1 / 28

Chapter 17 Electrochemistry The study of the interchange of chemical and electrical energy

Chapter 17 Electrochemistry The study of the interchange of chemical and electrical energy. Section 17.1: Galvanic Cells. Oxidation-reduction reaction Involves the transfer of electrons LEO the lion says GER Oxidation- Loss of electrons. Reduction- Gain of electrons

tilden
Download Presentation

Chapter 17 Electrochemistry The study of the interchange of chemical and electrical energy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 17 Electrochemistry The study of the interchange of chemical and electrical energy

  2. Section 17.1: Galvanic Cells Oxidation-reduction reaction Involves the transfer of electrons LEO the lion says GER Oxidation- Loss of electrons. Reduction- Gain of electrons half-reactions- Show the oxidation or reduction of a REDOX reaction Examples: MnO4- + Fe2+↔ Mn2+ + Fe3+

  3. How can redox reactions be used to obtain useful work instead of just heat? We can separate the two half-reactions and connect them with a wire. Why is a salt bridge or a porous disk needed between two solutions undergoing a redox reaction? The circuit is not complete and the electrons stop flowing. The bridge allows the electrons to keep moving.

  4. Galvanic cell - Where chemical energy is converted to electrical energy Cathode – Electrode compartment where reduction occurs Anode – Electrode compartment where oxidation occurs

  5. Cell potential – Ecell – The driving force on the electrons to keep them moving. -Also called electromotive force (emf) Volt – The unit of electrical potential (1 Joule/coulomb)

  6. Section 17.2 Standard Reduction Potentials • Standard hydrogen electrode - platinum electrode in contact with 1M H+ and H2 at 1 atm. • Standard reduction potentials – half reactions with all solutes at 1M and all gases at 1 atm. (Appendix 5.5) • All based on the standard hydrogen electrode (Eo=0) • All written as reduction reactions

  7. The reaction in a galvanic cell is always a REDOX reaction that can be broken down into two ½ reactions 2H+ + Zn(s) ↔ Zn2+ + H2 Anode: Zn(s) ↔Zn2+ + 2e- Cathode: 2H+ + 2e- ↔H2 Eocell = Eo(2H+↔H2) + Eo(Zn ↔ Zn2+)

  8. What must you do to obtain a balanced oxidation-reduction reaction from combining 2 half-reactions? • One of the reduction ½ reactions must be reversed. • 2. The ½ reaction with the largest positive potential will run as written. • Eocell = Eo(reduction) + Eo(oxidation) • Eocell = Eo(cathode) + Eo(anode) • 3. The number of electrons lost must equal the number gained we must multiply by integers to make them equal. (BUT DO NOT MULTIPLY Eo) • Example: Fe3+ + Cu(s) ↔ Cu2+ + Fe2+ • In-Class Practice: p. 831 #27

  9. If all substances involved in the reaction are aqueous, what is used as an electrode? • Electrodes are composed of the pure metal which is the same metal that is in solution. If a metal cannot be used as an electrode a non-reacting conductor must be used, such as Platinum • How do you determine which half-reaction runs in reverse? • A cell will always run spontaneously in the direction that produces a positive cell potential.

  10. Line notation: Fe3+ + Cu(s) ↔ Cu2+ + Fe2+ Fe3+ + 2e-↔Fe2+ (reduced) (cathode) Cu(s) ↔ Cu2+ + 2e- (oxidized) (anode) ANODE // CATHODE Cu(s)/Cu2+//Fe3+,Fe2+/Pt(s)

  11. Line Notation • Example: Cr3+ + Cl2↔ Cr2O72- + Cl- • Anode: • Cathode: • Line Notation: • Assignment: Worksheets

  12. Complete descriptions of galvanic cells include what four items? 1. The cell potential and the balanced cell reaction. 2. The direction of the flow, given by the half-reactions and using the direction that gives a positive Ecell. 3. Designation of the anode and cathode. 4. The nature of the electrode and the ions present in each compartment (line notation).

  13. Description of Galvanic Cell • Fe2+ +2e-→ Fe (s) Eo= -0.44 V • MnO4- + 5e- + 8H+ → Mn2+ + 4H2O Eo=1.51 V • Balanced Reaction and Cell Potential • Direction of flow • Anode and Cathode • Line Notation

  14. Section 17.3: Cell Potential, electrical work, and Free Energy The work that can be accomplished by transferring electrons through a wire depends on: -The thermodynamic driving force (VOLTS). volts = work (J)/charge (C) or E= -w/q (*negative because work is flowing OUT) –wmax = q Emax or q = nF Where n = moles and F = one Faraday (96,485 C/mol e- ) The maximum cell potential is directly related to the free energy difference between the reactants and the products. ∆Go = -nFEo

  15. Ex. Calculate the change in free energy for Cu2+ + Fe(s)→Cu(s) + Fe2+ • Ex. pg 831 #44 • Remember: • The higher the reduction potential, the better oxidizing agent it will be (easily reduced). • The lower the reduction potential, the better the reducing agent it will be (easily oxidized). • HW: pg. 831 #37, 42, 43

  16. Section 17.4: Dependence of Cell Potential on Concentration: How does Le Chatelier’s principle apply to electrochemical cells? Cells will look to reach equilibrium by increasing or decreasing the concentrations. Fe2+ + Ag+→ Ag (s) + Fe3+

  17. Concentration cell A cell in which both compartments have the same components, but the concentrations are different. The half-rxn for both sides is Ag+ + e-→ Ag (Eo= 0.80 V) If concentrations on both sides were the same, Eocell= 0.80 V-0.80 V=0 V Since the left side is lower, the rxn will shift to make more Ag+ in the left container. Therefore, the anode will lose more electrons to make more Ag+.

  18. How is the cell potential related to the concentrations and free energy? ΔG = ΔGo + RTln(Q) ΔG = -nFE and ΔGo = -nFEo Substitution yields -nFE = -nFEo + RTln(Q) Divide by –nF E = Eo – (RT/nF) ln(Q) (Nernst equation) Or E = Eo – 0.0591/n log(Q) (only @ 25oC) In a concentration cell Q = [ions in anode]/[ions in cathode]

  19. Nernst Equation • Ex. For 2Al(s) + 3Mn2+→ 2Al3+ + 3Mn(s), Eocell= 0.48 V. If [Mn2+]=0.50 M and [Al3+]=1.50 M at 25°C, what is the potential of the galvanic cell?

  20. When does a cell no longer have the ability to do work? When ΔG = O; the components of the cell have the same free energy. (dead battery) How can you use the Nernst equation to calculate equilibrium constants for redox reactions? At equlibrium Ecell = 0 and Q = K Therefore; 0 = Eo – 0.0591/n log(K) Or log(K) = nEo/0.0591 (@ 25oC) In a concentration cell Q = [ions in anode]/[ions in cathode]

  21. Calculating K • What is the Ksp for PbSO4 given the following information? • PbSO4 + 2e-→ Pb(s)+ SO42-Eo= -0.35 V • Pb2+ + 2e- → Pb(s)Eo= -0.13 V • HW: pg. 832-833 (#51, 53, 57, 59, 71, 73)

  22. Sections 17.5-17.6 • 17.5 Batteries • 17.6 Corrosion • Skipped for the sake of time

  23. Section 17.7: Electrolysis Electrolytic cell: Uses electrical energy to produce a chemical reaction Electrolysis: Forcing current through a cell to produce a chemical change. The cell potential is, therefore, negative.

  24. Notice in this picture a few things: • The anode and cathode are reversed for each cell type. • The voltage must be greater than Eo for the electrolytic cell to run. • The sign on the E is reversed to negative for the electrolytic cell.

  25. We can use stoichiometry to answer the question: “How much chemical change occurs with the flow of a given current for a specified time?” Ampere: 1 Coulomb of charge per second Example: How many electrons flow in a current of 10.0 A for 30 minutes?

  26. Plating • Depositing the neutral metal on the electrode by reducing the metal ions in solution. • These problems require the following steps: • Example: What mass of Cu plated when a current of 10.0 A is passed for 30 minutes through a solution of Cu2+? Moles of Electrons Moles of Metal Current and Time Quantity of charge in Coulombs Grams of Metal

  27. In a mixture of metal ions, what factor decided the metals order of plating? The standard reduction potentials Example: Ag+ + e-→ Ag Eo = 0.80 Cu2+ + 2e-→Cu Eo = 0.34 Zn2+ + 2e-→Zn Eo = -0.76 The more positive the Eo, the more easily the reaction occurs in that direction, therefore the oxidizing ability is Ag+ > Cu2+> Zn2+ This means that Ag will plate out first and so on. HW: pg. 834 (#77, 79)

  28. 17.8 Commercial Electrolytic Process • Skipped due to time constraints

More Related