)(ii), the response earns both points for the correct determination of the concentration of the Zn2+ assuming that Zn is the limiting reactant. Part (b) was answered correctly. No points are earned in part (c) because of the use of the incorrect number of moles of electrons (n) and the lack of units (these were common errors). The response earns 1 point in part (d): the shortened version of the Nernst equation is used, but the substitution for Q is done incorrectly. Part (e) is answered correctly.

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2002 Q&A

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ap 2005 chemistry form b free-response questions = 173 coulombs h2 × 1 mol 1 mol h ( )2 ... 2006...

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