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Chapter 20 Electrochemistry

Chapter 20 Electrochemistry. Oxidation and Reduction. Oxidation (loss of e - ) Na Na + + e - Reduction (gain of e - ) Cl + e - Cl -

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Chapter 20 Electrochemistry

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  1. Chapter 20 Electrochemistry Chapter 20

  2. Oxidation and Reduction Oxidation (loss of e-) Na Na+ + e- Reduction (gain of e-) Cl + e- Cl- Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is needed to make them occur (nonspontaneous). Chapter 20

  3. Chapter 20

  4. Oxidation-Reduction (Redox) Reactions BOTH reduction and oxidation must occur. A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced). A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized). Chapter 20

  5. Oxidation-reduction (Redox) Reactions Is it a redox reaction?? Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Quick hint: Chapter 20

  6. Oxidation Number Guidelines • Atoms in elemental form, oxidation number is zero. • (Cl2, H2, P4, Ne are all zero) • Monoatomic ion, the oxidation number is the charge on the ion. (Na+: +1; Al3+: +3; Cl-: -1) • O is usually -2. But in peroxides (like H2O2 and Na2O2)it has an oxidation number of -1. • H is +1 when bonded to nonmetals and -1 when bonded to metals. • (+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3) • The oxidation number of F is -1. • The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule). Chapter 20

  7. Example Determine the oxidation state of all elements in ammonium thiosulfate (NH4)2(S2O3). (NH4)2(S2O3) Chapter 20

  8. Determining Oxidation States What is the oxidation state of Mn in MnO4-? Answer: +7 Chapter 20

  9. Balancing oxidation-reduction equations We know: Balancing chemical equations follows law of conservation of mass. AND, for redox reactions, gains and losses of electrons must also be balanced. Chapter 20

  10. Half-Reactions Separate oxidation and reduction processes in equation, Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) Oxidation: Reduction: Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction. Chapter 20

  11. Balancing Equations by the Method of Half-Reactions: Acidic Consider : MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g) Unbalanced half-reactions: MnO4-(aq) Mn2+(aq) C2O42-(aq) CO2(g) First, balance everything EXCEPT hydrogen and oxygen. Deal with half-reactions SEPARATELY. (acidic) Chapter 20

  12. Balancing Equations by the Method of Half-Reactions: Acidic MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g) To balance O: Add 4H2O to products to balance oxygen in reactants. To balance H: Add 8H+ to reactant side to balance the 8H in water. Chapter 20

  13. Balancing Equations by the Method of Half-Reactions: Acidic Balance charge: Add up charges on both sides. Add 5 electrons to reactant side. 5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l) Mass balance of C in oxalate half-reaction. C2O42-(aq) 2CO2(g) Balance charge by adding two electrons to the products. Last step: Cancel electrons and add reactions together. Chapter 20

  14. Balancing Equations by the Method of Half-Reactions: Acidic 5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l) C2O42-(aq) 2CO2(g) + 2e- Top reaction times 2. Bottom reaction times 5. ALL PARTS!! Chapter 20

  15. Balancing Equations by the Method of Half-Reactions: Acidic Summary 1. Divide equation into two incomplete half-reactions. 2. Balance each half-reaction (a) balance elements other than H and O. (b) balance O atoms by adding H2O. (c) balance H atoms by adding H+ (basic conditions will require further work at this step). (d) balance charge by adding e- to the side with greater overall positive charge. Chapter 20

  16. Balancing Equations by the Method of Half-Reactions: Summary • Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction. 4. Add the two half-reactions and cancel out all species appearing on both sides of the equation. 5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!! Chapter 20

  17. Balancing Equations by the Method of Half-Reactions: Basic Balancing process is started using H+ and H2O, then adjusting with OH- to uphold reaction conditions. (H+ does not exist in basic solutions.) Balance the following reaction: H2O2(aq) + ClO2(aq) ClO2-(aq) + O2(g) (basic) Chapter 20

  18. Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Split into two half-reactions. H2O2(aq) O2(g) ClO2(aq) ClO2-(aq) Balance elements, then oxygen by adding H2O. Then, add H+ to balance H, just like an acidic redox reaction. H2O2(aq) O2(g) + 2H+ ClO2(aq) ClO2-(aq) Chapter 20

  19. Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Add OH- to both sides, enough to neutralize all H+ (basic reactions cannot support H+). Combine H+ and OH- to form H2O. Chapter 20

  20. Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Balance charge by adding e-. 2OH- + H2O2(aq) O2(g) + 2H2O + + ClO2(aq) ClO2-(aq) Multiply each reaction so both have same e-. Then add them together, cancelling where possible. 2OH- + H2O2(aq) + 2ClO2(aq) O2(g) + 2H2O + 2ClO2-(aq) Double-check your answer! Chapter 20

  21. Balance this redox equation Cu(s) + NO3-(aq) Cu2+(aq) + NO2(g) (acidic) Ans: Cu(s) + 2NO3-(aq) + 4H+(aq) Cu2+(aq) + 2NO2(aq) + 2H2O(l) Chapter 20

  22. Balance this redox equation NO2-(aq) + Al(s) NH3(aq) + Al(OH)4-(aq) (basic) 2Al(s) + NO2-(aq) +OH-(aq) + 5H2O(aq) 2Al(OH)4-(aq) + NH3(aq) Chapter 20

  23. Voltaic Cells Voltaic (aka galvanic) cells: electrochemical reactions in which electron transfer occurs via an external circuit. The energy released in a voltaic cell reaction can be used to perform electrical work. Reactions are spontaneous. Chapter 20

  24. Voltaic Cells: Components • For example • Anode: Zn(s)  Zn2+(aq) + 2e- ( half-reaction) • Cathode: Cu2+(aq) + 2e- Cu(s) ( half reaction) • Salt bridge: cations move from anode to cathode, anions move from cathode to anode. • External circuit (wire): electrons move from anode to cathode (between two solid metal electrodes). Electron transfer can naturally occur in OTHER forms besides a wire (direct electron transport between solution and metal). Chapter 20

  25. Voltaic Cells: Ion Flow Anions and cations move through a porous barrier or salt bridge. Cations move into the cathodic compartment to neutralize the excess negatively charged ions. Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation. Chapter 20

  26. Simplified Voltaic Cell In any voltaic cell, the electrons flow from the anode through the external circuit to the cathode. Anode labeled with negative sign (-) and cathode with positive sign (+). Anions migrate toward anode. Cations toward the cathode. Chapter 20

  27. Voltaic Cells: Electron Flow As oxidation occurs, Zn(s) is converted to Zn2+ and 2e-. And Cu2+ is converted to Cu(s). Chapter 20

  28. One-Pot Voltaic Cells If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves, forming Zn2+. Zn is spontaneously oxidized to Zn2+ by Cu2+ (Cu is the oxidizing agent). The Cu2+ is spontaneously reduced to Cu0 by Zn (Zn is the reducing agent). The entire process is spontaneous. Chapter 20

  29. The Atomic Level A Cu2+(aq) ion comes into contact with a Zn(s) atom on the surface of the electrode. Two electrons are directly transferred from the Zn(s), forming Zn2+(aq), to the Cu2+(aq) forming Cu(s). Chapter 20

  30. Cell Electromotive Force (EMF) Electromotive force (emf): the force required to push electrons through the external circuit. Potential difference: difference per electrical charge between two electrodes. Units = Volts (V). One volt is the potential required to impart one joule of energy to a charge of one coulomb: Chapter 20

  31. Cell EMF Cell potential (Ecell): the emf of a cell. Aka cell voltage.Positive for spontaneous cell reactions. Ecell strictly depends on reactions that occur at the cathode and anode, the concentration of reactants and products, and the temperature and is described by the equation: Ecell = Ered(cathode) - Ered(anode) For 1M solutions at 25˚C and 1 atm (standard conditions), the standard emf (standard cell potential), Ecell, is written as E˚cell. Chapter 20

  32. Standard Reduction Potentials Potential associated with each half-reaction is chosen to be the potential for reduction to occur at that electrode. Hence, Standard Reduction Potentials. Standard reduction potentials, E˚red, are measured relative to the standard hydrogen electrode (SHE). Chapter 20

  33. Standard Hydrogen Electrode (SHE) SHE is assumed as the cathode (for consistency). Pt electrode in a tube containing 1M H+ solution. H2(g) is bubbled through the tube and equilibrium is established. Chapter 20

  34. Determining Standard Reduction Potentials For the SHE: 2H+(aq, 1M) + 2e- H2(g, 1 atm) E˚red = . Standard reduction potentials can then be calculated using the SHE (E˚= 0V) as E˚red(cathode). Each calculated E˚ is rewritten as a reduction and tabulated. Chapter 20

  35. Finding the Standard Reduction Potential E˚cell is measured, 0V (SHE) is used for E˚(cathode), and the reduction potential for Zn is found. Chapter 20

  36. Zn Standard Reduction Potential We measure E˚cell relative to the SHE: E˚cell = E˚red(cathode) - E˚red(anode) 0.76V = 0V - E˚red(anode). Therefore, E˚red(anode) = -0.76V And we find that -0.76V can be assigned to reduction of zinc. Chapter 20

  37. Zn Standard Reduction Potential Since E˚red = -0.76 V (negative!) we conclude that the reduction of Zn2+ in the presence of the SHE is notspontaneous. However, the oxidation of Zn with the SHE is spontaneous. Reactions with E˚red > 0 are spontaneous reductions relative to the SHE. E˚red < 0 are spontaneous oxidations. Changing the stoichiometric coefficient does not affect E˚red. 2Zn2+(aq) + 4e-2Zn(s) E˚red = -0.76 V Chapter 20

  38. Full list in Appendix E Chapter 20

  39. EMF Trends The larger the difference between E˚red values, the larger E˚cell. Spontaneous voltaic cell: E˚red(cathode) is more positive than E˚red(anode), resulting in a positive E˚cell. Recall Chapter 20

  40. Calculating E˚red from E˚cell Zn(s) + Cu2+(aq, 1M) Zn2+(aq, 1M) + Cu(s) E˚cell = 1.10V Given: Zn2+ + 2e- Zn(s) E˚red = -0.76V Calculate the E˚red for the reduction of Cu2+ to Cu. What about E˚red for the oxidation of Cu(s) – reverse reaction? Chapter 20

  41. Oxidizing and Reducing Agents We can use E˚red values to understand aqueous reaction chemistry. e.g. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Since Cu2+ is responsible for the oxidation of Zn(s), Cu2+ is called the . Since Zn(s) is responsible for the reduction of Cu2+, Zn(s) is called the . E˚red for Cu2+ (0.34V) indicates that, compared to E˚red for Zn (-0.76V), it will be reduced. Chapter 20

  42. Example Problem: Cell emf Calculate the standard emf for the following: Ni(s) + 2Ce4+(aq) Ni2+(aq) + 2Ce3+(aq) Ask yourself…What are the half-reactions? Which one is oxidized? Reduced? Ni2+(aq) + 2e- Ni(s) E˚red = -0.28V Ce4+(aq) + 1e- Ce3+(aq) E˚red = 1.61V is the oxidizing agent. Chapter 20

  43. Example Problem: Cell emf Which reaction will undergo reduction? What reaction occurs at the anode? Sn4+(aq) + 2e- Sn2+(aq) E˚red = 0.154V MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) E˚red = 1.51V Chapter 20

  44. Oxidizing and Reducing Agents: EMF The more positive E˚red, the greater the tendency for the reactant in the half-reaction to be reduced. Therefore, the stronger the oxidizing agent. The more negative E˚red, the greater the tendency for the product in the half-reaction to be oxidized. This means the product is tends to be a reducing agent. Chapter 20

  45. A species at higher left of the table of standard reduction potentials will spontaneously oxidize a species that is lower right in the table. F2 will oxidize H2 or Li. Ni2+ will oxidize Al(s). Chapter 20

  46. Example: Redox Agents Which is the strongest reducing agent? Oxidizing agent? Ce4+, Br2, H2O2, or Zn? E˚red = 1.61V for Ce4+ (aq) 1.065V for Br2 (l) 1.776 for H2O2 (aq) -0.763V for Zn(s) Chapter 20

  47. Spontaneity of Redox Reactions In a spontaneous voltaic cell, E˚red(cathode) must be more positive than E˚red(anode). A positive E˚cell indicates a spontaneous voltaic cell process. A negative E˚cell indicates a non-spontaneous process. It all relates back to Gibbs Free Energy. Chapter 20

  48. Spontaneity of Redox Reactions • Cell emf and free-energy change are related by • G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. G units are J (assumed per mole). • 1F = 96,500 C/mol = 96,500 J/Vmol • If standard conditions: G˚ = -nFE˚cell • If Ecell > 0 then , both of which indicate spontaneous processes. Chapter 20

  49. Example: Cell spontaneity Calculate ΔG˚ for the following reaction. Is it spontaneous? Cl2(g) + 2I-(aq) 2Cl-(aq) + I2(s) E˚cell = 0.823V Chapter 20

  50. Batteries Defn: Self-contained electrochemical power source with one (or more) complete voltaic cell. When multiple cells or multiple batteries are connected in series, greater emfs can be achieved. Cathode labeled with a plus sign; the anode with a minus sign. Chapter 20

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