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Electrochemistry

Electrochemistry. The study of the interchange between chemical and electrical energy. Oxidation-Reduction Reactions. “Redox” reactions involve the transfer of electrons (e-) Reduction : gain e- Oxidation : lose e- “LEO the lion says, ‘GER’”

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Electrochemistry

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  1. Electrochemistry The study of the interchange between chemical and electrical energy

  2. Oxidation-Reduction Reactions • “Redox” reactions involve the transfer of electrons (e-) • Reduction: gain e- • Oxidation: lose e- • “LEO the lion says, ‘GER’” • “OIL RIG” • Use oxidation states to keep track of the e-

  3. Leo says Ger My name is Leo. Grr-rrrr… “Lose electron oxidation” Zn  2e-+ Zn2+ “Gain electron  reduction” 2e- + Cu2+  Cu

  4. Assigning Oxidation States Specific rules for assigning Ox #’s • Usually the same charge assigned by the PT • H is almost always +1 • O is almost always -2 • F is always -1 in compounds • For elements (H2, O2, F2, Ca, K, etc ) the oxidation state always = 0 Some exceptions do exist!

  5. Assigning Oxidation Numbers • Overall charge = sum of the oxidation states of all atoms in it Neutral Compounds (e.g. H2O, CO2, CH4) • H2O : The overall charge is 2(1) + -2 = 0 • CO2: What is the oxidation state of C? Since C + 2 (O) = 0… C + 2(-2) = 0, thus… • CH4: Is C still +4? H is always +1 To remain neutral… 4(1) + C = 0 C must = - 4 and O = -2 H = +1 • C = +4

  6. Assigning Oxidation Numbers • Charged compounds (e.g. NO3-, CO32-) NO3- or (NO3)- : What is the oxidation # of N? O is -2, and the overall charge is -1 So N + 3(O) = -1 or N + 3(-2) = -1 N = + 5 (CO3)2-: What is the oxidation # of C? O is -2, and the overall charge is -2 So C + 3(O) = -2 or C + 3(-2) = -2 C = +4 The oxidation # of ions = charge of ions Mn3+ has an oxidation # of +3 S2- has an oxidation # of -2

  7. Assigning Oxidation # Practice Assign oxidation numbers to each atom Cl2 Fe2+ ClO3- ClO4- IO2- CrO42- Fe3(PO4)2 CoSO4 H2CO3 Cl: 0 (element) Fe: 2+ (ion) O: 2-, 3(2-) + Cl = 1-…Cl: 5+ O: 2-, 4(2-) + Cl = 1-…Cl: 7+ O: 2-, 2(2-) + I = 1-…I: 3+ O: 2-, 4(2-) + Cr = 2-…Cr: 6+ Fe: 2+ (ion) PO4:3- (ion)….O:2-, 4(2-) + P = 3-, P: 5+ Co: 2+ (ion) SO4:2- (ion)….O:2-, 4(2-) + S = 2-, S: 6+ H: 1+ (ion) CO3:2- (ion)….O:2-, 3(2-) + C = 2-, C: 4+

  8. Assigning Oxidation Numbers Review • Try these…MnO4-, Cr2O72-, C2O42- • (MnO4)- O = -2, so [4(-2) + Mn = -1] Mn = +7 • (Cr2O7)2- O = -2, so [7(-2) + 2Cr = -2] 2Cr = 12, therefore… • (C2O4)2- O = -2, so [2C + 4(-2) = -2] 2C = 6, therefore… • Cr = +6 C = +3

  9. Oxidation-Reduction Reactions • Two separate reactions occurring simultaneously • Oxidation:oxidation # of an atom increases • e.g. Fe(s) → Fe3+(aq) • Reduction: oxidation # of an atom is“reduced” • e.g. O2(g) → O2-(aq) When occurring together… • Fe(s) + O2(g) → Fe3+(aq) + O2-(aq) • This is the redox reaction responsible for rust! But, how do we balance this? (ox # goes from 0 → +3) • (oxidation # goes from 0 → -2)

  10. Balancing by Half-Reactions*in acidic solution • Assign oxidation states for each element. • Write separate half-reactions for the reduction/oxidation reactions. • Balance all the atoms EXCEPT O and H. • Balance the oxygen with water (H2O). • Balance the hydrogen with hydrogen ions (H+). • Balance the charge with electrons. • Multiply each half-reaction by an appropriate number to make the electrons equal. • Combine both reactions into one and cancel the e -

  11. Balancing by Half-Reactions*in acidic solution CH3OH (aq) + Cr2O72-(aq) → CH2O(aq) + Cr3+(aq) 1. Assign oxidation states. C-2H4+O2- + (Cr26+O72-)2- → C0H2+O2- + Cr3+ 2. Write separate half-reactions for the reduction and oxidation reactions. (only keep charges that are changing…) Ox:C-2H4O→ C0H2O (C is going from -2 to 0) Red: (Cr26+O7)2- → Cr3+ (Cr is being reduced from +6 to +3)

  12. Balancing the Oxidation… Ox: C2-H4O→ C0H2O 3. For each half reaction, balance all the atoms EXCEPT O and H. 4. Balance the oxygen by adding water (H2O). • Balance the hydrogen by adding hydrogen ions (H+) • Balance the charge by adding electrons.…use the oxidation state as a guide + 2H+ + 2e- Carbon is already balanced! Oxygen is already balanced!

  13. On to the reduction… (Cr26+O7)2- → Cr3+ • Balance all elements except H and O • Balance O by adding H2O, if necessary • Balance H by adding H+,if necessary • Balance charge by adding e- Remember, you only care about the charges that are changing… 6e- + 2 + 7H2O 14H+ +

  14. Adding Half-Reactions*in acidic solution Now add the two reactions together… Ox:CH4O→ CH2O + 2H+ + 2e- Red:6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O 7. Multiply each half-reaction by an appropriate number to make the electrons equal. CH4O→ CH2O + 2H+ + 2e- 6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O 3CH4O→ 3CH2O + 6H+ + 6e- ) 3 (

  15. Adding Half-Reactions*in acidic solution 6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O 3CH4O→ 3CH2O + 6H+ + 6e- 3CH4O + + Cr2O72- → 3CH2O + 2Cr3+ + 7H2O …and the reaction is now balanced! 8. Combine both reactions into one and cancel. 8H+

  16. Practice Balancing Redox Reactions Unbalanced reaction (in acid): MnO4 + Fe2+ Mn2+ + Fe3+ Balanced Reduction half-reaction: 8H+ + MnO4 + 5e Mn2+ + 4H2O Balanced Oxidation half-reaction: Fe2+ Fe3+ + e Balanced overall reaction: 8H+ + MnO4 + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O ) 5(

  17. Balancing by Half-Reactions*in basic solution • Assign oxidation states. • Write separate half-reactions for the reduction/oxidation reactions. • Balance all the atoms EXCEPT O and H. • Balance the oxygen by adding water (H2O). • Balance the hydrogen by adding H+. • Balance the charge by adding electrons. • Multiply each half-reaction by an appropriate number to make the electrons equal. • Combine both reactions into one and cancel. • Add OH- to both sides to cancel out H+ and create H2O. Simplify further, if necessary.

  18. Balancing by Half-Reactions(in basic solution) Let’s balance a previous example in basic solution Remember, it is all the same steps up to this point 3CH4O + 8H+ + Cr2O72- → 3CH2O + 2Cr3+ + 7H2O 3CH4O + + Cr2O72- → 3CH2O + 2Cr3+ + 7H2O + 8OH- 3CH4O + H2O + Cr2O72- → 3CH2O + 2Cr3+ + 8OH- + 8OH- + 8OH- 8H2O

  19. Practice Balancing Basic Redox Rxns Unbalanced reaction: ClO + Zn Cl- + Zn2+ Balanced Reduction half-reaction: 2e- +2H+ + ClO- Cl- + H2O Balanced Oxidation half-reaction: Zn  Zn2+ + 2e- Balanced overall reaction (acidic): 2H+ + ClO + Zn  Zn2+ + Cl- + H2O Balanced overall reaction (basic): H2O + ClO + Zn  Zn2+ + Cl- + 2OH-

  20. Redox Vocabulary ClO + Zn Cl- + Zn2+ Oxidized species (atom, ion, molecule, or compound) whichever species is being oxidized (↑ oxidation #) Reduced species whichever species is being reduced (↓ oxidation #) Oxidizing agent whichever species CAUSES oxidation to occur—in other words, the oxidizing agent IS the reduced species Reducing agent whichever species CAUSES reduction to occur—in other words, the reducing agent IS the oxidized species

  21. Oxidizing Agent Reducing Agent ClO- ClO + Zn Cl- + Zn2+ Zn Oxidized species Reduced species Notice that these terms only ever apply to reactants

  22. Electrochemistry The study of the interchange of chemical and electrical energy • Sample electrochemical processes: • 1. Corrosion • e.g. 4 Fe(s) + 3 O2(g)⇌ 2 Fe2O3(s) • 2. Biological processes • e.g. C6H12O6 + 6 O2⇌ 6 CO2 + 6 H2O • 3. Batteries (Galvanic or Voltaic cells) • Electrochemical cells that produce a current (flow of electrons) • as a result of a redox reaction • 4. Electrolytic cells • Electrical energy is used to produce chemical change • Used to prepare or purify metals (such as sodium, aluminum, copper)

  23. Harnessing the Energy e- K+ NO3- “anode” and “oxidation” begin with vowels “cathode” and “reduction” begin with consonants • Separate the half-reactions • Creates a galvanic or voltaic cell Luigi Galvani Alessandro Volta Red Cat salt bridge − (produces electrons) + (attracts electrons) Cu Ag KNO3(aq) Anode Cathode Cu2+ SO42- Ag+ NO3- 1 M CuSO4 Cu(s)  Cu2+(aq) + 2e- 1 M AgNO3 Ag+(aq) + e-  Ag(s) Reduction Oxidation

  24. Standard Reduction Potentials The cell potential, ℰ°cell ,can be determined from the standard reduction potentials (ℰ°red) for the half-reactions: Reduction potential = tendency for reduction to happen Positive ℰ°red  spontaneous reduction reaction Negative ℰ°red non-spontaneous reduction (use reverse reaction) Standard (o) = standard conditions (1 M solutions)

  25. Standard Reduction Potentials Half-Reaction ℰ° (V) F2 + 2e- 2F- 2.87 Au3+ + 3e- Au 1.50 Ag+ + e- Ag 0.80 Cu2+ + 2e- Cu 0.337 2H+ + 2e- H2 0.00 Ni2+ + 2e- Ni -0.28 Zn2+ + 2e- Zn -0.763 Al3+ + 3e- Al -1.66 Li+ + e- Li -3.05 ℰ° > 0 Spontaneous reduction (Oxidizing Agents!)  ℰ° = 0 Standard Hydrogen Electrode Ni  Ni2+ + 2e- +0.28 Zn  Zn2+ + 2e- +0.763 Al  Al3+ + 3e- +1.66 Li  Li+ + e- +3.05 ℰ° < 0 Non-Spontaneous reduction (Reducing Agents!) Spontaneous oxidation (reverse rxn) Remember: an oxidation CANNOT happen without a reduction!

  26. Cell Potential ℰ°cell = ℰ°reduction + ℰ°oxidation Ag+(aq) + e- Ag(s)ℰ° = 0.80 V Cu2+(aq) + 2 e-  Cu(s)ℰ° = 0.34 V Reduction reaction: 2(Ag+(aq) + e- Ag(s)) ℰ° = +0.80 V Oxidation reaction: Cu(s)  Cu2+(aq) + 2 e-ℰ° = - 0.34 V Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)ℰ°cell = +0.46 V The ℰ°cell MUST be + and thus spontaneous for Galvanic cells

  27. Brain Warmup Half-Reaction ℰ° (V) Ag+ + e- Ag 0.80 Cu2+ + 2e- Cu 0.34 Zn2+ + 2e- Zn -0.76 Al3+ + 3e- Al -1.66 What is ℰ° for each of the following reactions? Which reaction(s) are spontaneous? ℰ° 2.46 V 1.10 V -0.90 V Spontaneous? Y Y N 3 Ag+(aq) + Al(s) 3 Ag(s) + Al3+(aq) Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) 2 Al3+(aq) + 3 Zn(s) 2 Al(s) + 3 Zn2+(aq) Zn can reduce Cu2+, but not Al3+

  28. Line Notation for Galvanic Cells e- K+ NO3- Cu Ag Anode (−) Cathode (+) Cu2+ SO42- Ag+ NO3- 1 M CuSO4 Cu(s)  Cu2+(aq) + 2e- 1 M AgNO3 Ag+(aq) + e-  Ag(s) Reduction Oxidation Cu(s)Cu2+ (1 M)Ag+ (1 M)Ag(s) Anode always on the left Cathode always on the right

  29. Practice Time Given the following information, draw a galvanic cell. Fe(s)Fe2+(1 M)Au3+(1 M)Au(s) Be sure to include the following: Anode/Cathode reactions Balanced overall reaction with potential Complete circuit (external wire with e- flow direction, salt bridge) Label all parts of the cell (solution, electrode, etc.)

  30. Fe(s)Fe2+(1 M)Au3+(1 M)Au(s) e- anions cations Fe Au Anode Cathode Fe2+ Au3+ 1 M Fe2+ Fe(s)  Fe2+(aq) + 2e- 1 M Au3+ Au3+(aq) + 3e-  Au(s) Reduction Oxidation 3Fe(s) + 2Au3+(aq) → 3Fe2+(aq) + 2Au(s) ℰ°cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) = 1.94 V

  31. Practice Time Given the following information, draw a galvanic cell. C(gr)Cr2+(1 M), Cr3+(1 M)Cl-(1M)Cl2(g)Pt(s) Be sure to include the following: Anode/Cathode reactions Balanced overall reaction with potential Complete circuit (external wire with e- flow direction, salt bridge) Label all parts of the cell (solution, electrode, etc.)

  32. C(gr)Cr2+(1 M), Cr3+(1 M)Cl-(1M)Cl2(g)Pt(s) e- anions cations C Pt Anode Cathode Cl- Cl2 Cr2+ Cr3+ 1 M Cr2+/Cr3+ Cr2+  Cr3+ + e- 1 M Cl-, 1 atm Cl2(g) Cl2(g) + 2e-  2Cl - Reduction Oxidation

  33. The First Battery Alessandro Volta (1745 – 1827) Generated electricity by putting a layer of cardboard soaked in brine between discs of copper and zinc – a voltaic cell. When he made a ‘pile’ of these cells, he increased the amount of electricity generated. This was the first battery – a collection of cells. Zinc disc Copper disc Newmark, CHEMISTRY, 1993, page 46

  34. Batteries Car Battery (Lead storage battery) Anode:Pb + HSO4- PbSO4 + H+ + 2e- Cathode:PbO2 + HSO4- + 3H+ + 2e- PbSO4 + 2H2O Cell:Pb + PbO2 + 2H+ + 2 HSO4- 2 PbSO4 + 2H2O 2 volts per cell, 6 cells to a battery  12 volt battery Alkaline Battery Anode: Zn  Zn2+ + 2e- Cathode: 2 MnO2 + H2O + 2e- Mn2O3 + 2OH- ℰ°cell = 1.5 volts Lemon Battery Anode: Zn  Zn2+ + 2e- Cathode: Cu2+ + 2e- Cu Cell reaction: Zn + Cu2+ Zn2+ + Cu ℰ°cell = 1.1 volts (if all goes well) Want more volts? Link cells in series...

  35. Onion Battery…? • According to YouTube, you can charge your iPod with just an onion and Gatorade • As Michael Scott (from the Office) points out, anyone can put anything on the Internet, so you know the information is good…

  36. One Cell of a Lead Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 595

  37. Mercury Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 596

  38. Common Alkaline Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 596

  39. Cathodic Protection of an Underground Pipe Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 598

  40. Polishing Silver with Aluminum Foil 3 Ag+(aq) + Al(s) 3 Ag(s) + Al3+(aq) ℰ° = 2.46 V 3 Ag2S + 2 Al + 3 H2O  6 Ag + Al2O3 + 3 H2S  Stinky!! (Add NaHCO3to neutralize H2S)

  41. Reaction Quotient The reaction quotient (Q) sets up a ratio of products and reactants For a reaction, A + 2B → 3C + 4D [C]3[D]4 [A]1[B]2 Only include concentrations (aq) OR pressures (g) Solids (s) and liquids (l) are not included Q =

  42. Reaction Quotient practice • Write the Q expression for the following reaction CH4(g) + O2(g) → CO2(g) + H2O(g) Reaction must be balanced first… CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) (CO2)(H2O)2 (CH4)(O2)2 Q =

  43. Reaction Quotient practice Write the Q expression for the following reaction Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) (Cu2+)(Ag)2 (Cu)(Ag+)2 Is this correct? NO: Solids aren’t included in the equation! (Cu2+) (Ag+)2 Q = Q =

  44. Under Non-standard conditions:The Nernst Equation ℰcell = ℰ°cell - If we don’t have 1 M concentration or 1 atm pressure, we must take a different approach… ΔG = ΔG° + RT lnQ ΔG = -nFℰ ΔG° = -nFℰ° -nFℰ = -nFℰ° + RT lnQ R = 8.3145 J/mol K T = temperature (in K) n = moles of e- transferred F = Faraday’s constant 96,485 C/mol e- Walther Nernst

  45. Practice with the Nernst Equation Cu(s)  Cu2+(aq) + 2e- Ag+(aq) + e-  Ag(s) Cu Ag ℰcell ℰcell = ℰ°cell - Ag+ NO3- Cu2+ SO42- What will be the cell potential ℰ of a Cu/Ag cell using 0.10 M Cu2+ and 1.0 M Ag+ solutions at 25°C? Cu(s)Cu2+ (0.10 M)Ag+ (1.0 M)Ag(s) 2( ) Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) ℰcell = 0.46 V – (-0.03 V) ℰcell= 0.49 V

  46. More Practice… ℰcell = ℰ°cell - What will be the cell potential ℰ of a Al/MnO4-,Mn2+ cell using 0.5 M Al3+, 1.5 M Mn2+, 1.0 M MnO4- and 2.0 M H+ solutions at 15°C? Al(s)Al3+ (0.5 M)Mn2+ (1.5 M), MnO4- (1.0 M), H+ (2.0 M)Pt(s) Al(s)  Al3+(aq) + 3e- MnO4- + 8H+ + 5e-  Mn2+ + 4 H2O 3( ) 5( ) 5Al(s) + 3MnO4- + 24H+ 3Mn2++ 5 Al3+ + 12 H2O = 6.285 x 10-9 ℰcell Al Pt ℰcell = 3.17 V – (-0.03 V) ℰcell= 3.20 V H+ MnO4-, Mn2+ Al3+

  47. Le Chatelier’s principle When a system at equilibrium is disturbed, it shifts to a new equilibrium that balances the disturbance… N2(g) + 3 H2(g) 2 NH3(g) Disturbance Equilibrium Shift Add more N2………………….. “ “ H2………………….. “ “ NH3………………… Remove NH3………………….. Add a solid/liquid……………… no shift Remove either reactant………. Fritz Haber

  48. I wish I had sweat glands. In a chicken… CaO + CO2 CaCO3 (eggshells) In summer, [ CO2 ] in a chicken’s blood due to panting. -- shift ; eggshells are thinner How could we increase eggshell thickness in summer? -- give chickens carbonated water [ CO2 ] , shift -- put CaO additives in chicken feed [ CaO ] , shift -- air condition the chicken house TOO much $$$ -- pump CO2 gas into the chicken house would kill all the chickens!

  49. Other applications with LeChatelier’s Principle …favors the endothermic reaction (the reverse reaction) in which the rise in temperature is counteracted by the absorption of heat. Raising the temperature… …favors the forward reaction in which 4 mol of gas molecules is converted to 2 mol. Increasing the pressure… N2 + 3 H2 2 NH3 + heat Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 532

  50. Light-Darkening Eyeglasses AgCl + energy Ago + Clo (clear)(dark) Go outside… Sunlight more intense than inside light; “energy” shift to a new equilibrium: GLASSES DARKEN Then go inside… “energy” GLASSES LIGHTEN shift to a new equilibrium:

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