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+3Pu+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+33Sb+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 + 3Zn3Zn + 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