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Electrochemistry

Explore the study of the interchange between chemical and electrical energy through examples like forest fires and car combustion. Learn about redox reactions, oxidation, reduction, rules for assigning oxidation states, and identifying substances' roles. Understand balancing reactions using half-reaction methods in acidic solutions.

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Electrochemistry

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  1. Electrochemistry • Chapter 17

  2. Electrochemistry • The study of the interchange of chemical and electrical energy.

  3. What are some examples of redox reactions? forest fire rusting steel combustion in auto engine metabolism of food in the body What everday uses depend on redox reactions? starting a car calculator digital watch portable radio portable CD player What are some examples of Redox Reactions?

  4. Review of Terms • oxidation-reduction (redox) reaction: involves a transfer of electrons from the reducing agent to the oxidizing agent. • oxidation: loss of electrons • reduction: gain of electrons

  5. OIL RIG • Oxidation Is Loss. • Reduction Is Gain.

  6. Redox • Oxidation is the loss of electrons--oxidation number becomes more positive. • Reduction is the gain of electrons--oxidation number becomes more negative.

  7. Redox • metals nonmetals • oxidation reduction • reducing agents oxidizing agents • metal ions nonmetal ions • reduction oxidation • oxidizing agents reducing agents

  8. Redox Reactions • Loss and gain of electrons must be simultaneous. • Loss and gain of electrons must be equal. • Why must the loss and gain of electrons be equal? Law of Conservation of Matter

  9. Redox Reactions • Redox reactions are reactions in which electrons are transferred. • Decomposition and synthesis reactions may be redox. • Single replacement reactions are always redox. • Double replacement reactions are never redox. • Combustion reactions are always redox.

  10. Identifying Oxidation & Reduction in a Reaction • Identify the element which is oxidized and the one which is reduced. • 1. 2Mg(s) + O2(g) ---> 2MgO(s) • 2. 2Al(s) + 3I2(s) ---> 2AlI3(s) • 3. 2Cu(s) + O2(g) ---> 2CuO(s) • 4. 2Cs(s) + F2(g) ---> 2CsF(s)

  11. Half-Reactions • The overall reaction is split into two half-reactions, one involving oxidation and one reduction. • 8H+ + MnO4 + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O • Reduction: 8H+ + MnO4 + 5e Mn2+ + 4H2O • Oxidation: 5Fe2+ 5Fe3+ + 5e

  12. Rules for Assigning Oxidation States • 1. Oxidation state of an atom in an element = 0 • 2. Oxidation state of monatomic element = charge • 3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1) • 4. H = +1 in covalent compounds • 5. Fluorine = 1 in compounds • 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

  13. Charges & Oxidation States • Oxidation states are written as +2. • Charges are written 2+.

  14. Determining Oxidation States • SF6 [NO3]- • +6-6 = 0 +5 -6 = -1 (-1 for each F) (-2 for each O) The most electronegative element is assigned a negative oxidation number--see electronegativity chart on page 354.

  15. Determining Oxidation States oxidation number rule 0 1 +1, -1 2 +4, -2 3 +1, -1 3 -3, +1, -1 5, 4, 7, & 2 • substance • Na(s) • NaF(s) • SO2(g) • H2O2 • NH4Cl(s)

  16. Redox • Oxidizing agent is the electron acceptor--usually a nonmetal. • Reducing agent is the electron donor--usually a metal. • CH4(g) + 2O2(g) ----> CO2(g) + 2HOH(g) • Carbon is oxidized. • Oxygen is reduced. -4 +1 0 +4 -2 +1-2+1 CH4 is the reducing agent. O2 is the oxidizing agent.

  17. Redox Reactions • Identify the substance oxidized and the substance reduced as well as the oxidizing and reducing agents. • PbO(s) + CO(g) ---> Pb(s) + CO2(g) • oxidized • reduced • oxidizing agent • reducing agent +2 -2 +2 -2 0 +4 -2 Carbon Lead PbO CO

  18. Redox Reactions • Identify the substance oxidized and the substance reduced as well as the oxidizing and reducing agents. • 2PbS(s) +3O2(g) ---> 2PbO(s) + 2SO2(g) • oxidized • reduced • oxidizing agent • reducing agent +2 -2 0 +2 -2 +4 -2 sulfur oxygen O2 PbS

  19. Balancing by Half-Reaction Method • 1. Write separate reduction, oxidation reactions. • 2. For each half-reaction: • - Balance elements (except H, O) • - Balance O using H2O • - Balance H using H+ • - Balance charge using electrons

  20. Balancing by Half-Reaction Method (continued) • 3. If necessary, multiply by integer to equalize electron count. • 4. Add half-reactions & cancel identical species. • 5. Check that both elements and charges are balanced.

  21. Balancing By Half-Reaction Acidic Solution • H+(aq) + Cr2O72-(aq) + C2H5OH(l) ---> Cr3+(aq) + CO2(g) + HOH(l) • Red Cr2O72-(aq) ---> Cr3+(aq) • Ox C2H5OH(l)---> CO2(g) • Red 2(6e- + 14H+(aq) + Cr2O72-(aq) ---> 2Cr3+(aq) + 7HOH(l)) • Ox C2H5OH(l) + 3HOH(l) ---> 2CO2(g) + 12H+(aq) + 12e- • 16H+(aq) + 2Cr2O72-(aq) + C2H5OH(l) ---> 4Cr3+(aq) + 11HOH(l) + 2CO2(g) • 12+ = 12+

  22. Redox Reactions • Always add electrons to the side of the half-reaction with excess positive charge!

  23. Balancing By Half-Reaction Acidic Solution • Cu(s) + HNO3(aq) ---> Cu(NO3)2(aq) + NO(g) + HOH(l) • Ox Cu(s) + HNO3(aq) ---> Cu(NO3)2(aq) • Red HNO3(aq)---> NO(g) • Ox 3(Cu(s) + 2HNO3(aq) ---> Cu(NO3)2(aq) + 2H+(aq) + 2e-) • Red 2(3e- + 3H+(aq) +HNO3(aq)---> NO(g) + 2HOH(l) ) • 3Cu(s) + 8HNO3(aq) ---> 3Cu(NO3)2(aq) + 4HOH(l) + 2NO(g) • 0 = 0

  24. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution

  25. Half-Reaction Method - Balancing in Base • 1. Balance as in acid. • 2. Add OH that equals H+ ions (both sides!) • 3. Form water by combining H+, OH. • 4. Check elements and charges for balance.

  26. Balancing By Half-Reaction Basic Solution • Ag(s) + CN-(aq) + O2(g) ---> Ag(CN)2-(aq)(Basic) • Ox CN-(aq) + Ag(s) ---> Ag(CN)2-(aq) • Red O2(g) ---> • Ox 4(2CN-(aq) + Ag(s) ---> Ag(CN)2-(aq) + e-) • Red O2(g) + 4H+(aq) + 4e- ---> 2HOH(l) • 8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) ---> 4Ag(CN)2-(aq) + 2HOH(l)

  27. Balancing By Half-Reaction Basic Solution • 8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) + 4OH-(aq) • ---> 4Ag(CN)2-(aq) + 2HOH(l) + 4OH-(aq) • 8CN-(aq) + 4Ag(s) + O2(g) + 4HOH(l) • ---> 4Ag(CN)2-(aq) + 2HOH(l) + 4OH-(aq) • 8- = 8- + 2HOH(l)

  28. The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution

  29. Galvanic Cell • A device in which chemical energy is changed to electrical energy.

  30. Galvanic Cell • A device in which chemical energy is changed to electrical energy. The basic parts are: • Anode (where oxidation occurs) • Cathode (where reduction occurs) • electrochemical solution • porous disk or salt bridge (allows ions to flow without mixing solutions)

  31. Anode and Cathode • OXIDATION occurs at the ANODE. • REDUCTION occurs at the CATHODE. • AN OXRED CAT

  32. Voltaic Cell: Cathode Reaction

  33. Voltaic Cell: Anode Reaction

  34. Electrons are transferred at the interface between the electrodes and the solution. Porous disk allows ion flow.

  35. A galvanic cell (Daniell Cell) involving Zn and Cu electrodes. This cell was the energy source for telegraphy during the War Between the States.

  36. Cu-Zn (Daniell Cell) on the microscopic level.

  37. Zinc electrode compared to a Standard Hydrogen Electrode (SHE). The Zn has a potential of 0.76 V.

  38. Electrochemical Half-Reactions in a Galvanic Cell

  39. Standard Reduction Potentials • The E values corresponding to reduction half-reactions with all solutes at 1M and all gases at 1 atm. • Cu2+ + 2e Cu E = 0.34 V vs. SHE • SO42 + 4H+ + 2e H2SO3 + H2O E = 0.20 V vs. SHE SHE = Standard Hydrogen Electrode

  40. Cell Potential • Cell Potential or Electromotive Force (emf): The “pull” or driving force on the electrons. • Units are in Volts (V)

  41. Cell Potential Calculations • To Calculate cell potential using Standard Reduction Potentials: • 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive. • 2. The half-reactions must often be multiplied by an integer to balance electrons--this is not done for the cell potentials.

  42. Cell Potential Calculations Continued • Fe3+(aq) + Cu(s) ----> Cu2+(aq) + Fe2+(aq) • Fe3+(aq) + e- ----> Fe2+(aq) Eo= 0.77 V • Cu2+(aq) + 2 e- ----> Cu(s)Eo = 0.34 V • Reaction # 2 must be reversed.

  43. Cell Potential Calculations Continued • 2 (Fe3+(aq) + e- ----> Fe2+(aq))Eo= 0.77 V • Cu(s)----> Cu2+(aq) + 2 e-Eo = - 0.34 V • 2Fe3+(aq) + Cu(s) ----> Cu2+(aq) + 2Fe2+(aq) • Eo = 0.43 V

  44. Example #2 • A galvanic cell is based on the reaction: • MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn2+(aq) + H2O(l) • The half reactions are: • MnO4-+ 5e- + 8H+ Mn2+ + 4H2O Eo=1.51 V • ClO4- + 2H+ + 2e-  ClO3- + 4H2O Eo=1.19 V • Give the balanced cell reaction and calcualteEo for the cell.

  45. Example #2 (cont) • 2(MnO4-+ 5e- + 8H+ Mn2+ + 4H2O) Eo=1.51 V • 5(ClO3- + 4H2O  ClO4- + 2H+ + 2e-)Eo= -1.19 V • Reduction potential does not change because we multiplied the equation!

  46. Example #2 (cont) • 2(MnO4-+ 5e- + 8H+ Mn2+ + 4H2O) Eo=1.51 V • 5(ClO3- + 4H2O  ClO4- + 2H+ + 2e-)Eo= -1.19 V • 2MnO4- + 6H+ + 5ClO3- 2 Mn2+ + 3H2O + 5 ClO4-Eo=0.32 V

  47. Line Notation • Used to describe electrochemical cells. • Anode components are listed on the left. • Cathode components are listed on the right. • Separated by double vertical lines. • The concentration of aqueous solutions should be specified in the notation when known. • Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s) • Mg → Mg2+ + 2e–(anode) • Al3+ + 3e–→ Al(cathode)

  48. Galvanic Cells • A cell will always run spontaneously in the direction that produces a positive cell potential!

  49. emf and Work

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