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Chapter 20 Electrochemistry

Chapter 20 Electrochemistry. Chemistry, The Central Science , 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall. Which species is oxidized and which is reduced in the following reaction:

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Chapter 20 Electrochemistry

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  1. Chapter 20Electrochemistry Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Troy Wood University of Buffalo Buffalo, NY  2006, Prentice Hall

  2. Which species is oxidized and which is reduced in the following reaction: Zn(s) + 2 H+(aq)  Zn2+(aq) + H2(g) Zn, oxidized; H+, reduced H+, reduced; Zn, oxidized Zn2+, oxidized; H2, reduced H2, oxidized; Zn2+, reduced

  3. Correct Answer: Zn, oxidized; H+, reduced H+, reduced; Zn, oxidized Zn2+, oxidized; H2, reduced H2, oxidized; Zn2+, reduced The oxidation state of Zn goes from 0 to +2 while the oxidation state of H goes from +1 to 0.

  4. Balance the following oxidation-reduction reaction that occurs in acidic solution: C2O42 + MnO4 Mn2+ + CO2 • 8 H+ + 5 C2O42 + MnO4 Mn2+ + 4 H2O + 10 CO2 • 16 H+ + 2 C2O42 + 2 MnO4 2 Mn2+ + 8 H2O + 4 CO2 • 16 H+ + 5 C2O42 + 2 MnO4 2 Mn2+ + 8 H2O + 10 CO2 • C2O42+ MnO4 Mn2+ + 2 CO2 + 2O2

  5. Correct Answer: • 8 H+ + 5 C2O42 + MnO4 Mn2+ + 4 H2O + 10 CO2 • 16 H+ + 2 C2O42 + 2 MnO4 2 Mn2+ + 8 H2O + 4 CO2 • 16 H+ + 5 C2O42 + 2 MnO4 2 Mn2+ + 8 H2O + 10 CO2 • C2O42+ MnO4 Mn2+ + 2 CO2 + 2O2 Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

  6. Balance the following oxidation-reduction reaction that occurs in basic solution: CN + MnO4 CNO + MnO2 • CN + MnO4 + 2 OH CNO + MnO2 + H2O • 2 CN + 2 MnO4 + 2 OH 2 CNO + 2 MnO2+ 4 OH • 2 CN + MnO4 2 CNO + MnO2 + O2 • 3 CN + 2 MnO4 3 CNO + 2 MnO2 + 2 OH

  7. Correct Answer: • CN + MnO4 + 2 OH CNO + MnO2 + H2O • 2 CN + 2 MnO4 + 2 OH 2 CNO + 2 MnO2+ 4 OH • 2 CN + MnO4 2 CNO + MnO2 + O2 • 3 CN + 2 MnO4 3 CNO + 2 MnO2 + 2 OH Conservation of mass and charge must be maintained on both reactants’ and products’ side; practice using the method of half-reactions.

  8. Calculate the emf of the following cell: Zn(s)|Zn2+(aq, 1 M)|| H+(aq, 1 M)|H2(g, 1 atm)|Pt E° (Zn/Zn2+)= 0.76 V. +0.76 V +1.52 V 0.76 V 1.52 V

  9. Correct Answer: +0.76 V +1.52 V 0.76 V 1.52 V E°cell = E°cathodeE°anode Zn is the anode, hydrogen at the Pt wire is the cathode. E°cell = E°cathodeE°anode = 0.00 V  (0.76 V) E°cell = +0.76 V

  10. Calculate the emf produced by the following voltaic cell reaction: Zn+ 2 Fe3+  Zn2+ + 2 Fe2+ Zn2+ + 2 e Zn E° = 0.76 V Fe3+ + e Fe2+ E° = 0.77 V +0.01 V +0.78 V 0.78 V +1.53 V

  11. Correct Answer: +0.01 V +0.78 V 0.78 V +1.53 V E°cell = E°cathodeE°anode Zn is being oxidized at the anode and Fe3+ is being reduced at the cathode. Thus, E°cell = E°cathodeE°anode = 0.77 V  (0.76 V) E°cell = +1.53 V

  12. As written, is the following oxidation-reduction equation spontaneous or non-spontaneous? Zn2+ + 2 Fe2+Zn+ 2 Fe3+ Zn2+ + 2 e Zn E° = 0.76 V Fe3+ + e Fe2+ E° = 0.77 V Spontaneous Nonspontaneous

  13. Correct Answer: In this case, the reduction process is Zn2+ Zn while the oxidation process is Fe2+ Fe3+. Thus: Spontaneous Nonspontaneous E° = E°red (reduction)E°red (oxidation) E° = 0.76 V − (0.77 V) = 1.53 V A negative E° indicates a nonspontaneous process.

  14. Calculate the emf produced by the following voltaic cell reaction. [Zn2+] = 1.0 M, [Fe2+] = 0.1 M, [Fe3+] = 1.0 M Zn+ 2 Fe3+  Zn2+ + 2 Fe2+ Zn2+ + 2 e Zn E° = 0.76 V Fe3+ + e Fe2+ E° = 0.77 V +1.47 V +1.53 V +1.59 V

  15. Correct Answer: (0.0592) = - ° E E log Q n [ ] [ ] +1.47 V +1.53 V +1.59 V 2 + + 2 2 (0.0592) Fe Zn = - E 1.53 log [ ] 2 2 + 3 Fe [ ] [ ] 2 (0.0592) 0.1 1.0 = - E 1.53 log [ ] 2 2 1.0 (0.0592) = - = + = E 1.53 log (0.01) 1.53 0.0592 1.59 2

  16. A primary battery cannot be recharged. Which of the following batteries fits this category? • Lead-acid battery • Nickel-cadmium • Alkaline battery • Lithium ion

  17. Correct Answer: • Lead-acid battery • Nickel-cadmium • Alkaline battery • Lithium ion In this list, only the alkaline battery is a primary battery and is thus nonrechargeable.

  18. Based on the standard reduction potentials, which metal would not provide cathodic protection to iron? Magnesium Nickel Sodium Aluminum

  19. Correct Answer: In order to provide cathodic protection, the metal that is oxidized while protecting the cathode must have a more negative standard reduction potential. Here, only Ni has a more positive reduction potential (0.28 V) than Fe2+ (0.44 V) and cannot be used for cathodic protection. Magnesium Nickel Sodium Aluminum

  20. Ni2+ is electrolyzed to Ni by a current of 2.43 amperes. If current flows for 600 s, how much Ni is plated (in grams)? (AW Ni = 58.7 g/mol) 0.00148 g 0.00297 g 0.444 g 0.888 g

  21. Correct Answer:   i t FW = mass  n F 0.00148 g 0.00297 g 0.444 g 0.888 g ( )   2.43 A (600. s) (58.7 g/mol) = mass  (2 96,500 C/mol)

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