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Electrochemistry

Electrochemistry. Chapter 20. Half-reaction method…remember?. Example Al (s) + Cu 2+ (aq)  Al 3+ (aq) + Cu (s) Oxidation: Al (s)  Al 3+ (aq) + 3e - Reduction: 2e - + Cu 2+ (aq)  Cu (s) Use lowest common multiple to make both equivalent in number of electrons

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Electrochemistry

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  1. Electrochemistry Chapter 20

  2. Half-reaction method…remember? • Example Al(s) + Cu2+(aq) Al3+(aq) + Cu(s) • Oxidation: Al(s)  Al3+(aq) + 3e- • Reduction: 2e- + Cu2+(aq) Cu(s) • Use lowest common multiple to make both equivalent in number of electrons • Oxidation  multiply by 2 • 2Al(s)  2Al3+(aq) + 6e- • Reduction  multiply by 3 • 6e- + 3Cu2+(aq) 3Cu(s) • Collate (electrons cross out) • Net reaction: 2Al(s) + 3Cu2+(aq) 2Al3+(aq) + 3Cu(s) DOES EVERYTHING BALANCE? (Make sure to balance after every step!)

  3. In acidic milieu Fe2+(aq) + MnO4-(aq)  Fe3+(aq) + Mn2+(aq) • Oxidation: Fe2+(aq)  Fe3+(aq) + e- • Reduction: 8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq) • What did I do in the above half-rxn? • Is it fully balanced? 5e- + 8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq) • Balance both half-reactions: 5Fe2+(aq)  5Fe3+(aq) + 5e- (multiply by 5; why?) 5e- + 8H+(aq) + MnO4-(aq)  4H2O(l) + Mn2+(aq) • Collate • Net rxn: 5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  5Fe3+(aq) +4H2O(l) + Mn2+(aq)

  4. Solve VO2+(aq) + Zn(s) VO2+(aq) + Zn2+(aq)

  5. Answer VO2+(aq) + Zn(s) VO2+(aq) + Zn2+(aq) • Oxidation: Zn(s) Zn2+(aq) + 2e- • Reduction: e- +2H+(aq) + VO2+(aq)  VO2+(aq) + H2O(l) • Balancing both half-reactions: Zn(s) Zn2+(aq) + 2e- 2e- +4H+(aq) + 2VO2+(aq)  2VO2+(aq) + 2H2O(l) • Collate • Net reaction: Zn(s) +4H+(aq) + 2VO2+(aq)  2VO2+(aq) + 2H2O(l) +Zn2+(aq)

  6. In basic milieu I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) • Oxidation: I-(aq)  I2(aq) + e- 2I-(aq)  I2(aq) + 2e- • Reduction: MnO4-(aq)  2OH-(aq) + MnO2(s) 2H+(aq) + 2OH-(aq) + MnO4-(aq)  2OH-(aq)+ MnO2(s) 2H2O(l) + MnO4-(aq)  2OH-(aq)+ MnO2(s) 2H2O(l) + MnO4-(aq)  4OH-(aq)+ MnO2(s) 3e- + 2H2O(l) + MnO4-(aq)  4OH-(aq)+ MnO2(s) • Balance both half-reactions: 6I-(aq)  3I2(aq) + 6e- 6e- + 4H2O(l) + 2MnO4-(aq)  8OH-(aq)+ 2MnO2(s) • Collate • Net rxn: 6I-(aq) + 4H2O(l) + 2MnO4-(aq)  3I2(aq) + 8OH-(aq)+ 2MnO2(s)

  7. Solve • Al(s) + H2O(l) Al(OH)4-(aq) + H2(g)

  8. Answer • Al(s) + H2O(l) Al(OH)4-(aq) + H2(g) • Oxidation: Al(s) + 4OH-(aq)  Al(OH)4-(aq) + 3e- • Reduction: 2e- + 2H2O(l) 2OH-(aq) + H2(g) • Balance each half-reaction: 2Al(s) + 8OH-(aq)  2Al(OH)4-(aq) + 6e- 6e- + 6H2O(l) 6OH-(aq) + 3H2(g) • Collate • Net-reaction: 2Al(s) + 2OH-(aq) + 6H2O(l)  2Al(OH)4-(aq) +3H2(g)

  9. Electricity • Movt of electrons • Movt of electrons through wire connecting 2 half-reactions  electrochemical cell • Also called voltaic or galvanic cell • Cell produces current from spontaneous rxn • Example: copper in solution of AgNO3 is spontaneous • On the other hand, an electrolytic celluses electrical current to drive a non-spontaneous chemical rxn

  10. Solid Zn in zinc ion solution = half-cell Likewise, Cu/Cu-ion solution Wire attached to each solid Salt bridge = 1. contains electrolytes, 2. connects 2 half-cells, 3. anions flow to neutralize accumulated cations at anode and cations flow to neutralize accumulated anions at cathode (completes circuit) “An Ox” = anode oxidation Has negative charge because releases electrons “Red Cat” = reduction cathode Has positive charge because takes up electrons Voltaic cell

  11. Electrical current • Measured in amperes (A) • 1 A = 1 C/s • Coulomb = unit of electric charge • e- = 1.602 x 10-19 C • 1 A = 6.242 x 1018 e-/s • Electric current driven by difference in potential energy per unit of charge: J/C • Potential difference (electromotive force or emf) = volt (V) • Where 1 V = 1 J/C

  12. More… • In the voltaic cell, potential difference (emf) between cathode and anode is referred to as • Cell potential (Ecell) • Under standard conditions (1 M, 1 atm, 25°C), cell potential is • Standard cell potential = E°cell • Cell potential = measure of overall tendency of redox rxn to occur spontaneously • Thus, the higher the E°cell, the greater the spontaneity

  13. Electrochemical notation • Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s) • Notation describes voltaic cell • An ox on left • Red cat on right • Separated by double vertical line (salt bridge) • Single vertical line separates diff phases

  14. Electrochemical notation • Some redox rxns reactants & products in same phase • Mn doesn’t precipitate out  uses Pt at cathode • Pt is inert, but provides area for electron gain/loss • Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq), Mn2+(aq)|Pt(s) • Write out net reaction

  15. Answer • Fe(s)|Fe2+(aq)||MnO4-(aq), H+(aq), Mn2+(aq)|Pt(s) • Oxidation: Fe(s) Fe2+(aq) + 2e- • Reduction: 5e- + MnO4-(aq) + 8H+(aq)Mn2+(aq) + 4H2O(l) • Net-reaction: 5Fe(s) + 2MnO4-(aq) + 16H+(aq) 5Fe2+(aq) + 2Mn2+(aq) + 8H2O(l)

  16. Standard reduction potentials • One half-cell must have a potential of zero to serve as reference • Standard hydrogen electrode (SHE) half-cell • Comprises Pt electrode in 1 M HCl w/ H2 bubbling at 1 atm: • 2H+(aq) + 2e- H2(g); E°red = 0.00 V

  17. Example • Throw zinc into 1M HCl • Zn(s)|Zn2+(aq)||2H+(aq)|H2(g) • E°cell = E°ox + E°red= 0.76 V • If E°red = 0.00 V (as the reference) • Then E°ox = 0.76 V (= oxid of Zn half-rxn) • Reduction of Zn-ion • Is = -0.76 V (non-spontaneous)

  18. Problem • Cr(s)|Cr3+(aq)||Cl-(aq)|Cl2(g) • What is the std cell pot (E°cell) given oxid of Cr = 0.73 V and Cl red = 1.36V? • Hint: standard electrode potentials are intensive properties; e.g., like density • Stoichiometry irrelevant!

  19. Solution • E°cell = E°ox + E°red = 0.73V + 1.36V = 2.09V

  20. Appendix M, pages A-33-35 • Standard reduction potentials in aqueous solution @ 25°C • Also, pg. 967, Table 20.1 (gives increasing strengths of ox/red agents) • Let’s take a look at it • Does increasing strengths of ox/red agents make sense? • What happens to oxidizing agent, reducing agent?

  21. Problem • Calculate the standard cell potential for the following: Al(s) + NO3-(aq) + 4H+(aq) Al3+(aq) + NO(g) + 2H2O(l)

  22. Answer • Oxidation Al(s) Al3+(aq) + 3e-; E°ox = 1.66V • Reduction NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l); E°red = 0.96V • E°cell = E°ox + E°red =1.66V + 0.96V = 2.62V

  23. Predicting the spontaneous direction of a redox rxn • Generally, any reduction half-rxn is spontaneous when paired w/reverse of half-rxn below it in table of standard reduction potentials • Let’s look at table • Predict the exact value and spontaneity for the following: Fe(s) + Mg2+(aq) Fe2+(aq) + Mg(s)

  24. Answers Fe(s) + Mg2+(aq) Fe2+(aq) + Mg(s) • Oxidation Fe(s) Fe2+(aq) + 2e-; E°ox = 0.45V • Reduction Mg2+(aq) + 2e- Mg(s); E°red = -2.37V • E°cell = E°ox + E°red =0.45V + -2.37V = -1.92V nonspontaneous

  25. Will metal X dissolve in acid? • Metals whose reduction half-rxns lie below reduction of proton to hydrogen gas will dissolve in acids • Why? • Just look at the table! • Nitric acid is exception • Let’s take a look

  26. E°cell, G°, K • What must the values for E°cell, G°, & K be in order to have a spontaneous rxn? • G°<0 • E°cell>0 • K>1 • Product-favored

  27. Relationship between G° & E°cell • Faraday’s Constant (F) = 96,485 C/mol e- • G° =-ne-FE°cell • Problem: • CalculateG° for I2(s) + 2Br-(aq) 2I-(aq) + Br2(l) Is it spontaneous?

  28. Solution: it’s nonspontaneous! I2(s) + 2Br-(aq) 2I-(aq) + Br2(l) • Oxidation 2Br-(aq) Br2(l) + 2e-; E°ox = -1.09V • Reduction I2(s) + 2e- 2I-(aq); E°red = 0.54V • E°cell = -1.09V + 0.54V = -0.55V

  29. Problem 2Na(s) + 2H2O(l) H2(g) + 2OH-(aq) + 2Na+(aq) Is it spontaneous?

  30. Solution: it’s spontaneous! • Oxidation 2Na(s) 2Na+(aq) + 2e-; E°ox = 2.71V • Reduction 2H2O(l) + 2e-  H2(g) + 2OH-(aq) E°red = -0.83V • E°cell = 2.71V + -0.83V = 1.88V

  31. Relationship between E°cell & K

  32. Problem • Calculate K for 2Cu(s) + 2H+(aq) Cu2+(aq) + H2(g)

  33. Solution: is it product-favored? • Oxidation 2Cu(s) Cu2+(aq) + 2e-; E°ox = -0.34V • Reduction 2H+(aq)+ 2e- H2(g); E°red = 0.00V E°cell = -0.34V

  34. Cell potential & concentration: Nernst Equation • Concentration ≠ 1M • Non-standard conditions • Under standard conditions, Q = 1 •  Ecell = E°cell

  35. Problem • Compute the cell potential, given Cu(s) Cu2+(aq, 0.010 M) + 2e- MnO4-(aq, 2.0 M) + 4H+(aq, 1.0M) + 3e-  MnO2(s) + 2H2O(l)

  36. Solution • Balance the equation! • Oxidation 3Cu(s) 3Cu2+(aq)+ 6e-; E°ox = -0.34V • Reduction 2MnO4-(aq)+ 8H+(aq) + 6e-  2MnO2(s) + 4H2O(l) ; E°red = 1.68V

  37. To summarize • If Q<1, rxn goes to products • Ecell > E°cell • If Q>1, rxn goes to reactants • Ecell < E°cell • If Q = K, @ eq., • E°cell = 0 (& Ecell = 0) • Explains why all batteries die

  38. Concentration cells • Voltaic cells can be constructed from two similar half-rxns where difference in concentration drives current flow Cu(s) + Cu2+(aq, 2.0M)  Cu2+(aq, 0.010M) + Cu(s) • E°cell = 0 since both half-rxns are the same • However, using Nernst equation, different concentrations yield 0.068V • Let’s take a look • Flow is from lower Cu-ion concentration half-cell to higher one • Down the concentration gradient • The electrons will flow to the concentrated cell where they dilute the Cu-ion concentration • Results in  Cu-ion concentration in dilute cell &  Cu-ion concentration in concentrated cell

  39. Batteries • Dry-cell batteries • Don’t contain large amounts of water • Anode Zn(s) Zn2+(aq) + 2e- • Cathode 2MnO2(s) + 2NH4+(aq) + 2e- Mn2O3(s) + 2NH3(g) + H2O(l) • Cathode is carbon-rod immersed in moist (acidic) paste of MnO2 that houses NH4Cl • 1.5 V

  40. Batteries • More common dry-cell type • Alkaline battery • Anode Zn(s) + 2OH-(aq) Zn(OH)2(s) + 2e- • Cathode 2MnO2(s) + 2H2O(l) + 2e- 2MnO(OH)(s) + 2OH-(aq) • Longer shelf-life, “live” longer • Cathode in basic paste

  41. Car Batteries • Lead-acid storage batteries • 6 electrochemical cells (2V) in series • Anode Pb(s) + HSO4-(aq) PbSO4(s) + H+(aq) + 2e- • Cathode PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e-  PbSO4(s) + 2H2O(l) • In 30% soln of sulfuric acid • If dead due to excess PbSO4 covering electrode surfaces • Re-charge (reverse rxn)  converts PbSO4 to Pb and PbO2

  42. Rechargeable batteries • Ni-Cd • Anode Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2e- • Cathode 2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq) • KOH, usually, used • 1.30 V • Reverse rxn recharges battery • Excess recharging  electrolysis of water • EXPLOSION!!! • Muhahahaha!

  43. Rechargeable batteries • Since Cd is toxic • Developed safer alternative • Ni-MH • Hybrid car batteries: high energy density • Same cathode rxn as previous • Anode MH(s) + OH-(aq) M(s) + H2O(l) + e- • Commonly, M = AB5, where A is rare earth mixture of La, Ce, Nd, Pr, and B is Ni, Co, Mn, and/or Mn • Very few use AB2, where A = Ti and/or V

  44. Rechargeable batteries • Anode made of graphite w/incorporated Li-ions between carbon layers • Ions spontaneously migrate to cathode • Cathode = LiCoO2 or LiMn2O4 • Transition metal reduced • Used in laptop computers, cell phones, digital cameras • Light weight and high E density

  45. Fuel cell • Reactants flow through battery • Undergo redox rxn • Generate electricity • Hydrogen-oxygen fuel cell • Anode 2H2(g) + 4OH-(aq) 4H2O(l) + 4e- • Cathode O2(g) + 2H2O(l) + 4e-  4OH-(aq) • Used in space-shuttle program • And Arnold’s Hummah

  46. Electrolysis • Electrical current used to drive nonspontaneous redox rxn • In electrolytic cells • Used in • Electrolysis of water • Metal plating: silver coated on metal, jewelry, etc.

  47. Electrolytic cells: using electricity to run a rxn • Anode is “+”  gives electrons, connected to positive terminal of power source • Cathode is “-”  takes electrons, connected to negative terminal of power source • Opposite scheme of voltaic cell!

  48. Predicting the products of electrolysis • Pure molten salts • Anion oxidized/cation reduced • Obtain 2Na(s) and Cl2(g) from electrolysis of NaCl • Mixture of cations or anions • K+/Na+ and Cl-/Br- present • Look at page 967 & compare half-cell potentials • Cation/anion preferably reduced that has least negative, or most positive, half-cell potential

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