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Electrochemistry

Electrochemistry. The study of the interchange of chemical and electrical energy. AP Chemistry Ch. 17. Oxidation-Reduction Reactions (REDOX). Oxidation: An increase in oxidation state (a loss of electrons). Reduction: A decrease in oxidation state (a gain of electrons). Oxidizing Agent:

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Electrochemistry

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  1. Electrochemistry The study of the interchange of chemical and electrical energy. AP Chemistry Ch. 17

  2. Oxidation-Reduction Reactions (REDOX) • Oxidation: • An increase in oxidation state (a loss of electrons). • Reduction: • A decrease in oxidation state (a gain of electrons). • Oxidizing Agent: • Electron acceptor • A reactant that accepts electrons from another. • Reducing Agent: • Electron donor • A reactant that donates electrons to another substance to reduce the oxidation state of one of its atoms.

  3. Assigning Oxidation States Rules for Assigning Oxidation States The Oxidation State of… • An atom in an element is zero • Na(s), O2(g), Hg(l) • A monatomic ion is the same as its charge • Na+, Cl- • Fluorine is -1 in its compounds • HF, PF3 • Oxygen is usually -2 in its compounds • H2O, CO2 • Exception: Peroxides (containing O22-) in which oxygen is -1 • Hydrogen is +1 in its covalent compounds • H2O, HCl, NH3

  4. Recall… Zinc reacts with HCl: Zn(s) + H+(aq) + Cl-(aq)  Zn2+(aq) + Cl-(aq) + H2(g) • Half-Reactions

  5. Balancing a Redox Reaction that occurs in Acidic Solution • Steps: Acidic • 1. Write separate equations for the oxidation and reduction half-reactions. • 2. For each half-reaction • Balance all the elements except hydrogen and oxygen. • Balance oxygen using H2O. • Balance hydrogen using H+. • Balance the charge using electrons (e-). • 3. If necessary, multiply one or both balanced half-reactions by an integer equalize the number of electrons transferred in the two half-reactions. • 4. Add the half-reactions, and cancel identical species. • 5. Check that the elements and charges are balanced.

  6. Practice Complete and balance the following equations that occur in acidic solutions. a.) MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) b.) Mn2+(aq) + NaBiO3(s)  Bi3+(aq) + MnO4-(aq)

  7. Balancing a Redox Reaction that occurs in Basic Solution • Steps: Basic • The same as acidic and then… • 6. Add a number of OH- ions to both sides so that you just balance excess H+ ions. • 7. H+ and OH- will form H2O on the side with excess H+. Free OH- will appear on one side of the equation. • 8. Double check atoms and charges!

  8. Practice • Complete and balance the following equations that occur in basic solutions. a.) NO2-(aq) + Al(s)  NH3(aq) + AlO2-(aq) b.) Cr(OH)3(s) + ClO-(s)  CrO42-(aq) + Cl2(g)

  9. Voltaic (or galvanic) cells • Voltaic (galvanic) cell: • A device in which chemical energy from a spontaneous redox reaction is changed to electrical energy that can be used to do work. • Uses a spontaneous redox reaction to produce a current that can be used to do work. • Electrodes • a conductor through which electricity enters or leaves an object, substance, or region. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) • Cathode (RED CAT) • Anode (AN OX)

  10. Salt Bridge (or porous barrier): a U-tube containing an electrolyte that connects the two compartments of a galvanic cell, allowing ion flow without extensive mixing of the different solutions

  11. Analyze the Animation • Additional Video (Demo)

  12. Cell EMF Under Standard Conditions (E°) Electromotive Force (EMF) or Cell Potential (Ecell): • The driving force in a galvanic cell that pulls electrons from the reducing agent in one compartment to the oxidizing agent in the other. Voltage • Unit of electrical potential (V) • 1 joule of work per coulomb of charge transferred

  13. Standard Reduction Potentials (E°red) • Cell potential: E°cell = E°red (cathode) - E°red (anode) • For all spontaneous reactions at standard conditions, E°cell > 0. 2 H+(aq, 1 M) + 2 e- H2(g, 1 atm) E°red = 0.00 V (standard hydrogen electrode)

  14. Calculating Cell EMF • Determine the E°cellfor the reaction Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) • Electrons lost must equal electrons gained! • Determine the E°cellfor the reaction Cu(s) + Fe3+(aq)  Cu2+(aq) + Fe2+(aq)

  15. Example Consider a galvanic cell based on the following reactions. Give the balanced cell reaction and calculate the standard emf (E°) for the cell. • a. Al3+(aq) + Mg(s)  Al(s) + Mg2+(aq) • b. MnO4-(aq) + H+(aq) + ClO3-(aq)  ClO4-(aq) + Mn2+(aq) + H2O(l)

  16. Complete Description of a Galvanic Cell • Consider a galvanic cell based on the following half-reactions: Fe2+ + 2 e- Fe E° = -0.44 V MnO4- + 5 e- + 8 H+ Mn2+ + 4 H2O E° = 1.51 V • In a working galvanic cell, one of these reactions must run in reverse. Which one? • What is the balanced cell reaction? • What is the E°cell?

  17. Example Continued… • On the diagram below, draw what ions must be present in each solution and what metal is each electrode made out of. A chemically inert conductor (like platinum) is required if none of the substances participating in the half-reaction is a conducting solid.

  18. Example Problem • Describe completely (cell reaction, cell potential, and physical setup of cell) the galvanic cell based on the following half-reactions under standard conditions. Ag+ + e- Ag E° = 0.80 V Fe3+ + e- Fe2+ E° = 0.77 V

  19. Strengths of Oxidizing and Reducing Agents • The more positive the E°red value for a half-reaction, the greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize another species. • Rank the following species from the strongest to the weakest reducing agent: I-(aq), Fe(s), Al(s).

  20. Free Energy and Redox Reactions • E° = E°red (reduction process) - E°red (oxidation process) • What is the relationship between a spontaneous process and the E° cell ?

  21. Example Problem • Using standard reduction potentials, determine whether the following reaction is spontaneous under standard conditions. Cu(s) + 2 H+(aq)  Cu2+(aq) + H2(g)

  22. Cell Potential, Electrical Work, and Free Energy • The work that can be accomplished when electrons are transferred through a wire depends on the “push” (the thermodynamic driving force) behind the electrons. This driving force (the emf) is defined in terms of a potential difference (in volts) between two points in the circuit. • Emf = potential difference (V) = work (J) / charge (C) E° = -W/q • (work is viewed from the point of view of the system…thus work flows out of the system and work has a – sign.) Wmax = -q E°max q = nF q is charge ; n is number of electrons; F is Faraday’s constant

  23. Cell Potential, Electrical Work, and Free Energy • Faraday’s Constant (F):quantity of electrical charge in 1 mole of electrons. 1 F = 96,485 C/mol e- = 96,485 J/V∙mol • EMF and ΔG Wmax = ΔG = -nFE°

  24. Example Problem • Using standard reduction potentials, calculate ΔG for the reaction Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq) • Is this reaction spontaneous?

  25. Example Problem • Using standard reduction potentials, calculate ΔG and the equilibrium constant, K, at 25°C for the reaction 4 Ag(s) + O2(g) + 4 H+(aq)  4 Ag+(aq) + 2 H2O(l)

  26. Cell EMF Under Nonstandard Conditions • Standard Conditions – 1.0 M solutions • Nonstandard Conditions: Cu + 2 Ce4+ Cu2+ + 2 Ce3+E°cell = 1.36 V What would happen to Ecell if we increase [Ce4+]? • Concentration Cells Ag+ + e- Ag E°cell = 0.80 V

  27. The Nernst Equation **Excluded in AP Curriculum**Qualitative Understanding ONLY! • Ecell = E°cell – (RT/nF)lnQ = E°cell – (2.303RT/nF)logQ at 25°C Ecell = E°cell – (0.0592/n)logQ

  28. Example Problem • Describe the cell based on the following half-reactions: VO2+ + 2 H+ + e- VO2+ + H2O E° = 1.00 V Zn2+ + 2 e- Zn E° = -0.76 V where: T = 25°C [VO2+] = 2.0 M [H+] = 0.50 M [VO2+] = 0.010 M [Zn2+] = 0.10 M

  29. Example Problem • Calculate the emf at 298K for the cell when [Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, and [Cr3+] = 1.0 x 10-5 M. Cr2O72-(aq) + 14 H+(aq) + 6 I-(aq)  2 Cr3+(aq) + 3 I2(s) + 7 H2O(l)

  30. Concentration Cell: • A galvanic cell in which both compartments contain the same components, but at different concentrations. • Although the standard emf for this cell is zero, the cell operates under nonstandard conditions because the concentration of Ni2+(aq) is different in the two compartments. Ni(s)  Ni2+(aq) + 2 e-E°red = -28.0 V anode Ni2+(aq) + 2 e- Ni(s) E°red = -28.0 V cathode • Calculate Ecell

  31. Batteries and Fuel Cells • Lead-Acid Battery (12-V car battery) • Alkaline Battery • Nickel-Cadmium Batteries

  32. Corrosion • Corrosion of Iron • Preventing the Corrosion of Iron

  33. Electrolysis • Electrolysis: • A process that involves forcing a current through a cell to cause a nonspontaneous chemical reaction to occur. • Electrolytic Cell: • A cell that uses electrical energy to produce a chemical change that would otherwise not occur spontaneously. • Stoichiometry of electrolytic processes • Coulombs = Amperes (C/s) x seconds q = I x t • 1 mole of electrons carries a charge of 1 faraday (96,485 C)

  34. Example Problems • Calculate the number of grams of aluminum produced in 1.00 hour by the electrolysis of molten AlCl3 if the electrical current is 1.00 A. • How long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g silver metal?

  35. Example Problem • A solution in an electrolytic cell contains the ions Cu2+, Ag+, and Zn2+. If the voltage is initially low and gradually turned up, in which order will the metals be plated out onto the cathode? The standard reduction potentials: Ag+ + e- Ag E° = 0.80 V Cu2+ + 2 e- Cu E° = 0.34 V Zn2+ + 2 e- Zn E° = -0.76 V

  36. Example Problem • Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 103 kg of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V. 1 W = 1 J/s 1 kWh = 3.6 x 106 J

  37. Example Problem • An unknown metal M is electrolyzed. It took 74.1 s for a current of 2.00amp to plate out 0.107 g of the metal from the solution containing M(NO3)3. Identify the metal.

  38. Additional Practice • Textbook problems: • Beginning on page 830 • # 13, 15, 16, 25, 27, 29, 31, 33, 43, 49, 53 (conceptually only), 65, 73, 81, and 87 • Unit 8 Review Packet

  39. Final Exam Analysis • 25minutes individual review…..for half credit (based on reflection)! • Learning is a PROCESS! • For credit you must correct your answer and provide a brief reflection of why your answer choice was incorrect and how you corrected it. (3 to 4 sentences per question)

  40. RAFT – Final Project • R – Role (Recruiter/Motivator) • A- Audience (Future AP Chemistry Students) • F – Format (Your CHOICE!... Get CREATIVE) • T –Topic (Your CHOICE!... Any of the 8 Units) • See Rubric • Due May 21st for Presenting!

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