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Oxidation-Reduction

Oxidation-Reduction. Larry J. Scheffler Lincoln High School. LEO. LEO says. GER!. GER!. LEO says. L oss of Electrons = Oxidation Gain of Electrons = R eduction. Oxidation Numbers. Oxidation is the loss of electrons; Reduction is the gain of electrons

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Oxidation-Reduction

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  1. Oxidation-Reduction Larry J. Scheffler Lincoln High School

  2. LEO

  3. LEO says GER!

  4. GER! LEO says Loss of Electrons = Oxidation Gain of Electrons =Reduction

  5. Oxidation Numbers • Oxidation is the loss of electrons; Reduction is the gain of electrons • Oxidation and reduction go together. Whenever a substance loses electrons and another substance gains electrons • Oxidation Numbers is a system that we can use to keep track of electron transfers

  6. Oxidation Numbers

  7. Practice Assigning Oxidation Numbers

  8. Practice Assigning Oxidation Numbers

  9. Using Oxidation Numbers • Careful examination of the oxidation numbers of atoms in an equation allows us to determine what is oxidized and what is reduced in an oxidation-reduction reaction.

  10. An increase in the oxidation number indicates that an atom has lost electrons and therefore oxidized. A decreasein the oxidation number indicates that an atom has gained electrons and therefore reduced Example Zn + CuSO4 ZnSO4 + Cu 0 +2 +6-2 +2+6-2 0 Zn: 0+ 2Oxidized Cu: +20Reduced Using Oxidation Numbers

  11. Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2 Cu + HNO3 Cu(NO3)2+ NO2 + H2O HNO3 + I2 HIO3 + NO2

  12. Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2 0 +1-1 +1-10 Br increases from –1 to 0 -- oxidized Cl decreases from 0 to –1 -- Reduced K remains unchanged at +1

  13. Exercise For each of the following reactions find the element oxidized and the element reduced Cu + HNO3 Cu(NO3)2+ NO2 + H2O 0 +1+5-2 +2 +5-2 +4 –2 +1-2 • Cu increases from 0 to +2. It is oxidized • Only part of the N in nitric acid changes from +5 to +4. It is reduced • The nitrogen that ends up in copper nitrate remains unchanged

  14. Exercise For each of the following reactions find the element oxidized and the element reduced HNO3 + I2 HIO3 + NO2 1 +5 -2 0 +1+5-2+4-2 • N is reduced from +5 to +4. It is reduced. • I is increased from 0 to +5 It is oxidized • The hydrogen and oxygen remain unchanged.

  15. Oxidation-Reduction Reactions • All RedOx reactions have one element oxidized and one element reduced. • The compound that supplies the electrons (is oxidized) is the reducing agent. • The compound that accepts the electrons (is reduced) is the oxidizing agent. • Occasionally the same element may undergo both oxidation and reduction. This is known as an auto-oxidation reduction

  16. Balancing Redox Reactions • Many chemical reactions involving oxidations and reductions are complex and very difficult to balance by the “guess and check” methods we learned earlier. • For complicated reactions, a more systematic approach is required.

  17. Balancing Redox Reactions There are several basic steps • Assign oxidation numbers to the species in the reaction to find which atom is oxidized and which is being reduced. (Just learned this.) • Write half reactions for the oxidation and reduction Take compounds where “action” took place, split them and write them as individual reactions. There will be 2 half reactions. • Balance the atoms that change in the half reaction. If there are atoms of an element on one side, there must be EQUAL number of those same atoms on the other side. Use coefficients in front of atoms/compounds. (this does NOT include H or O) 4. Balance each half-equation for Oxygen by adding as many H2O to the opposite side that need Oxygens.

  18. Balancing Redox Reactions There are several basic steps (cont.) • Balance each half-equation for Hydrogen by adding as many H+ needed to the opposite side from water was added. • Balance each half-equation for charge by adding as many e- needed to get equal charges on each side of the half reactions. • The number of electrons in EACH half-reaction must be equal (and on opposite sides). Multiply any half-equations to get the # of electrons to be equal. • Combine the half reactions, cancelling out anything that is the same on both sides, including electrons. • Check your work. Make sure that both the atoms and charges balance

  19. Exercise Balancing Redox Equations 1 Cu + HNO3 Cu(NO3)2+ NO + H2O Balancing Redox Equations 2 HNO3 + I2 HIO3+ NO2 + H2O

  20. Metal Displacement Reactions • The electrochemical cell potentials form the basis for predicting which metals will react with salt solution of other metals • This order of reactivity of metals in single replacement reactions is called the activity series • The solid of more reactive metals will displace ions of a less reactive metal from solution. **(Higher metal will replace the ions lower )** 20

  21. Metal Displacement Reactions **(Higher metal will replace the ions lower)** • Oxidizing and reducing agents are not the same strength • The relative reactivity of metals is based on potentials of half reactions (reductions). • Elements with very different potentials react most vigorously. More reactive the metal, the stronger reducing agents (more easily oxidized). • They show the potential difference, in volts, between the electrodes of an electrochemical cell. • A positive valueindicates a spontaneous reaction (indicates that the direction is positive.) 21

  22. Standard Electrode Potentials in Aqueous Solution at 25°C Reference Point 22

  23. Electrochemical Cells The Metal Activity Series (most reactive to least reactive) Oxidized (-0.76V) Reduced (+0.34V) 23

  24. The Activity Series • Elements with highly negative reduction potentials are not easily reduced but they are easily oxidized. • Since metals react by being oxidized, the more negative the reduction potential, the more reactive the element. • Elements higher in the table (more negative potential) can displace any ions lower (more positive potential). • So Zn + CuCl2 ZnCl2 + Cu Cu + ZnCl2 No Reaction The activity Series is really a reduction potential table arranged from negative to positive 24

  25. Calculating Cell Potentials From Standard Reduction Potenials From the standard reduction table: Zn+2 + 2 e-  Zn - 0.763 v Ag+ + e- Ag + 0.799v • Calculate the cell potential for a cell made from silver and zinc electrodes. Since there must be one oxidation and one reduction, the direction of one of two half reactions above must be reversed. Reversing the zinc half reaction making it the oxidation would yield a positive cell potential Zn  Zn+2 + 2 e- (+0.763 V) Ag+ + e- Ag (+0.799 V) Zn(s) + Ag+(aq) Zn+2(aq) + Ag(s) Cell potential = +1.562 volts 25

  26. Application Question 2 Mg2+ + 2e-  Mg E°= - 2.37 V Zn2+ + 2e-  Zn E°= - 0.76 V • Does Zn react with Mg2+? • Does Mg react with Zn2+? If Mg is oxidized Mg  Mg2+ + 2e- Eo = +2.37 v Combining this with the reduction of Zn2+Eo = - 0.76 v Leaves an overall positive cell potential + 1.66 v Therefore Mg reacts with Zn2+. Zn does not react with Mg2+ 26

  27. Electrochemical Cells • Two types of cells • Voltaic, Galvanic • Electrolytic • One cell uses chemistry to make electrical energy. • Voltaic (battery) • One cell uses electrical energy to make chemistry. • Electrolytic

  28. Voltaic Cells Remember: to be spontaneous, must have a positive cell potential (based on Standard Potential Table). In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

  29. Voltaic Cells • A typical VOLTAIC cell looks like this. • The oxidation occurs at the anode.(OIL) • The reduction occurs at the cathode. (RIG)

  30. Voltaic Cells • Electrons will not flow unless there is a completed circuit. • Salt bridge is used, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. • Cations move toward the cathode. • Anions move toward the anode.

  31. Voltaic Cells • Reduction happens at the Cathode and is PAWSitive (positive) • Oxidation happens at the Anode. (negative) • Memory Device: • Red Cat. has Paws • An Ox Electrons still flow from anode to cathode.

  32. Voltaic Cells • At Anode, electrons leave atoms, become ions and moves into solution (erodes electrode) • At Cathode, electrons join ions (in solution) and comes metal atoms (plates onto electrode) e- are released from atom, becomes ion Ion comes close and gains e-, becomes metal

  33. Shorthand Notation for Electrochemical cells • The shorthand representation of an electrochemical cell showing the two half-cells connected by a salt bridge or porous barrier, such as: Zn(s)/ZnSO4(aq)//CuSO4(aq)/Cu(s) anodecathode The electrodes are shown on the ends and the electrolytes for each side are shown in the middle. 33

  34. Batteries Are Applications of Electrochemical Cells Batteries • device that converts chemical energy into electricity Primary Cells • non-reversible electrochemical cell • non-rechargeable cell Secondary Cells • reversible electrochemical cell • rechargeable cell 34

  35. A Common Dry Cell 35

  36. A 9 Volt Dry Cell 36

  37. Nickel-Cadmium (Ni-Cad) Cathode NiO2(s)+ 2 H2O (l) +2e- Ni(OH)2 (s) + 2OH- (aq) Anode Cd (s) + 2OH- (aq) Cd(OH)2 (s) +2e- Overall reaction Cd(s) + 2 Ni(OH)3(s) = Cd(OH)2(s) + 2 Ni(OH)2(s) E NiCad = 1.25 v/cell 37

  38. Lead-Acid (Car Battery) 38

  39. Lead-Acid (Car Battery) Cathode PbO2(s) + SO42- (aq) + 4H+ (aq) +2e- PbSO4 (s) + 2 H2O (l) Anode Pb (s) + SO42 (aq) PbSO4 (s) +2e- Overall reaction Pb (s) + PbO2 (s) + 2 H2SO4 (aq) = 2 PbSO4 (s) + 2 H2O (l) E = 2.0- volts per cell 39

  40. Electrolysis 40

  41. Electrolytic Cells • Electrolytic cells use electrical energy to bring about a non-spontaneous reaction. • An electrolytic cell is made up of three parts. • Source of direct current (Battery) • Two electrodes • Cathode (negative pole) – reduction *NOTE: no longer PAWS-itive • The Red Cat has lost its PAWS. • Anode (positive pole) – oxidation • An electrolytic (ionic compound) that balances the flow of electrons through the circuit. • Electrons still flow from anode to cathode. *Note: electrons are “pushed” and “pulled” by the battery

  42. Cell Construction vessel - + battery power source e- e- conductive medium (-) (+) inert electrodes Sign or polarity of electrodes

  43. What chemical species would be present in a vessel of molten sodium chloride, NaCl (l)? Na+ Cl- Let’s examine the electrolytic cell for molten NaCl.

  44. Molten NaCl Observe the reactions at the electrodes - + battery Cl2 (g) escapes Na (l) NaCl (l) Na+ Cl- Na+ Cl- (-) (+) electrode half-cell electrode half-cell Cl- Na+ Na+ + e- Na 2Cl- Cl2 + 2e-

  45. Molten NaCl At the microscopic level - + battery e- NaCl (l) cations migrate toward (-) electrode anions migrate toward (+) electrode Na+ Cl- Na+ e- Cl- (-) (+) anode cathode Cl- Na+ 2Cl- Cl2 + 2e- Na+ + e- Na

  46. A Typical Electrolytic Cell

  47. Molten NaCl Electrolytic Cell cathode half-cell (-) REDUCTION Na+ + e- Na (Red Cat has lost its PAWS-itive) anode half-cell (+) OXIDATION 2Cl- Cl2 + 2e- overall cell reaction 2Na+ + 2Cl- 2Na + Cl2 X 2 Non-spontaneous reaction!

  48. Corrosion • Corrosion of metals is a common oxidation-reduction process in nature. • The rusting of iron can be thought of as a form of an electrochemical cell. 48

  49. Rusting of Iron 49

  50. Rusting Iron O2(g) + 4 H+(aq) + 4 e- 2 H2O(l) Eo = 1.23V Rusting Process Fe(s) Fe+2(aq) + 2 e- Eo = 0.44 V O2(g) + 4 H+(aq) + 4 e- 2 H2O(l) Eo = 1.23 V --------------------------------------------------------- -------------- 2 Fe(s) + O2(g) + 4 H+(aq)2 H2O(l) + Fe+2(aq) Eo = 1.67 V 50

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