Houston Community College System
EXAM # 3
Chemistry 1412
Instructor : Dr. Komala Krishnaswamy
CHEM 1412 EXAM # 3
Name:
Score:
Part I-Please DO NOT circle. Write the correct answer in space provided. (3 points each)
_____1. Given the following standard reduction potentials in acid solution
O2 + 4H+ + 4e– → 2H2O
Sn4+ + 2e– → Sn2+
Zn2+ + 2e– → Zn(s)
E° = +1.23 V
E° = +0.13 V
E° = –0.76 V
Choose the strongest reducing agent.
A. O2
2+
D. Sn
B. Sn4+
E. Zn2+
C. Zn
_____2. Consider the following standard reduction potentials in acid solution:
Cr3+ + 3e– → Cr
E° = –0.74 V
2+
–
Co + 2e → Co
E° = –0.28 V
4+
–
2+
Sn + 2e → Sn
E° = +0.13 V
The strongest oxidizing agent listed above is
A. Cr+3
B. Sn4+
2+
D. Sn
E. Co2+
C. Cr
_____3. Consider an electrochemical cell based on the following cell diagram:
Pt | Pu3+(aq), Pu4+(aq) || Cl2(g), Cl– (aq) | Pt
Which of the following statements is incorrect?
A. The anode and the cathode are passive
B. Pt is oxidized and reduced
C. The oxidizing agent is Cl2
D. The reaction occurring at the anode is: Pu+3Pu+4 + eE. The product of reduction is Cl_____4. Consider the following reduction half reactions:
Pb2+ + 2e– → Pb
Cu2+ + 2e– → Cu
E° = –0.22 V
E° = +0.34 V
What will be the standard EMF, Eo, of a voltaic cell using Pb and Cu electrodes?
A. 0.56V
D. 1.12 V
B. -1.12V
E. 0.24V
C. 0.12V
_____5. For the reaction, 2Cr2+ + Cl2(g) → 2Cr3+ + 2Cl–, E°cell is 1.78 V. Calculate the standard free
energy change.
A. -85.9 kJ
D. -1370 kJ
B. -8240 kJ
E. -172 kJ
C. -343 kJ
_____6. The overall reaction 2Co3+(aq) + 2Cl–(aq) → 2Co2+(aq) + Cl2(g) has the standard cell
voltage E°= 0.46 V. Calculate the equilibrium constant for the cell reaction at 25oC.
A. 1.66 × 10-8
D. 3.63 × 1015
B. 2.76 × 10-16
E. 0
C. 6.02 × 107
_____7. According to the following cell diagram, which chemical species undergoes reduction?
Sn | Sn2+ || NO3– (acid soln), NO(g) | Pt
A. Sn
D. NO
B. Sn+2
E. Pt
C. NO3-
_____8. Ar-40 forms from the positron emission of which of the following radioactive nuclide?
A. Cl-40
D. K-40
B. Cl-39
E. Ar-41
C. K-41
_____9. Identify the missing target particle of the following nuclear transmutation: ?(18O, 4n) 263Sg
A. No-245
D. No-249
B. Ra-259
E. Cf-249
C. Cf-245
_____10. In the following nuclear transmutation reaction, 10B (X, )7Li, the particle X is a
A. An alpha paricle
D. positron
B. beta particle
E. proton
C. neutron
_____11.What is the product of electron capture by Lu-167?
A. Yb-167
D. Hf-167
B. Lu-166
E. Ta-171
C. Tm-163
_____12. When Th-234 decays to Pa-234, the mode of decay is
A, electron capture
D. A & C
B. alpha decay
E. beta decay
C. Positron emission
_____13. At-192 undergoes alpha decay to form which nucleus?
A. Tl-188
D. Bi-188
B. Tl-190
E. Po-192
C. Bi-190
_____14. Calculate the amount of energy released when Po-210 decays as follows:
210
Po 4 + 206Pb. The masses are: Po-210 = 209.982857 amu; He-4 = 4.002603 amu;
Pb-206 = 205.974449 amu & c = 3.00 × 108 m/s.
A. 4.14 × 109 kJ/mol
D. 4.66 × 109 kJ/mol
B. 7.20 × 1011 kJ/mol
E. 6.43 × 1012 kJ/mol
C. 5.22 × 108 kJ/mol
_____15. The active ingredient in Tylenol is the following compound,
Which of the following classifications is COMPLETELY correct?
A. It is an alcohol & an amine
B. It is a Ketone, alcohol & an amine
C. It is a phenol & an amide
D. It is a nitrile and an alcohol
E. It is a cycloalkane, an amine, an alcohol & a ketone
_____16. Ethyl methyl ether and isopropyl alcohol are constitutional isomers. Which of the
following statements is correct?
A.
B.
C.
D.
E.
They have identical physical properties
They have identical chemical properties
They have identical molar masses
They have identical structures
Two of these
_____17. Which structure is neopentane?
A
B
C
D
E
_____18. Which of the following are structural isomers of n-heptane?
i.
iv.
Cycloheptane
2-methylhexane
A. iii. only
D. ii. Only
ii. 3,3-dimethylpentane
iii. 2-methylheptane
B. i & iii
E. i only
C. ii & iv
_____19. Name the compound according to IUPAC nomenclature.
CH3
CH CH CH CH3
CH3 CH2
A. 3,5-Dimethyl- 4- ethylheptane
C. 2,3,4-Triethylpentane
E. 4-methy-2,3-diethylhexane
CH2 CH2 CH3
CH3
B. 4-Ethyl-3,5-dimethylheptane
D. 2,3- Diethyl-4-methylhexane
_____20. Which of these organic compounds contains an ether linkage:
A.
B.
D.
E. All of them
C.
Part II-Essay questions(Please show all your work for complete credit.
21. a. Complete and balance the following redox equation using the set of smallest whole-number
coefficients.
BrO3–(aq) + Sb3+(aq) → Br–(aq) + Sb5+(aq) (acidic solution)
b. Identify the reducing agent & the oxidizing agent.
22. a. Balance the equation for this reaction carried out in a basic solution:
PO33-(aq) + MnO4-(aq) → PO43-(aq) + MnO2(s)
b. Identify the reducing agent & the oxidizing agent
23. Calculate the nuclear binding energy per nucleon, in joules, for Mg-25 (nuclear mass 24.985839
amu). Additional data: mass of proton= 1.007825 amu, mass of neutron= 1.008665 amu;
1 g = 6.022 × 1023 amu; c = 3.00 × 108 m/s.
24. How old is a bottle of wine if the tritium (3H) content is 25% that of a new wine? The half-life
of tritium is 12.5 years.
25. An electroplating solution is made up of NiSO4. How much time would it take to
deposit 500 mg of nickel using a current of 3.00 A?
26. (10 Points) Consider a voltaic cell based on the following reduction half-reactions:
Zn2+ + 2e– → Zn(s)
E° = –0.76 V
3+
–
Cr + 3e → Cr
E°red= –0.74 V
A. What is the overall cell reaction?
B. Calculate the associated standard free energy change.
C. Calculate the equilibrium constant
D. Calculate the free energy change when [Zn+2] = 0.0500M & [Cr+3] = 0.0750 M
E. Is the reaction spontaneous under these conditions?
Part I (3 points each)
1. C
2. B 3. B 4. A 5. C 6. D 7. C 8. D 9. E 10. C
11. A
12. E 13. D 14. C 15. C 16. C 17. B 18. C 19. B 20. A
Part II (8 points each)
21. a.
BrO3- + 6H+ + 6 e- Br- + 3H2O: reduction half-reaction
3Sb+33Sb+5 + 6e-: oxidation half-reaction
Net reaction: BrO3- + 6H+ +3Sb+3 Br- + 3H2O + 3Sb+5
b. Reducing agent: Sb+3; Oxidizing agent: BrO322. a.
3PO3-3 + 3H2O  3PO43-+ 6H+ + 6e-: oxidation
2MnO4- + 8H+ + 6e- 2MnO2 + 4H2O: reduction
Net reaction in acidic solution: 3PO3-3 +2MnO4- + 2H+3PO43-+2MnO2 + H2O
Net reaction in basic solution: 3PO3-3 +2MnO4- + 2H++ 2OH-3PO43-+2MnO2 + H2O+ 2OHwhich simplifies to 3PO3-3 +2MnO4- + H2O  3PO43-+2MnO2 + 2OH-
b. Reducing agent: PO3-3; Oxidizing agent: MnO423.
Mg-25 has 12 protons, 13 neutrons = 25 nucleons
∆m= (12 ×1.007825) + (13 ×1.008665)- 24.985839 amu = 0.220706 amu =
(0.220706/(6.022 × 1023) g = 3.66 × 10-25g = 3.66 × 10-28kg
∆E = ∆m × c2 = 3.66 × 10-28kg × 9.00 × 1016m2/s2 = 3.30 × 10-11J = total binding energy
Therefore binding energy per nucleon = 3.30 × 10-11J/25 nucleons = 1.32 × 10-12J/nucleon
24. ln(25) = ln(100)-(0.693/12.5) × t
(0.693/12.5) × t = ln(100)-ln(25) = 1.39
t = 12.5 × 1.39/0.693 yr = 25 yr
25.
Ni+2 + 2e-  Ni
1 mol Ni needs 2 mol e500 mg Ni = 0.500 g Ni = (0.500/58.69) mol Ni = 0.00852 mol Ni
Mol e- needed = 0.00852 mol Ni × 2 mol e-/1 mol Ni = 0.0170 mol e- = n
s = nF/A = 0.0170 × 96485/3.00 s = 548 s
26. (10 Points) Consider a voltaic cell based on the following reduction half-reactions:
3Zn2+ + 6e– → 3Zn(s)
E° = –0.76 V
3+
–
2Cr + 6e → 2Cr
E°red= –0.74 V
+3
+2
F. 2Cr + 3Zn3Zn + 2Cr
G. Eo = 0.02 V; ∆G = -6 × 96485 × 0.02 J = -1.16 × 104J
H. K= 106
I. Q = [0.05]3/[0.075]2 = 0.0222; E = 0.02- (0.0592/6) × log(0.0222) = 0.036 V
J. Yes, spontaneous