1 / 6

Cell potentials and Reduction potentials

Cell potentials and Reduction potentials. The light at the end of the tunnel . Read 17.7 (pg. 716), do PE 7, 8 - use Ex. 17.8 (ignore Ex. 17.7). Also 17.62, 17.64 (731) Note: you need to multiply equations so that e – cancel out. However, unlike Hess’s law problems, DO NOT also multiply E°

ailis
Download Presentation

Cell potentials and Reduction potentials

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. Cell potentials and Reduction potentials

  2. The light at the end of the tunnel Read 17.7 (pg. 716), do PE 7, 8 - use Ex. 17.8 (ignore Ex. 17.7). Also 17.62, 17.64 (731) Note: you need to multiply equations so that e– cancel out. However, unlike Hess’s law problems, DO NOT also multiply E° PE 7 - NiO2 has the greater reduction potential, thus it is reduced and Fe is oxidized … NiO2 + 2H2O + 2e– Ni(OH)2 + 2OH– Fe + 2OH– Fe(OH)2 + 2e– NiO2 + 2H2O + Fe  Ni(OH)2 + Fe(OH)2 E°cell = E°reduced - E°oxidized = 0.49 V - -0.88 V = 1.37 V

  3. PE 8 PE 8 - MnO4– has the greater reduction potential, thus it is reduced and Cr is oxidized … MnO4– + 8H+ + 5e– Mn2+ + 4H2O Cr  Cr3+ + 3e– Electrons cannot exist in isolation (they must cancel out), so first x 3 and second x 5 3MnO4– + 24H+ + 15e– 3Mn2+ + 12H2O 5Cr  5Cr3+ + 15e– 3MnO4– + 24H+ + 5Cr  3Mn2+ + 12H2O + 5Cr3+ E°cell = E°reduced - E°oxidized = 1.49 V - -0.74 V = 2.23 V

  4. 17.62 17.62 We’re looking for half cells that contain: NO3– …  NO … and Fe2+  Fe3+ In table 17.1 we find: NO3– + 4H+ + 3e– NO + 2H2O E° = 0.96 V Fe2+  Fe3+ + e– E° = 0.77 V NO3– + 4H+ + 3e– NO + 2H2O E° = 0.96 V 3Fe2+  3Fe3+ + 3e– E° = 0.77 V NO3– + 4H+3Fe2+ 3Fe3+ + NO + 2H2O E°cell = E°reduced - E°oxidized = 0.96 V - 0.77 V = 0.19 V

  5. 17.64 BrO3– has the greater reduction potential, thus it is reduced and I– is oxidized … BrO3– + 6H+ + 6e– Br– + 3H2O 2I– I2 + 2e– Electrons cannot exist in isolation (they must cancel out), so second x 3 BrO3– + 6H+ + 6e– Br– + 3H2O 6I– 3I2 + 6e– BrO3– + 6H+ + 6I–  Br– + 3H2O + 3I2 E°cell = E°reduced - E°oxidized = 1.44 V - 0.54 V = 0.90 V

  6. The End For more lessons, visit www.chalkbored.com

More Related