1 / 21

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:

lan
Télécharger la présentation

Chapter 20 Electrochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related