1 / 60

Electrochemistry

Electrochemistry. Chapter 20. Oxidation-Reduction Reactions. Electrochemistry study conversion of chemical reactions to electrostatic work It must involve oxidation-reduction reactions, for example: Zn + Cu 2+  Zn 2+ + Cu - D G

teresa
Download Presentation

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. Electrochemistry Chapter 20 Chapter 20

  2. Oxidation-Reduction Reactions • Electrochemistry study conversion of chemical reactions to electrostatic work • It must involve oxidation-reduction reactions, for example: Zn + Cu2+ Zn2+ + Cu -DG • If the reaction take place in solution, electron transfer occurs in the solution, the energy is released in term of heat, does not generate electricity Chapter 20

  3. Oxidation-Reduction Reactions Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g). • The oxidation number of Zn has increased from 0 to 2+. • The oxidation number of H has reduced from 1+ to 0. • Therefore, Zn is oxidized to Zn2+ while H+ is reduced to H2. • H+ causes Zn to be oxidized and is the oxidizing agent. • Zn causes H+ to be reduced and is the reducing agent. • Note that the reducing agent is oxidized and the oxidizing agent is reduced. Chapter 20

  4. Oxidation-Reduction Reactions • However, if the reactions are separated into two half-reactions,: • Zn  Zn2+ + 2e • Cu2+ + 2e  Cu • and connect the half reactions with a wire to transfer electrons (outside of the solution), then the reaction causes electric current, makes electric work. • The free energy change is the maximum non-volume work: Chapter 20

  5. Oxidation-Reduction Reactions • If the reactions are reversible • Electric work: • For n mole electrons, the total charge • Faraday constant is the magnitude of electric charge per mole of electrons. • Then Chapter 20

  6. Balancing Oxidation-Reduction Equations • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. • Conservation of charge: electrons are not lost in a chemical reaction. • In complicated redox reactions, we need to look at the transfer of electrons carefully. Chapter 20

  7. Balancing Oxidation-Reduction Equations • Half-Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions. • The half-reactions for • Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq) • are • Sn2+(aq)  Sn4+(aq) +2e- • 2Fe3+(aq) + 2e- 2Fe2+(aq) • Always use +e Chapter 20

  8. Balancing Oxidation-Reduction Equations • For acidic solution • 1. Write down the two half reactions. • 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. Then balance charge by adding electrons • 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. Chapter 20

  9. Balancing Oxidation-Reduction Equations Write oxidation-reduction equation for the following reaction in acidic solusion MnO4- + C2O42- => Mn2+(aq) + CO2(g) Chapter 20

  10. Balancing Oxidation-Reduction Equations 1. The half reactions are MnO4-(aq)  Mn2+(aq) C2O42-(aq)  2CO2(g) 2. Adding water and H+ if necessary 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O C2O42-(aq)  2CO2(g) Chapter 20

  11. Balancing Oxidation-Reduction Equations 3. Adding electrons : 5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O C2O42-(aq)  2CO2(g) + 2e- 4. To balance electrons, multiplying 2 and 5 gives: 10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O 5C2O42-(aq)  10CO2(g) + 10e- 5. Adding gives: 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) balanced! Chapter 20

  12. Balancing Oxidation-Reduction Equations • Balancing in Basic Solution • Use OH- and H2O rather than H+ and H2O. • Others are the same. Chapter 20

  13. Voltaic Cells • The energy released in a spontaneous redox reaction is used to perform electrical work. • Voltaic cells are devices in which the electron transfer of redox reactions occurs via an external circuit. • The oxidation side is anode, the reduction side is cathode. Anode - > produce electron Cathode -> consume electron • Need a bridge to allow ions flow so that the solution is charge neutral. Chapter 20

  14. Voltaic Cells Chapter 20

  15. Voltaic Cells Chapter 20

  16. Voltaic Cells • A Molecular View of Electrode Processes • Metals consist metal ions and electrons • In contact with solution, depends on Gibbs free energy difference for an ion in solution and in the metal, ions may enter solution from solid or enter the metal solid from solution • The ions are attracted to the surface due to electrostatic interactions (intend to diffuse due to thermal motions) • Form electrostatic double layers Chapter 20

  17. Voltaic Cells • A Molecular View of Electrode Processes • At equilibrium, the double layer cause electrostatic potential. • If electrons are allowed to move, then the equilibrium moves, continue redox reactions generate electric currency • The currency generates work. Chapter 20

  18. Voltaic Cells • A Molecular View of Electrode Processes • Consider the spontaneous redox reaction between Zn(s) and Cu2+(aq). • At anode, Zn2+ enter solution, leave 2e on the surface • At cathode, Cu2+enter the metal, leave X- in the solution. • This process build up the electric potential • Once the circuit is open, electron flows, the redox reactions continue. Chapter 20

  19. Cell EMF • Potential difference is a sum of three parts: near the anode, in the solution body, and near the cathode: • The salt bridge keeps the center as zero • The difference in electrical potential. Measured in volts. • One volt is the potential difference required to impart one joule of energy to a charge of one coulomb: Chapter 20

  20. Cell EMF • Electromotive force (emf) is the potential difference between anode and cathode. • Cell potential: Ecell is the emf of a cell. • For 1M solutions at 25 C (standard conditions), the standard emf (standard cell potential) is called Ecell. Chapter 20

  21. Cell EMF • Standard Reduction Potentials • The potential is called reduction potential, the free energy of accepting electron • Convenient tabulation of electrochemical data. • Standard reduction potentials, Ered are measured relative to the standard hydrogen reduction potential. • The emf of a cell can be calculated from standard reduction potentials: • Ecell = Ered(cathode) - Ered(anode) • Note Chapter 20

  22. Cell EMF Standard Reduction Potentials Chapter 20

  23. Cell EMF • Standard Reduction Potentials • Hydrogen reduction potential is measured using standard hydrogen electrode (SHE) • SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H+ solution. H2 is bubbled through the tube. • For the SHE, we assign • 2H+(aq, 1M) + 2e- H2(g, 1 atm) • Ered of zero. Chapter 20

  24. Cell EMF • Standard Reduction Potentials • Consider Zn(s)  Zn2+(aq) + 2e-. We measure Ecell relative to the SHE (cathode): • Ecell = Ered(cathode) - Ered(anode) • 0.76 V = 0 V - Ered(anode). • Therefore, Ered(anode) = -0.76 V. • Standard reduction potentials must be written as reduction reactions: • Zn2+(aq) + 2e- Zn(s), Ered = -0.76 V. • Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. Chapter 20

  25. Cell EMF • Standard Reduction Potentials • The reverse, oxidation of Zn with the SHE is spontaneous. • Reactions with Ered > 0 are spontaneous reductions relative to the SHE. • Reactions with Ered < 0 are spontaneous oxidations relative to the SHE. • Note changing the stoichiometric coefficient does not affect Ered. • 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V. Chapter 20

  26. Cell EMF Oxidizing and Reducing Agents Chapter 20

  27. Cell EMF Standard Reduction Potentials Ecell = Ered(cathode) - Ered(anode) • The larger the difference between Ered values, the larger Ecell. • A positive Ecell indicates a spontaneous process (galvanic cell). • A negative Ecell indicates a nonspontaneous process. Chapter 20

  28. Spontaneity of Redox Reactions • Consider the displacement of silver by nickel: Ni(s) + 2Ag+(aq)  Ni2+(aq) + 2Ag(s) has E = Ered(Ag+/Ag) - Ered(Ni2+/Ni) = (0.80 V) - (-0.28 V) = 1.08 V, which indicates a spontaneous process. Chapter 20

  29. Effect of Concentration on Cell EMF • The Nernst Equation • The Nernst equation relates emf to concentration using • and noting that Chapter 20

  30. Effect of Concentration on Cell EMF • The Nernst Equation • This rearranges to give the Nernst equation: • The Nernst equation can be simplified at fixed temperature of 298 K: • (Note that change from natural logarithm to base-10 log.) • Remember that n is number of moles of electrons. Chapter 20

  31. Nernst equation for electrode potential • Reduction potential is defined as • Mn+ + ne-M, Ered • Since • The higher E, the easier to be reduced, the stronger oxidizing power • At any concentrations

  32. Effect of Concentration on Cell EMF • Concentration Cells • We can use the Nernst equation to generate a cell that has an emf based solely on difference in concentration. • One compartment will consist of a concentrated solution, while the other has a dilute solution. • Example: 1.00 M Ni2+(aq) and 1.00 10-3M Ni2+(aq). • The cell tends to equalize the concentrations of Ni2+(aq) in each compartment. • The concentrated solution has to reduce the amount of Ni2+(aq) (to Ni(s)), so must be the cathode. Chapter 20

  33. Effect of Concentration on Cell EMF • Concentration Cells • Since the two half-reactions are the same, Eº will be zero. Chapter 20

  34. Effect of Concentration on Cell EMF • Cell EMF and Chemical Equilibrium • A system is at equilibrium when G = 0. • From the Nernst equation, at equilibrium: Chapter 20

  35. Batteries • Lead-Acid Battery • A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. • (-) Pb | H2SO4 | PhO2 (+) • Cathode: PbO2 on a metal grid in sulfuric acid: • PbO2(s) + HSO42-(aq) + 3H+(aq) + 2e- PbSO4(s) + 2H2O(l) • +1.685V • Anode: Pb: • Pb(s) + HSO42-(aq)  PbSO4(s) + H + 2e- • -0.356 V Chapter 20

  36. Batteries • Lead-Acid Battery • The overall electrochemical reaction is • PbO2(s) + Pb(s) + HSO42-(aq) + 2H+(aq)  2PbSO4(s) + 2H2O(l) • It is reversible, at charge, the above reaction reverses • To be thermodynamically reversible, the charge and discharge must be infinitely slow. Chapter 20

  37. Batteries • Alkaline Battery (nonrechargable) • Anode: Zn cap: • Zn(s) + 2OH- Zn(OH)2(s) + 2e- • Cathode: MnO2: • 2MnO2(s) + 2H2O(l) + 2e- 2MnO(OH) (s) + 2OH-(aq) Chapter 20

  38. Batteries Alkaline Battery Chapter 20

  39. Batteries • Lithium-ion Battery (rechargable) • Anode: • LinC  Lin-1C + Li+ + e- • Cathode: • Lin-1(CoO2)n + e- + Li+ nLiCoO2 Chapter 20

  40. Batteries • Fuel Cells • Direct production of electricity from fuels occurs in a fuel cell. • On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity. • Cathode: reduction of oxygen: • 2H2O(l) + O2(g) + 4e- 4OH-(aq) • Anode: oxidization of hydrogen • 2H2(g) + 4OH-(aq)  4H2O(l) + 4e- • 2H2(g) + O2(g)  2H2O(l) Chapter 20

  41. Batteries Fuel Cells Chapter 20

  42. Batteries • Proton exchange membrane fuel cells • Cathode: reduction: • O2(g) + 4H+ + 4e- 2H2O(g) • Anode: of hydrogen • 2H2(g)  4H+ + 4e- • Net: • 2H2(g) + O2(g)  2H2O(l) Chapter 20

  43. Nernst equation for electrode potential • Reduction potential is defined as • Mn+ + ne-M, Ered • Since • The higher E, the easier to be reduced, the stronger oxidizing power • At any concentrations

  44. Corrosion • Corrosion of Iron • Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen. • Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l). 1.23V • Anode: Fe(s)  Fe2+(aq) + 2e-. -0.44V • Dissolved oxygen in water usually causes the oxidation of iron. • Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s). • Oxidation occurs at the site with the greatest concentration of O2. Chapter 20

  45. Nernst equation for electrode potential • Reduction potential is defined as • Mn+ + ne-M, Ered • Since • The higher E, the easier to be reduced, the stronger oxidizing power • At any concentrations

  46. Corrosion Corrosion of Iron Chapter 20

  47. Corrosion • Preventing the Corrosion of Iron • Corrosion can be prevented by coating the iron with paint or another metal. • Galvanized iron is coated with a thin layer of zinc. • Zinc protects the iron since Zn is the anode and Fe the cathode: • Zn2+(aq) +2e- Zn(s), Ered = -0.76 V • Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V • With the above standard reduction potentials, Zn is easier to oxidize than Fe. Chapter 20

  48. Corrosion Preventing the Corrosion of Iron Chapter 20

  49. Corrosion • Preventing the Corrosion of Iron • To protect underground pipelines, a sacrificial anode is added. • The water pipe is turned into the cathode and an active metal is used as the anode. • Often, Mg is used as the sacrificial anode: • Mg2+(aq) +2e- Mg(s), Ered = -2.37 V • Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V Chapter 20

  50. Corrosion Preventing the Corrosion of Iron Chapter 20

More Related