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Reduction- Oxidation Reactions (1)

Reduction- Oxidation Reactions (1). 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6 th edition. By the end of the lecture the student should be able to:. Define oxidation, reduction, oxidizing agent, and reducing agent.

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Reduction- Oxidation Reactions (1)

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  1. Reduction- Oxidation Reactions (1) 213 PHC 10th lecture (1) Gary D. Christian, Analytical Chemistry, 6th edition.

  2. By the end of the lecture the student should be able to: • Define oxidation, reduction, oxidizing agent, and reducing agent. • Understand the principals of electrochemical cell and electrode potential • Calculate the electrode potential by Nernest equation.

  3. It is a reaction occurs between an oxidizing agent and a reducing agent. What is a Redox reaction?

  4. OXIDATION REDUCTION A loss of electrons to give a higher oxidation state Fe2+ Fe3+ + e- A gain of electrons to give a lower oxidation state Ce4+ + e- Ce3+

  5. Oxidizing Agent Reducing Agent • Take on electrons • Gets reduced Ce4+ + e- Ce3+ • Give up electrons • Gets oxidized Fe2+ Fe3+ + e-

  6. Fe2+ + Ce4+  Fe3+ + Ce3+     reducing oxidizing oxidized reduced agent agent form form (Reduced form)(oxidized form)

  7. The reducing or oxidizing tendency of a substances will depend on its reduction potential.

  8. Electrode Potential (Eo) • Each half-reaction will generate a potential that adopted by an inert electrode dipped in the solution. • Individual electrode potential can’t be measured. • The difference between 2 electrode potentials can be measured. • The standard hydrogen electrode is used to measure the potential of any half reaction because it’s potential is zero.

  9. The more +veEo= (oxidation). • The more -veEo= (reduction).

  10. Fe3+ + e- Fe2+Eo= 0.771 V Sn4+ + 2e-  Sn2+Eo= 0.154 V 2Fe3+ + Sn2+  2Fe2+ + Sn4+

  11. Questions?

  12. Example Fe3+ + e- = Fe2+Eo = 0.771 V I3- + 2e- = 3I-Eo = 0.535 V 2Fe3+ + 3I- = 2Fe2+ + I3- Eocell= 0.771 – 0.535 = 0.235 V

  13. Quiz • What is the overall cell reaction and the cell potential for the two half-reactions? A) Cu2+ + 2e = Cu Eo = 0.34 V Zn2+ + 2e = Zn Eo = -0.76 V B) Fe3+ + e = Fe2+Eo = 0.77 V Ti4+ + e = Ti3+Eo= 0.15 V

  14. The Nernst equation

  15. The Nernst equation Describes the dependence of potential on concentration.

  16. E = Eo – 2.3026 RT/ n F log [Red] / [Ox] E = reduction potential at specific conc. Eo=standard potential. n = no. of electrons R = gas const. (8.3143) T = absolute temp. F = Faraday const. (96.487)

  17. At 25oC E = Eo – (0.05916/ n) log [Red] / [Ox]

  18. Questions?

  19. Summary: • Definition of oxidation-reduction reactions. • Electrochemical cells. • Electrode potential. • Nernast equation

  20. Thank you

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