Department of Chemistry, Geosciences and Environmental Sciences


Exam 3-A-Key

Chemisty 1084: Section 010 Fall 2006




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


Gas Constant: R = 8.314 J/mol-K Faraday Constant: F = 96485 C/mol


Nernst Equation:

Cell potential and the equilibrium constant:


E = mc2 c = 2.998 108 m/s

1 amu = 1.661 10-27 kg



Part 1 (40 points):_____________________



Part 2 (60 points):_____________________



Total (100 points):_____________________



Dont 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 total entropy of the universe is ____________________.

(a) constant (c) continuously decreasing

(b) continuously increasing (d) zero

(e) equal to the energy of the universe


Answer: B

This is a statement of the 2nd Law of Thermodynamics.


2. Which phase has the smallest amount of entropy?

(a) solid (c) liquid

(b) gas


Answer: A

Gases have more entropy than liquids. Liquids have more entropy than solids.


3. A reaction is spontaneous at higher temperatures but becomes non-spontaneous at lower temperatures when DH is ___________ and DS is ______________.

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

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


Answer: A

Derived based on DG = DH TDS and the fact that a spontaneous reaction has a negative DG.


4. A spontaneous redox reaction will have a _______________ value for DG and a ______________ value for E.

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

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


Answer: B

Spontaneous reactions have a negative DG and a positive value for E.


5. Which reactant is the reducing agent in the reaction below?

Cr2O72-(aq) + 3 Ni(s) + 14H+(aq) 2 Cr3+(aq) + 3 Ni2+(aq) + 7 H2O(l)

(a) Cr2O72-(aq) (d) Ni2+(aq)

(b) Ni(s) (e) Cr3+(aq)

(c) H+(aq) (f) H2O(l)


Answer: B

The reducing agent gets oxidized. Ni(s) is being oxidized to Ni2+ in the reaction above.


6. The electrode at which the oxidation takes place is called the _______________.

(a) oxidizing agent (c) reducing agent

(b) cathode (d) anode

(e) voltaic cell


Answer: D

Answer questions 7 and 8 based on the following table of standard reduction potentials:


Standard Reduction Potentials at 298 K


Redox reaction

E (V)

Cr3+(aq) + 3e Cr(s)


Fe2+(aq) + 2e Fe(s)


Fe3+(aq) + e Fe2+(aq)


Sn4+(aq) + 2e Sn2+(aq)



7. The value of E for the redox reaction shown below is ______________ V.

Sn2+(aq) + 2 Fe3+(aq) Sn4+(aq) + 2 Fe2+(aq)

(a) +0.46 (c) +1.39

(b) +0.617 (d) 0.46

(e) +1.21


Answer: B

E = E(cathode) E(anode) = 0.771 V 0.154 V = 0.617 V



8. Which species is the strongest reducing agent?

(a) Cr(s) (c) Fe(s)

(b) Fe2+(aq) (d) Sn2+(aq)


Answer: A

The product in the half-reactions above with the most negative value of E is the strongest reducing agent.


9. Which of the following is the symbolic notation for an alpha particle in a nuclear reaction?

(a) (c)

(b) (d)


Answer: C


10. What is the daughter nucleus produced when 105Rh decays by beta emission?

(a) 101Tc (c) 105Pd

(b) 105Ru (d) 104Rh


Answer: C

Beta emission keeps the mass number the same but increases the atomic number by 1.



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) Given the thermodynamic data in the table below, calculate DH, DS, and DG (=DH TDS) at 298.15 K for the reaction shown below:

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


Thermodynamic Data
















DH = (2 mol H2O)(241.82 kJ/mol) + (2 mol SO2)(296.9 kJ/mol) (2 mol H2S)(20.17 kJ/mol) (3 mol O2)(0 kJ/mol)

= 483.64 kJ 593.8 kJ + 40.34 kJ

= 1037.1 kJ


DS = (2 mol H2O)(188.83 J/molK) + (2 mol SO2)(248.5 J/molK) (2 mol H2S)(205.6 J/molK) (3 mol O2)(205.0 J/molK)

= 377.66 J/K + 497.0 J/K 411.2 J/K 615.0 J/K

= 151.5 J/K


DG = DH TDS = 1037.1 kJ (298.15 K) = 1037.1 kJ + 45.18 kJ

= 991.9 kJ


All answers end in the tenths digit.

12. (15 points) A voltaic cell is constructed at 298.15 K which utilizes the following two redox half-reactions. The standard reduction potential for each half-reaction is given to the right of the reduction half-reaction:

Ce4+(aq) + e Ce3+(aq) E = +1.61 V

Cu2+(aq) + 2 e Cu(s) E = +0.337 V


(a) (2 points) Write the redox half-reaction that occurs at the anode.

The reverse of the half-reaction with the less positive E occurs at the anode:

Cu(s) Cu2+(aq) + 2 e


(b) (2 points) Write the redox half-reaction that occurs at the cathode.

The half-reaction with the more positive E occurs at the cathode:

Ce4+(aq) + e Ce3+(aq)


(c) (2 points) Write the balanced redox reaction that is associated with the voltaic cell.

When equalizing electrons gained and lost, the balanced redox reaction is the anode half-reaction plus twice the cathode half-reaction:

Cu(s) + 2 Ce4+(aq) Cu2+(aq) + 2 Ce3+(aq)


(d) (3 points) Calculate the standard cell potential, E, for this cell.

E = E(cathode) E(anode) = 1.61 V 0.337 V = 1.27 V


(e) (6 points) Calculate the cell potential, E, when [Cu2+] = 0.125 M, [Ce3+] = 0.175 M and [Ce4+] = 0.000525 M.

Use the Nernst equation:

E = 1.27 V (calculated in part 2), n = 2 (electrons gained and lost in balanced reaction)

Placing all the numbers into the Nernst equation:

13. (10 points) The value of DG is 2.25 kJ/mol for a reaction at 65C. What is the value of the equilibrium constant, K, for this reaction?


Use DG = RTlnK.

Performing the necessary conversions first:

DG = 2.25 kJ/mol(1000 J/kJ) = 2250 J/mol

T = 65C + 273 = 338 K


Place these values into the equation above and solve for K:



14. (10 points) Using the table of standard reduction potentials on page 3 of this exam, calculate the value of E and DG (in units of kJ/mol) for the following redox reaction:

2 Cr3+(aq) + 3 Fe(s) 2 Cr(s) + 3 Fe2+(aq)

The following reduction potentials come from the table on page 3:

Cr3+(aq) + 3 e Cr(s) E = 0.74 V = E(cathode)

Fe2+(aq) + 2 e Fe(s) E = 0.440 V = E(anode)


Calculate E for the cell: E = E(cathode) E(anode) = 0.74 V (0.440 V) = 0.30 V


DG can be calculated using DG = nFE. n = 6 (lowest common multiple of 2 and 3):

Convert to kJ/mol:

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


(a) SO42-(aq) + Zn(s) SO2(aq) + Zn2+(aq) (acidic solution)


The skeletal half-reactions are:

SO42- SO2

Zn Zn2+

Balance O atoms by adding H2Os:

SO42- SO2 + 2 H2O

Zn Zn2+

Balance H atoms by adding H+s:

SO42- + 4 H+ SO2 + 2 H2O

Zn Zn2+

Balance charge by adding es:

SO42- + 4 H+ + 2 e SO2 + 2 H2O

Zn Zn2+ + 2 e

Since electrons gained and lost are already equal, add the two half-reactions to get the balanced redox reaction:

SO42- + Zn + 4 H+ SO2 + Zn2+ + 2 H2O




(b) Bi(OH)3(s) + SnO22-(aq) Bi(s) + SnO32-(aq) (basic solution)



The skeletal half-reactions are:

Bi(OH)3 Bi

SnO22- SnO32-

Balance O atoms by adding H2Os:

Bi(OH)3 Bi + 3 H2O

SnO22- + H2O SnO32-

Balance H atoms by adding H+s:

Bi(OH)3 + 3 H+ Bi + 3 H2O

SnO22- + H2O SnO32- + 2 H+

Balance charge by adding es:

Bi(OH)3 + 3 H+ + 3 e Bi + 3 H2O

SnO22- + H2O SnO32- + 2 H+ + 2 e

Equalize the number of electrons gained and lost by multiplying the top half-reaction by 2 and the bottom by 3:

2 Bi(OH)3 + 6 H+ + 6 e 2 Bi + 6 H2O

3 SnO22- + 3 H2O 3 SnO32- + 6 H+ + 6 e

Add the two half-reactions:

2 Bi(OH)3 + 3 SnO22- 2 Bi + 3 SnO32- + 3 H2O

Since H+ ions do not appear in the balanced redox reaction, no conversion to basic solution is necessary.


(c) S2O32-(aq) + I2(s) I(aq) + S4O62-(aq) (acidic solution)



The skeletal half-reactions are:

S2O32- S4O62-

I2 I

Balance all the atoms besides O and H:

2 S2O32- S4O62-

I2 2 I

Since the O atoms are balanced on both sides, there is no need to place H2O molecules. Furthermore, since H atoms do not appear in either half-reaction, there is no need to place H+ ions. Balance the charge by adding es:

2 S2O32- S4O62- + 2 e

I2 + 2 e 2 I

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

2 S2O32- + I2 S4O62- + 2 I



(d) Mn2+(aq) + H2O2(aq) MnO2(s) + H2O(l) (basic solution)


The skeletal half-reactions are:

Mn2+ MnO2

H2O2 H2O

Balance O atoms by placing H2Os:

Mn2+ + 2 H2O MnO2

H2O2 2 H2O

Balance H atoms by placing H+s:

Mn2+ + 2 H2O MnO2 + 4 H+

H2O2 + 2 H+ 2 H2O

Balance charge by placing es:

Mn2+ + 2 H2O MnO2 + 4 H+ + 2 e

H2O2 + 2 H+ + 2 e 2 H2O

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

Mn2+ + H2O2 MnO2 + 2 H+

Convert to basic by adding 2 OH to each side. The 2 H+ are converted to 2 H2Os by the addition.

Mn2+ + H2O2 + 2 OH MnO2 + 2 H2O


(10 points extra credit) Define 2 out the 3 following terms. Credit will be given only for the complete definition. No partial credit will be given. Choose only 2 out of the three terms. If you give definitions for all three terms without clearly noting which ones you want graded, only the first two terms will be graded.



Oxidizing agent: a substance that causes another substance to be oxidized.





Reversible process: an equilibrium process





Electrolytic cell: an electrochemical cell that uses an electric current to drive a nonspontaneous redox reaction.