Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                     Exam 3-A-Key

                       

Chemisty 1084:  Section 010 & 030                                                                       Spring 2007

 

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.  Eutrophication of a lake is the process of ______________________.

(a)  a rapid decline in the pH of the lake due to acid rain.

(b)  dissolved oxygen being depleted by an overpopulation of fish.

(c)  a rapid increase in the amount of dead and decaying plant matter as a result of excessive plant growth.

(d)  restoration of the lake’s oxygen supply by aerobic bacteria.

(e)  stocking the lake with fish.

 

Answer:  C

 

2.  DS is positive for which of the following reactions?

       (a)  Ag⁺(aq)  +  Cl¯(aq)  ¾¾®  AgCl(s)                 (c)  H₂(g)  +  F₂(g)  ¾¾®  2 HF(l)

       (b)  CO₂(s)  ¾¾®  CO₂(g)                                       (d)  N₂(g)  +  3 H₂(g)  ¾¾®  2 NH₃(g)

                                                            (e)  H₂O(l)  ¾¾®  H₂O(s)

 

Answer:  B

Gases have more entropy than solids.

 

3.  A reaction that is spontaneous at lower temperatures but becomes non-spontaneous at higher temperatures has a _____________ value of DH and a ___________ value of DS.

       (a)  negative, positive                                                 (c)  negative, negative

       (b)  positive, positive                                                 (d)  positive, negative

                                                            (e)  negative, zero

 

Answer:  C

This is an application of DG = DH –TDS and which combination of DH and DS will give a negative value at lower temperatures and turn positive as the temperature is increased.

 

4.  A spontaneous redox reaction has:

       (a)  a negative value of DG and a positive value of E.

       (b)  a positive value of DG and a positive value of E.

       (c)  a negative value of DG and a negative value of E.

       (d)  a positive value of DG and a negative value of E.

 

Answer:  A

 

 

 

 

Table of Thermodynamic Data at 298.15 K

 

Substance	DHf° (kJ/mol)	S° (J/mol-K)	Substance	DHf° (kJ/mol)	S° (J/mol-K)
NO(g)	91	211	O2(g)	0	205
NO2(g)	33	240	H2O(l)	–286	70
HNO3(g)	–134	267	H2O(g)	–242	189
HNO3(aq	–207	146	H2S(g)	–21	206
			SO2(g)	–297	248

 

 

5.  Given the table of thermodynamic data above, calculate the value of DS° for the reaction:

2 H₂S(g)  +  3 O₂(g)  ¾¾®  2 H₂O(g)  +  2 SO₂(g)

       (a)  –391 J/K                                                              (c)  153 J/K

       (b)  –153 J/K                                                              (d)  391 J/K

                                                            (e)  462 J/K

 

Answer:  B

DS° = (2 mol)(189 J/mol·K) + (2 mol)(248 J/mol·K) – (2 mol)(206 J/mol·K) – (3 mol)(205 J/mol·K)

        = –153 J/K

 

6.  What is the value of DG° for the reaction shown above?

       (a)  –1036 kJ                                                              (c)  –990 kJ

       (b)  46 kJ                                                                    (d)  0 kJ

                        (e)  DG_f^° values are needed to calculate DG°

 

Answer:  C

DH° = (2 mol)(–242 kJ/mol) + (2 mol)(–297 kJ/mol) – (2 mol)(–21 kJ/mol) = –1036 kJ

 

 

Answer questions 7 and 8 using the Table of Standard Reduction Potentials given on the next page:

 

7.  Which of the following substances is the strongest reducing agent?

       (a)  Sn(s)                                                                     (c)  Fe²⁺(aq)

       (b)  Pb(s)                                                                    (d)  H₂(g) under acidic conditions

                                                            (e)  MnO₂(s) under basic conditions

 

Answer:  A

Reducing agents are products in the half-reactions shown in the table.  The reducing agent with the most negative reduction potential is the strongest.  Sn(s)  –0.136 V, Pb(s)  –0.126 V, Fe²⁺(aq)  +0.771 V, H₂ under acidic conditions 0.000 V, MnO₂(s) under basic conditions +0.59 V

 

8.  Which one of the values below is equal to the value of E° for the following reaction:

3 Sn⁴⁺⁽aq)  +  2 Al(s)  ¾¾®  3 Sn²⁺(aq)  +  2 Al³⁺(aq)

       (a)  –1.50 V                                                                (c)  7.94 V

       (b)  1.81 V                                                                  (d)  1.50 V

                                                            (e)  –1.81 V

 

Answer:  E

Oxidation:  Al(s)  ¾¾®  Al³⁺⁽aq) + 3 e¯     E_red = –1.66 V

Reduction:  Sn⁴⁺⁽aq)  + 2 e¯   ¾¾®  Sn²⁺(aq)   E_red = +0.154 V

E° =E_red(reduction) – E_red(oxidation) = +0.154 V – ( –1.66 V) = –1.81 V

 

 

Table of Standard Reduction Potentials

 

Half-reaction	E° (V)
Ag+(aq)  +  e¯  ¾¾® Ag(s)	+0.799
Al3+(aq)  +  3e¯  ¾¾®  Al(s)	–1.66
Cd2+(aq)  +  2e¯  ¾¾®  Cd(s)	–0.403
Ce4+(aq)  +  e¯  ¾¾®  Ce3+(aq)	+1.61
Cr3+(aq)  +  3e¯  ¾¾®  Cr(s)	–0.74
Cu2+(aq)  +  2e¯  ¾¾®  Cu(s)	+0.337
Cu2+(aq)  +  e¯  ¾¾®  Cu+(aq)	+0.153
Cu+(aq)  +  e¯  ¾¾®  Cu(s)	+0.521
Fe2+(aq)  +  2e¯  ¾¾®  Fe(s)	–0.440
Fe3+(aq)  +  e¯ ¾¾®  Fe2+(aq)	+0.771
2 H+(aq)  +  2 e¯  ¾¾®  H2(g)	0.000
MnO4¯(aq)  +  8 H+(aq)  +  5e¯  ¾¾®  Mn2+(aq)  +  4 H2O(l)	+1.51
MnO4¯(aq)  +  2 H2O(l) + 3e¯  ¾¾®  MnO2(s)  +  4 OH¯(aq)	+0.59
Pb2+(aq)  +  2e¯  ¾¾®  Pb(s)	–0.126
Sn2+(aq)  +  2e¯  ¾¾®  Sn(s)	–0.136
Sn4+(aq)  +  2e¯  ¾¾®  Sn2+(aq)	+0.154

 

9.  The products of electrolysis of brine (aqueous sodium chloride) are:

       (a)  H₂, Cl₂, and NaOH                                               (c)  H₂ and O₂

       (b)  Na and Cl₂                                                           (d)  H₂ and Cl₂

                                                            (e)  NaCl

 

Answer:  A

 

10.  The isotope, ⁶³Ni, is a radioactive isotope that decays by beta emission to which of the following isotopes?

       (a)  ⁵⁹Fe                                                                      (c)  ⁶³Cu

       (b)  ⁶³Co                                                                      (d)  ⁶²Ni

                                                            (e)  ⁶⁴Zn

 

Answer:  C

 

 

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 particular reaction has DG° = –21.48 kJ/mol at a temperature of 237°C.  What is the value of the equilibrium constant, K, for this reaction?

 

DG° = –RTlnK

Perform the necessary conversion:

Placing these values into the expression above and solving for K:

12.  (15 points)  A galvanic electrochemical cell utililzes the following redox reaction:

3 Ce⁴⁺⁽aq)  +  Al(s)  ¾¾®  3 Ce³⁺(aq)  +  Al³⁺(aq)

(a)  (2 points) Which balanced redox half-reaction occurs at the anode?

The oxidation half-reaction occurs at the anode:  Al(s)  ¾¾®  Al³⁺⁽aq)  +  3 e¯

 

(b)  (2 points)  Which balanced redox half-reaction occurs at the cathode?

The reduction half-reaction occurs at the cathode:  Ce⁴⁺⁽aq)  +  e¯  ¾¾®  Ce³⁺(aq)

 

(c)  (3 points)  What is the value for E° for this electrochemical cell?  (Use the table of standard reduction potentials given in part 1)

E° = E_red(reduction) — E_red(oxidation) = +1.61 V – (–1.66 V) = + 3.27 V

 

 

 

(d)  (8 points)  What is the value of E for this electrochemical cell when [Ce⁴⁺] = 0.15 M, [Al³⁺] = 0.075 M and [Ce³⁺] = 0.050 M?  Assume that the temperature is equal to 298.15 K.

Using the Nernst equation: 

13.  (10 points)  A certain reaction has DH = –35.4 kJ and DS = –112.5 J/K at 35°C.  What is the value of DG, in units of kJ, for this reaction?  Is this reaction spontaneous at 35°C?

 

DG = DH – TDS

Performing the necessary conversions:

Plugging into the equation:

Since DG is negative, this reaction is spontaneous at 35°C.

 

 

 

14.  (10 points)  Using the table of standard reduction potentials, calculate the value of DG°, in units of kJ/mol, and K for the following redox reaction:  Fe(s)  +  Cd²⁺(aq)  ¾¾®  Fe²⁺(aq)  +  Cd(s)

 

Calculate E° for this reaction:

  oxidation:  Fe(s)  ¾¾®  Fe²⁺(aq)  +  2e¯     E_red = –0.440 V

reduction:  Cd²⁺(aq) +  2 e¯  ¾¾®  Cd(s)  E_red =  –0.403 V

E^° = E_red(reduction) – E_red(oxidation)  =  –0.403 V – (–0.440 V) = 0.037 V

Calculate DG° from DG° = –nFE°:

DG° = –(2)(96485 C/mol)(0.037 J/C) = –7100 J/mol = –7.1 kJ/mol

Calculate K from :

 

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

 

(a)  NO₃¯(aq)  +  SO₂(g)  ¾¾®  NO₂(g)  +  HSO₄¯(aq)     (acidic solution)

The skeletal half-reactions are:

NO₃¯   ¾¾®  NO₂

SO₂  ¾¾®  HSO₄¯

Balance the oxygen atoms by adding H₂O:

NO₃¯   ¾¾®  NO₂  + H₂O

SO₂  +  2 H₂O  ¾¾®  HSO₄¯

Balance the H atoms by adding H⁺:

NO₃¯  +  2 H⁺  ¾¾®  NO₂  + H₂O

SO₂  +  2 H₂O  ¾¾®  HSO₄¯  +  3 H⁺

Balance the charge by adding e¯’s:

NO₃¯  +  2 H⁺  +  e¯   ¾¾®  NO₂  + H₂O

SO₂  +  2 H₂O  ¾¾®  HSO₄¯  +  3 H⁺  +  2e¯

Multiply the 1^st half-reaction by 2 and then add to get the balanced redox reaction:

2 NO₃¯  +  SO₂  +  H⁺  ¾¾®  2 NO₂  +  HSO₄¯ 

 

(b)  NiO₂(s)  +  Cl¯(aq)  ¾¾®  Ni(OH)₂(s)  +  Cl₂(g)     (basic solution)

The skeletal half-reactions are:

NiO₂   ¾¾®  Ni(OH)₂

Cl¯  ¾¾®  Cl₂

Balance the Cl atoms in the second half-reaction:

NiO₂   ¾¾®  Ni(OH)₂

2 Cl¯  ¾¾®  Cl₂

Balance the H atoms by adding H⁺:

NiO₂  +  2 H⁺   ¾¾®  Ni(OH)₂

2 Cl¯  ¾¾®  Cl₂

Balance the charge by adding e¯’s:

NiO₂  +  2 H⁺  +  2e¯  ¾¾®  Ni(OH)₂

2 Cl¯  ¾¾®  Cl₂  +  2e¯

Add the two half-reactions to get the balanced redox reaction in acidic solution:

NiO₂  +  2 H⁺  +  2 Cl¯  ¾¾®  Ni(OH)₂  +  Cl₂

Convert to basic solution by adding 2 OH¯ to each side.  The 2 H⁺’s on the reactant side convert to H₂O:

NiO₂  +  2 Cl¯  +  2 H₂O  ¾¾®  Ni(OH)₂  +  Cl₂  +  2 OH¯

 

(c)  Cr₂O₇^2-(aq)  +  Cl¯(aq)  ¾¾®  Cr³⁺(aq)  +  OCl¯(aq)     (acidic solution)

Write the skeletal half-reactions:

Cr₂O₇^2-    ¾®  Cr³⁺

Cl¯  ¾¾®  OCl¯

Balance the Cr atoms in the 1^st half-reaction:

Cr₂O₇^2-    ¾®  2 Cr³⁺

Cl¯  ¾¾®  OCl¯

Balance the O atoms by adding H₂O:

Cr₂O₇^2-    ¾®  2 Cr^{3+  +}  7 H₂O

Cl¯ +  H₂O  ¾¾®  OCl¯

Balance the H atoms by adding H⁺:

Cr₂O₇^2- +  14H⁺   ¾®  2 Cr³⁺  +  7 H₂O

Cl¯ +  H₂O  ¾¾®  OCl¯ + 2H⁺

Balance the charge by adding e¯’s:

Cr₂O₇^2- +  14H⁺  +  6e¯  ¾®  2 Cr³⁺  +  7 H₂O

Cl¯ +  H₂O  ¾¾®  OCl¯ + 2H⁺  +  2e¯

Multiply the second half-reaction by 3 and add to get the balanced redox reaction:

Cr₂O₇^2-  +  3 Cl¯  +  8 H⁺  ¾¾®  2 Cr³⁺  +  3 OCl¯  +  4 H₂O

 

 

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.

Galvanic cell:  An electrochemical cell in which a spontaneous redox reaction generates an electric current.

 

 

Oxidizing agent:  A substance that causes another substance to get oxidized.

 

 

Reversible process:  A process in equilibrium.