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Electrochemistry Lesson 7 The Standard Hydrogen Cell

Electrochemistry Lesson 7 The Standard Hydrogen Cell. The Standard Hydrogen Half Cell The zero point of the reduction chart is the hydrogen half-cell. E is the cell potential or voltage E 0 is the standard cell potential @ 25 o C solutions are 1.0 M gases are 101 KPa.

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Electrochemistry Lesson 7 The Standard Hydrogen Cell

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  1. Electrochemistry Lesson 7 The Standard Hydrogen Cell

  2. The Standard Hydrogen Half Cell The zero pointof the reduction chart is the hydrogen half-cell. Eis the cell potential or voltage E0 is the standard cell potential @ 25 oC solutions are 1.0 M gases are 101 KPa

  3. All half reactions are compared to H2 Ag+ reacts with H2 spontaneously Eo = +0.80 v Zn2+ is nonspontaneous with H2Eo = -0.76 v The voltage for any combination is the difference

  4. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right. Formula Agent Type E0 HClO4 oxidizing

  5. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right.  Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing

  6. HNO3 is both H+ and NO3- Take higher one Stronger Oxidizing Agent

  7. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right.  Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing 0.96 v HCl oxidizing

  8. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right.  Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing 0.96 v HCl oxidizing 0.00 v SO42- reducing

  9. The reaction is reversed so the E0 = -2.01 v Oxidation

  10. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right.  Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing 0.96 v HCl oxidizing 0.00 v SO42- reducing -2.01 v MnO4- Acid oxidizing

  11. Using the Reduction Potential Chart Look on your reduction chartto find the voltage of each agent. Oxidizing agentsare on the left and reducing agentsare on the right.  Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing 0.96 v HCl oxidizing 0.00 v SO42- reducing -2.01 v MnO4- Acid oxidizing 1.51 v MnO4- Alkaline oxidizing

  12. Using the Reduction Potential Chart Look on your reduction chart to find the voltage of each agent. Oxidizing agents are on the left and reducing agents are on the right. Formula Agent Type E0 ClO4- Acid oxidizing 1.39 v HNO3 oxidizing 0.96 v HCl oxidizing 0.00 v SO42- reducing -2.01 v MnO4- Acid oxidizing 1.51 v MnO4- Alkaline oxidizing 0.60 v

  13. Formula Agent Type E0 H2SO4 oxidizing

  14. H2SO4→ HSO4- + H+ Don’t have SO42-

  15. Formula Agent Type E0 H2SO4 oxidizing 0.00 v H2O Neutral oxidizing

  16. Formula Agent Type E0 H2SO4 oxidizing 0.00 v H2O Neutral oxidizing -0.41 v H2O Neutral reducing

  17. Formula Agent Type E0 H2SO4 oxidizing 0.00 v H2O Neutral oxidizing -0.41 v H2O Neutral reducing -0.82 v Fe2+ reducing

  18. Formula Agent Type E0 H2SO4 oxidizing 0.00 v H2O Neutral oxidizing -0.41 v H2O Neutral reducing -0.82 v Fe2+ reducing -0.77 v Fe2+ oxidizing

  19. Formula Agent Type E0 H2SO4 oxidizing 0.00 v H2O Neutral oxidizing -0.41 v H2O Neutral reducing -0.82 v Fe2+ reducing -0.77 v Fe2+ oxidizing -0.45 v

  20. Determine if the reaction below is spontaneous. Write a balanced equation for the reaction and calculate the E0. 1. MnO4- & Mn2+ Acid

  21. The top reaction is written forward Reduction The bottom reaction is reversed and the voltage is changed to negative Oxidation

  22. Determine if the reaction below is spontaneous. Write a balanced equation for the reaction and calculate the E0. 1. MnO4- & Mn2+ Acid 2(MnO4- + 8H+ + 5e-→ Mn2++ 4H2O) 1.51 v 5(Mn2+ + 2H2O→ MnO2(s)+ 4H+ + 2e-)-1.22 v 2MnO4-+16H++5Mn2++10H2O → 2Mn2++8H2O+5MnO2(s)+20H+ 2MnO4- + 3Mn2+ + 2H2O → 5MnO2(s) + 4H+ 0.29 v simplify add potentials 3 2 4 0.29 v positive voltage- spontaneous

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