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Electrochemistry szczepas@canisius.edu

Are you sure I can have that electron?. Na. Cl. Cl -. Na +. Electrochemistry szczepas@canisius.edu. I’m positive!. Useful Links. This presentation: www.canisius.edu/~szczepas Past Years’ Exams + Answers Google ACS Chemistry Olympiad

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Electrochemistry szczepas@canisius.edu

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  1. Are you sure I can have that electron? Na Cl Cl- Na+ Electrochemistryszczepas@canisius.edu I’m positive!

  2. Useful Links • This presentation: www.canisius.edu/~szczepas • Past Years’ Exams + Answers • Google ACS Chemistry Olympiad • http://www.acs.org/content/acs/en/education/students/highschool/olympiad/pastexams.html

  3. O O O O O H H H H H H H H H H e- Cl- H+ Cu+ Cu e- e-

  4. Voltaic (Galvanic) Cell Electrode Electrode Reduction Oxidation

  5. to the cathode to the anode

  6. 2008 Local 39. Which occurs at the anode of any voltaic cell? I. A metal electrode dissolves. SO32- + H2O → SO42- + 2 H+ + 2 e- II. A substance undergoes oxidation. Fe(s) → Fe2+ + 2 e- III. Positive ions are deposited from the solution. • I only • II only • I and II only • I and III only

  7. Standard Cell Potential T = 25°C Standard State  Gas (P = 1atm) Species in Solution (1 M Concentration) • Use Standard Reduction Potentials for the Reduction and Oxidation Half-Reactions • Note: No Multiplying Reduction Potential By Stoichiometry • Voltaic (Galvanic) Cell: Positive Ecell • Electrolytic Cell: Negative Ecell

  8. From 2012

  9. 2008 Local 41. What is the standard cell potential for the voltaic cell: Cr | Cr3+ || Pb2+ | Pb ? E0red / V Pb2+ + 2 e- → Pb -0.13 Cr3+ + 3 e- → Cr -0.74 • 1.09 • 0.61 • -0.61 • -1.09

  10. How many moles of electrons must pass through a cell to produce 5.00 kg of Aluminum from Al2O3? Al2O3 + 6 e- 2 Al(s) + 3 O2- 1. Calculate the number of moles of electrons needed. Using F = 96485 C/mol e-, and A=C/s, 2. How long will this take using a current of 33.5 A? Should also know that W = J/s.

  11. 2008 Local 42. During the electrolysis of AgNO3, what would happen to the mass of silver metal deposited if the current is doubled and the electrolysis time is decreased to ½ of its initial value? • It would stay the same. • It would increase to twice its initial value. • It would decrease to ¼ of its initial value. • It would decrease to ½ of its initial value.

  12. 2012 Local #42

  13. Oxidation Numbers (States) • Keeping Track of Electrons Gained or Lost 4 Rules for Assigning Oxidation Numbers: In order of Importance! • Atom in Elemental Form • Oxidation Number is Always Zero Examples: Fe Atom Ar Atom H Atom in H2 Molecule O Atom in O2 Molecule P Atom in P4 Molecule

  14. Oxidation Numbers (States) • Monatomic Ions • Oxidation Number = Charge • Examples: K+ Oxidation Number = +1 Mg2+ Oxidation Number = +2 Al3+ Oxidation Number = +3 N3- Oxidation Number = -3 S2- Oxidation Number = -2 FeCl2 ON(Fe) = +2 Oxidation Numbers are written with the sign before the number to distinguish them from Actual Charges

  15. Oxidation Numbers (States) • Nonmetals (Usually Negative Oxidation #’s, But Can Be Positive) Fluorine • Oxidation Number is -1 In All Compounds Hydrogen • Oxidation Number is +1 When Bonded to Non Metals • Oxidation Number is -1 When Bonded to Metals Oxygen • Oxidation Number is Usually -2 In Molecular and Ionic Compounds • In Peroxides (O22-) Oxidation Number is -1 for Each O Atom Other Halogens • Oxidation Number is -1 in Most Binary Compounds • Oxidation Number When Combined with Oxyanions Can Be Positive

  16. Oxidation Numbers (States) • Sum of the Oxidation Number of All Atoms in a Neutral Compound is Zero H2SO3 Oxidation Number of H = +1 (H Bonded to NonMetal) Oxidation Number of O = -2 Sum is Zero 0 = 2×ON(H) + 1×ON(S) + 3×ON(O) 0 = 2× (+1) + 1×ON(S) + 3× (-2) ON(S) = +4 Sum of the Oxidation Numbers in a Polyatomic Ion Equals the Charge of the Ion H2AsO4- -1 = 2×ON(H) + 1×ON(As)+ 4×ON(O) ON(As) = +5

  17. From 2012

  18. 2012

  19. Balancing Redox Reactions by Half-Reactions Reduction Half Reaction (Electrons Taken In) 2H+(aq) + 2e- H2(g) Another Example: Ag+(aq) + e- Ag(s) Oxidation Half Reaction (Electrons Given Off) Zn(s)  Zn2+(aq) + 2 e- Another Example: C2O42- (aq)  2 CO2(g) + 2 e-

  20. Half-Reactions To Balance Electrons, Reductions and Oxidations MUST Occur Simultaneously NO3-(aq) + 4 H+ + 3 e- NO(g) + 2 H2O(l) C2O42- (aq)  2 CO2(g) + 2 e- ID the Reducing Agent in the Unbalanced Reaction: ClO3- + Br-  Cl2 + Br2

  21. Balancing Redox Rxns In Acidic Solution: Cr2O72-(aq)+CH3OH(aq)HCO2H(aq)+Cr3+(aq) 1. Divide total reaction into two half reactions. 2. Balance each half a. All elements besides H and O b. Balance O by adding H2O c. Balance H by adding H+ d. Balance residual charge by adding e- 3. Multiply each half to least common multiple of electrons 4. Add half reactions and cancel 5. Check if balanced *To convert from acid to base conditions after steps 1-5, add enough OH- to both sides to neutralize all the H+ to H2O, then cancel out any excess. In Basic Solution: H2O2(aq)+ClO2(aq)ClO2-(aq)+O2(g)

  22. 2012

  23. A sample of copper metal is dissolved in 6 M nitric acid contained in a round bottom flask. This reaction yields a blue solution and emits a colorless gas which is found to be nitric oxide. Write a balanced equation for this reaction. Unbalanced

  24. Cell Potential • Voltaic Cell Spontaneous RedoxReaction (Ecell>0) Used to Perform Electrical Work Similar to a Waterfall (Water Falls from High to Low Potential Energy) Electrons Flow Spontaneously from High to Low Electric Potential Use Cell Potential (Cell EMF) (Ecell) • Volt  Difference in Potential Energy per Electrical Charge (1V = 1J/C) (e- charge = 1.60x10-19C) • Potential Difference Between 2 Electrodes

  25. Half Reaction E0 (V) Zn2+(aq) + 2e- Zn(s) -0.763 Cr3+(aq) + e- Cr2+(aq) -0.408 Tl+(aq) + e- Tl(s) -0.336 Cu2+(aq) + e- Cu+(aq) +0.161 Fe3+(aq) + e- Fe2+(aq) +0.769 Use the Standard Reduction Potentials to Find the Standard Cell Potential, E0cell, for the Reaction: Zn(s) + 2 Tl+(aq)  Zn2+(aq) + 2 Tl(s) 81Tl: Thallium

  26. Half Reaction E0 (V) Zn2+(aq) + 2e- Zn(s) -0.763 Cr3+(aq) + e- Cr2+(aq) -0.408 Tl+(aq) + e- Tl(s) -0.336 Cu2+(aq) + e- Cu+(aq) +0.161 Fe3+(aq) + e- Fe2+(aq) +0.769 Calculate the E0rxn based on the standard reduction potentials above. Which reaction(s) is(are) spontaneous?

  27. Ecell Non-Standard Conditions Nernst Equation

  28. Oxidation-Reduction Reactions +2 0 +1 0 Use Oxidation # to ID Oxidized and Reduced Species Zn is Oxidized (Reducing Agent) to Zn2+ H+ is Reduced (Oxidizing Agent) to H2

  29. Identify Half-Rxn (Ox or Red) +5 +5 VO3- VO2+ CrO2- CrO42- SO3 SO42- NO3 NO2- Neither: Lewis Acid/Base +3 +6 Oxidation +6 +6 Neither: Lewis Acid/Base +6 +3 Reduction

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