Department of Chemistry, Geosciences and Environmental Sciences

Exam 3-A-Key

Chemisty 1084:  Section 010 & 030                                                                       Spring 2006

Name:________________________________________________________

Read all directions and questions carefully!!  This exam consists of two parts.  The first part consists of 10 multiple choice questions worth four points each for a total of 40 points.  The second part consists of five numerical problems worth either 10 or 15 points per question for a total of 60 points.  Show all your work necessary for the numerical problems as partial credit will be given for those problems.

Possibly Useful Constants and Equations

∆G = ∆H – T∆S ∆G° = ∆H° – T∆S°

∆G° = –RTlnK

Nernst equation:                             ∆G° = –nFE°

                                                             ∆G = –nFE

Gas constant:  R = 8.314 J/mol·K                               Faraday constant:  F = 96485 C/mol

Score

Part 1 (40 points):_____________________

                                    Part 2 (60 points):_____________________

                                    Total (100 points):_____________________

Don’t forget to put your name on this test!

Good Luck!!

Part 1

Multiple Choice

Please indicate the answer to each question by putting your choice in the space provided.  There is only one correct answer for each question.  There will be 10 multiple choice questions worth 4 points each.

1.  The quality of fresh water found in rivers and lakes is generally determined by the amount of dissolved ___________ found in the water.

       (a)  CO₂                                                                      (c)  O₃

       (b)  O₂                                                                         (d)  H₂O

                                                            (e)  NaCl

Answer:  B

2.  Which one of the following processes will result in a decrease in entropy?

       (a)  boiling water to form steam.                                (c)  mixing of two gases in one container.

       (b)  dissolving solid KCl in water.                             (d)  freezing water to make ice.

Answer:  D

Going from liquid to gas phase (boiling) increases entropy.  Dissolving a solid into a solution increases entropy.  Mixing two gases increases entropy.  Going from liquid to solid phase (freezing) decreases entropy.

3.  For a spontaneous change,

       (a)  DS must be positive.                                             (c)  DH must be negative.

       (b)  DG must be negative.                                           (d)  All of the above.

Answer:  B

A spontaneous change required DS_universe to be positive and DG to be negative.  Choice (a) is incorrect because DS is the entropy change for the system not the universe.

4.  An endothermic reaction with an increase in entropy will be spontaneous ___________________.

       (a)  at all temperatures.                                               (c)  at higher temperatures only.

       (b)  at lower temperatures only.                                 (d)  never at any temperature.

Answer:  C

This question is based on the equation:  DG=DH–TDS.  DH is positive for an endothermic reaction. DS is positive when entropy is increased.  If DH is positive and DS is positive, DG will be negative at higher temperatures (T is always positive).

5.  The total entropy of the universe is:

       (a)  always increasing.                                               (c) constant.

       (b)  always decreasing.

Answer:  A

This is a statement of the Second Law of Thermodynamics

6.  Which substance is the reducing agent in the reaction below:

Pb  +  PbO₂  + 2 H₂SO₄ ¾¾®  2 PbSO₄  +  2 H₂O

       (a)  Pb                                                                         (c)  PbSO₄

       (b)  PbO₂                                                                    (d)  H₂SO₄

                                                            (e)  H₂O

Answer:  A

The reducing agent gets oxidized.  Pb goes from an oxidation state of 0 in Pb to an oxidation state of +2 in PbSO₄.  Pb is getting oxidized.  Pb goes from an oxidation state of 4 in PbO₂ to an oxidation state of +2 in PbSO₄.  PbO₂ is getting reduced.  All other elements do not change their oxidation numbers.  The reactant that is getting oxidized is Pb.

7.  Given the attached table of reduction potentials, which of the following is the strongest reducing agent?

       (a)  Na(s)                                                                    (c)  K(s)

       (b)  Al(s)                                                                    (d)  Li(s)

Answer:  D

See appendix E in your book for a table of reduction potentials.  Looking at the half-reactions in the table, the product with the most negative reduction potential is the strongest reducing agent.  This is Li.

8.  In which of the following species does nitrogen have the highest (most positive) oxidation number?

       (a)  NH₃                                                                      (c)  NaNO₃

       (b)  NO                                                                       (d)  N₂O₃

                                                            (e)  N₂

Answer:  C

The oxidation number of N in NH₃ is –3.  The oxidation number of N in NO is +2.  The oxidation number of N in NaNO₃ is +5.  The oxidation number of N in N₂O₃ is +3.  The oxidation number of N in N₂ is 0.

9.  Which of the following transformations could take place at the anode of an electrochemical cell?

       (a)  NO ¾¾®  NO₃¯                                              (c)  VO₂⁺ ¾¾®  VO²⁺

       (b)  CO₂ ¾¾®  C₂O₄^2-                                             (d) H₂AsO₄ ¾¾®  H₃AsO₃

                                                            (e)  O₂ ¾¾®  H₂O₂

Answer:  A

The anode is the electrode at which oxidation takes place.  The only choice above which involves an oxidation is choice A where the oxidation number of N increases from +2 in NO to +5 in NO₃¯.

10.  The formation of rust is slowed down by:

       (a)  the presence of salts               (c)  high pH conditions

       (b)  low pH conditions                 (d)  both the presence of salts and high pH conditions.

                        (e)  both the presence of salts and low pH conditions.

Answer:  C

Rust is slowed down by basic conditions which are high pH conditions.  The presence of salts and low pH (acidic) conditions accelerate the formation of rust.

Part 2

Numerical Problems

Solve the following problems, keeping track of significant figures where applicable.  Please show all the work necessary to obtain your answer in order to receive partial credit for possibly wrong answers.  Generally, full credit will not be given for the correct answer without any of the work performed to obtain the answer being shown on the paper.  Each question is worth either 10 or 15 points.

11.  (10 points)  A certain reaction has DG° = 3.76 kJ/mol at a temperature of 225°C.  Calculate the value of the equilibrium constant, K, for this reaction.

This problem involves the equation:  DG° = –RTlnK

Before using this equation, DG° must be converted to units of J/mol and the temperature must be converted to Kelvin:

DG° = (3.76 kJ/mol)(1000 J/kJ) = 3760 J/mol

T = 225°C + 273 = 498 K

Using the equation above:

12.  (15 points)  An electrochemical cell is constructed utilizing the following redox reaction:

2 Cu⁺(aq)  +  Sn²⁺(aq) ¾¾®   2 Cu(s)  +  Sn⁴⁺(aq)

(a)  (3 points)  Using the attached table of standard reduction potentials, what is the value of E° for this electrochemical cell?

See Appendix E in your book for a table of standard reduction potentials.  The following two half-reactions pertain to this reaction:

Cu⁺(aq)  +  e¯  ¾¾®  Cu(s) E° = +0.521 V

Sn⁴⁺⁽aq)  +  2e¯  ¾¾® Sn²⁺(aq) E° = +0.154 V

Since the second half-reaction is reversed in the redox reaction, that half-reaction occurs at the anode.  The E° for the cell is equal to:

E° = E°(cathode) – E°(anode) =  +0.521 V – 0.154 V = +0.367 V

(b)  (12 points)  What is the cell potential, E, at 298.15 K for this cell when [Cu⁺] = 0.0125 M, [Sn²⁺] = 0.00575 M, and [Sn⁴⁺] = 0.175 M?

This problem utilizes the Nernst equation: 

E° = +0.367 V as calculated in part (a).  The value of the reaction quotient can be obtained from the redox reaction and the given concentrations:

The value of “n” is the number of electrons transferred in the balanced redox reaction.  For this reaction this is equal to “2”.  (Looking at two half-reactions, the 1^st half-reaction would be multiplied by 2 and added to the reverse of the second half-reaction.  2e¯’s would cancel when this happens).

Plugging all these values into the Nernst equation: 

*13.  (10 points)  A certain reaction has DH = –173 kJ and DS = –367 J/K at a temperature of 225°C.  Show that this reaction is not spontaneous at 225°C.  Then, calculate the temperature, in °C, at which this reaction will become spontaneous.

This problem uses the equation:  DG = DH – TDS.  Before using this equation, convert the temperature into Kelvin and DS into units of kJ/K:

Calculating the value of DG:  DG = –173 kJ –(498 K)(–0.367 kJ/K) = –173 kJ + 183 kJ = +10. kJ

Since DG is positive, this reaction is not spontaneous at 225°C.  

To calculate the temperature at which the reaction will become spontaneous, calculate the temperature at which DG =0 assuming that DH and DS do not change their values:

Converting back to Celcius:  °C = 471 K -273 = 198°C

14.  (10 points)  A voltaic cell is constructed using the following redox half reactions:

Cd²⁺⁽aq)  +  2 e¯  ¾¾® Cd(s)

Co²⁺⁽aq)  +  2 e¯  ¾¾®  Co(s)

       Using the attached table of standard reduction potentials, answer the following questions:

(a)  (3 points)   Which redox reaction occurs at the anode?  The reverse of the half-reaction with the more negative reduction potential occurs at the anode. Looking at Appendix E, the first half-reaction (Cd²⁺/Cd) has E° = –0.403 V and the second half-reaction (Co²⁺/Co) has E° = –0.277 V. The more negative reduction potential is for the 1^st half-reaction. Therefore, the half-reaction that occurs at the anode is: Cd(s) ¾¾® Cd²⁺⁽aq) + 2 e¯

(b)  (3 points)  Which redox reaction occurs at the cathode? The other half-reaction occurs at the cathode:

Co²⁺⁽aq) + 2 e¯ ¾¾® Co(s)

(c)  (2 points)  What is the value of E° for this cell?

E° = E°(cathode) – E°(anode) = –0.277 V – (–0.403 V) = 0.126 V

(d)  (2 points)  Write the balanced redox reaction associated with this cell:

The balanced redox reaction obtained from the two half-reactions is equal to the sum of the half-reactions since the number of electrons gained and lost are equal:

Cd(s) + Co²⁺(aq) ¾¾® Cd²⁺(aq) + Co(s)

15.  (15 points)  Balance the following redox reactions:

(a)  H₅IO₆(aq)  +  Cr(s) ¾¾®  IO₃¯(aq)  +  Cr³⁺(aq) (acidic solution)

The skeletal half-reactions are:

H₅IO₆ ¾¾® IO₃¯

Cr ¾¾® Cr³⁺

All the non-H and non-O atoms are already balanced. Balance the O atoms in the first half-reaction by placing 3 H₂O molecules on the product side:

H₅IO₆ ¾¾® IO₃¯ + 3 H₂O

Cr ¾¾® Cr³⁺

Balance the H atoms in the first half-reaction by placing 1 H⁺ on the reactant side:

H₅IO₆ + H⁺ ¾¾® IO₃¯ + 3 H₂O

Cr ¾¾® Cr³⁺

Balance the charges on both sides of both half-reactions by placing 2 e¯ on the reactant side of the 1^st half-reaction and 3e¯ on the product side of the second half-reaction:

H₅IO₆ + H⁺ + 2 e¯ ¾¾® IO₃¯ + 3 H₂O

Cr ¾¾® Cr^{3+ +} 3 e¯

Equalize the number of electrons gained and lost by multiplying the 1^st half-reaction by 3 and the second half-reaction by 2, and then add the two half-reactions to obtain the balanced redox reaction:

3 H₅IO₆ + 3 H⁺ + 6 e¯ ¾¾® 3 IO₃¯ + 6 H₂O

2 Cr ¾¾® 2 Cr^{3+ +} 6 e¯

The balanced redox reaction is:

3 H₅IO₆ + 2 Cr + 3 H⁺ ¾¾® 3 IO₃¯ + 2 Cr³⁺ + 6 H₂O

(b)  Se(s)  +  Cr(OH)₃(s) ¾¾®  Cr(s)  +  SeO₃^2-(aq) (basic solution)

The skeletal half-reactions are:

Se ¾¾® SeO₃^2-

Cr(OH)₃ ¾¾® Cr

All the non-H and non-O atoms are already balanced. Balance the O atoms in the first half-reaction by placing 3 H₂O molecules on the reactant side. Balance the O atoms in the second half-reaction by placing 3 H₂O molecules on the product side:

Se + 3 H₂O ¾¾® SeO₃^2-

Cr(OH)₃ ¾¾® Cr + 3 H₂O

Balance the H atoms by placing 6 H⁺ on the product side in the 1^st half-rxn and 3 H⁺ on the reactant side in the 2^nd half-rxn:

Se + 3 H₂O ¾¾® SeO₃^{2- +} 6 H⁺

Cr(OH)₃ + 3 H⁺ ¾¾® Cr + 3 H₂O

Balance the charge on both sides by placing 4 e¯’s on the product side in the first half-reaction and 3 e¯’s on the reactant side in the second half-reaction:

Se + 3 H₂O ¾¾® SeO₃^{2- +} 6 H⁺ + 4 e¯

Cr(OH)₃ + 3 H⁺ + 3 e¯ ¾¾® Cr + 3 H₂O

The balanced redox reaction is 3 times the 1^st half-reaction plus 4 times the 2^nd half-reaction:

3 Se + 9 H₂O ¾¾® 3 SeO₃^{2- +} 18 H⁺ + 12 e¯

4 Cr(OH)₃ + 12 H⁺ + 12 e¯ ¾¾® 4 Cr + 12 H₂O

______________________________________

3 Se + 4 Cr(OH)₃ ¾¾® 3 SeO₃^2- + + 4 Cr + 6 H⁺ + 3 H₂O

Since this reaction is in basic solution, 6 OH¯ must be added to both sides of this reaction to convert it to basic solution. This will add 6 OH¯ ions to the reactant side. The 6 H⁺ ions are converted to 6 H₂O molecules which add to the 3 H₂O molecules to give:

3 Se + 4 Cr(OH)₃ + 6 OH¯ ¾¾® 3 SeO₃^2- + + 4 Cr + 9 H₂O

(c)  Br₂(l)  +  HClO₂(aq) ¾¾®  BrO₂¯(aq)  +  Cl₂(g) (acidic solution)

The skeletal half-reactions are:

Br₂ ¾¾® BrO₂¯

HClO₂ ¾¾® Cl₂

Balance the Br atoms in the 1^st half-reaction by placing a 2 in front of BrO₂¯. Balance the Cl atoms in the 2^nd half-rxn by placing a 2 in front of HClO₂:

Br₂ ¾¾® 2 BrO₂¯

2 HClO₂ ¾¾® Cl₂

Balance the O atoms by placing 4 H₂O’s as reactants in the 1^st half-reaction and 4 H₂O’s as products in the 2^nd half-reaction:

Br₂ + 4 H₂O ¾¾® 2 BrO₂¯

2 HClO₂ ¾¾® Cl₂ + 4 H₂O

Balance the H atoms by placing 8 H⁺ ions as products in the 1^st half-reaction and 6 H⁺ ions as reactants in the 2^nd half-reaction:

Br₂ + 4 H₂O ¾¾® 2 BrO₂¯ + 8 H⁺

2 HClO₂ + 6 H⁺ ¾¾® Cl₂ + 4 H₂O

Balance the charge by placing 6 e¯’s as products in the 1^st half-reaction and 6 e¯’s as reactants in the 2^nd half-reaction:

Br₂ + 4 H₂O ¾¾® 2 BrO₂¯ + 8 H⁺ + 6e¯

2 HClO₂ + 6 H⁺ + 6 e¯¾¾® Cl₂ + 4 H₂O

Since the number of electrons gained and lost are equal, the balanced redox reaction is the sum of the 2 half-reactions:

Br₂ + 2 HClO₂ ¾¾® 2 BrO₂¯ + Cl₂ + 2 H⁺

For 10 points extra credit:

Define 2 of the following 3 terms.  Please remember to choose only 2 of the choices, if you answer all 3 with no indication of which 2 you want graded, only the first two definitions will be graded.  Each complete definition is worth 5 points.  No partial credit will be given.

Voltaic cell: an electrochemical cell in which a spontaneous redox reaction generates electricity.

Irreversible process: a spontaneous process.

Reducing agent: a substance that causes another substance to get reduced.