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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

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

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  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

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