Electrochemistry AP Chem/Mrs. Molchany (0808)

2 out of 49 Drill Use AP Review Drill #75-77

3 out of 49 Objectives SWBAT Determine which electrode supports oxidation and which supports reduction. Write a half reaction for the anode and cathode of a galvanic cell. Explain the purpose of a salt bridge in a galvanic cell.

4 out of 49 Voltaic Cell (or Galvanic Cell) A voltaic cell is a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants. In a voltaic cell, chemical energy is changed to electrical energy. The energy released in a spontaneous redox reaction can be used to perform electrical work.

5 out of 49 Half Cells The voltaic cell is thought of as being comprised of two "half-cells." One cell is where oxidation occurs and the other is reduction.

6 out of 49 Half Cells Each half cell is made of a metal strip in contact with a solution of its ions. In the oxidation half cell, the metal electrode loses electrons. In the reduction half cell, the ions in solution gain electrons and create metal atoms which deposit onto the electrode.

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8 out of 49 Anode/Cathode The two solid metals that are connected by the external circuit are called electrodes. The electrode at which oxidation occurs is called the anode. This electrode strongly attracts negative ions in the solution, these ions are then called anions. The electrode at which reduction occurs is called the cathode. Cations will be deposited onto this electrode as a metal.

9 out of 49 Half Reactions Anode (oxidation half-reaction): Zn (s)→Zn +2 (aq) + 2e- Cathode (reduction half-reaction): Cu +2 (aq) +2e- →Cu (s) Determining half reactions will be shown later.

10 out of 49 OIL – RIG Remember the acronym OIL – RIG Oxidation Is Loss of electrons Reduction Is Gain of electrons

11 out of 49 Red-Ox Reaction In a red-ox reaction, one substance must be oxidized and another substance must be reduced. The substance that is “oxidized” is the “reducing agent”. The substance that is “reduced” is the “oxidizing agent”.

12 out of 49 Cell Operation (using the Zn/Cu cell) Electrons become available when the zinc metal is oxidized at the anode. The electrons flow through the external circuit to the cathode, where they are consumed as Cu +2 is reduced. Because zinc is oxidized in the cell, the zinc electrode loses mass, and the concentration of the Zn +2 solution increases as the cell operates. At the same time, the Cu electrode gains mass, and the Cu +2 solution becomes less concentrated as the Cu +2 is reduced to Cu (s).

13 out of 49 Cell Operation As the voltaic cell operates, oxidation of Zn introduces additional Zn +2 ions into the anode compartment. Unless a means is provided to neutralize this positive charge, no further oxidation can take place. At the same time, the reduction of Cu +2 at the cathode leaves an excess of negative charge in solution in that compartment.

14 out of 49 Salt Bridge Electrical neutrality of the system is maintained by a migration of ions through the porous glass disc (salt bridge) that separates the two compartments. A salt bridge consists of a U-shaped tube that contains an electrolyte solution whose ions will not react with other ions in the cell or with the electrode materials.

15 out of 49 Salt Bridge As oxidation and reduction proceed at the electrodes, ions from the salt bridge migrate to neutralize charge in the cell compartments. The cell diagram slide shows the Zn +2 ions migrating into the salt bridge from the left cell and the SO 4 -2 ions migrating into the salt bridge from the right cell.

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17 out of 49 Salt Bridge The salt bridge is required for completing the circuit. It allows the movement of ions from one solution to the other so that a charge does not build up in either of the cells.

18 out of 49 Ion Migration Anions migrate toward the anode and cations toward the cathode. No measurable electron flow will occur through the external circuit unless a pathway is provided for ions to migrate through the solution from one electrode compartment to another, completing the circuit.

19 out of 49 Electron Flow In any voltaic cell the electrons flow from the anode through the external circuit to the cathode. Because the negatively charged electrons flow from the anode to the cathode, the anode in a voltaic cell is labeled with a negative sign and the cathode with a positive sign.

20 out of 49 Oxidation Numbers Remind yourself how to determine oxidation numbers. Oxidation numbers show what the charge of each atom would be (in a molecule or ion), if each atom were an ion. Go to the text and revisit the Rules for Assigning Oxidation Numbers section.

21 out of 49 Balancing Red-Ox Equations Obey the Law of Conservation of Mass Gain and loss of electrons must be balanced

22 out of 49 Half Reactions Cu + 2Ag + → 2 Ag + Cu +2 Determine the oxidation number for each substance in the reaction Write two half reactions based on one substance being oxidized and one being reduced.

23 out of 49 Half Reactions Cu → Cu e-oxidation 2e- + 2Ag + → 2 Agreduction Half reactions show the number of electrons gained or lost by the substance, standard redox equations do not. Go to ChemReview packet to practice half reaction writing.

24 out of 49 Review Sample Ex: 20.4 Try Practice Exercises

25 out of 49 Cell EMF What is the driving force that pushes the electrons through an external circuit in a voltaic cell? An oxidizing agent in one compartment pulls electrons through a wire from a reducing agent in the other compartment. The pull (or driving force) on the electrons is called the cell potential ( E cell ) or electromotive force (emf) of the cell. Or, electrons flow from the anode to the cathode (in a voltaic cell) because of a difference in potential energy.

26 out of 49 Potential Energy of Electrons The potential energy of electrons is higher in the anode than in the cathode. Therefore, electrons spontaneously flow through an external circuit from the anode to the cathode. For any cell reaction that proceeds spontaneously (i.e. voltaic cell), the cell potential will be positive.

27 out of 49 Standard Conditions Standard Conditions: * 1M concentrations for reactants and products in solution * 1 atm pressure (for gases) * 25 ˚C Under Standard Conditions the EMF is called the “standard emf” or the standard cell potential (E ˚ cell ) Keep in mind that the superscript ˚ denotes standard-state conditions. Note: Read the symbol “ ˚ ” as the word not

28 out of 49 Cell Potential The cell potential of a voltaic cell depends on the particular cathode and anode half-cells involved. Standard potentials have been assigned to each individual half-cell. You can look them up in your AP Packet. Then use the half-cell potentials to determine E ˚ cell. The cell potential is the difference between two electrode potentials (anode and cathode).

29 out of 49 Cell Potential The potential associated with each electrode is chosen to be the potential for reduction to occur at that electrode. Therefore, your AP Packet only shows Standard Reduction Potential reactions and EMF values.

30 out of 49 Standard Reduction Potentials Standard electrode potentials are tabulated for reduction reactions, so they are called standard reduction potentials ( E ˚ red ). E ˚ cell = E ˚ red (cathode) - E ˚ red (anode) Or E ˚ cell = E ˚ red (cathode) + E ˚ ox (anode)

31 out of 49 Standard Reduction Potentials I will use the following equation: E ˚ cell = E ˚ red (cathode) + E ˚ ox (anode) When using this equation: * Determine which half reaction is the oxidation reaction. * “Flip” the reaction by writing it backward. * Remember to change the sign of the EMF value.

32 out of 49 Standard Reduction Potentials The hydrogen half-reaction was used as a reference for all of the other half-reactions. The standard reduction potential for the hydrogen half-reaction is assigned a standard reduction potential of exactly 0 V. It is also called the Standard Hydrogen Electrode. (SHE)

33 out of 49 Calculating EMF We will use standard reduction potentials to calculate the EMF of a voltaic cell. Because electrical potential measures potential energy per electrical charge, standard reduction potentials are intensive properties. Changing the stoichiometric coefficient in a half- reaction does not affect the value of the standard reduction potential. Do Not multiply the Standard Reduction Potential value by the number of moles !!!

34 out of 49 FYI In general, the more active the metal, the lower its potential.

35 out of 49 Metal Activity The most active metal, Li, has the lowest tendency to gain electrons (to be reduced). You will see on the chart that it’s reduction potential value is extremely low. Li has the strongest tendency of the metals listed to lose electrons (to be oxidized).

36 out of 49 Metal Activity The least active metal, Au, is at the bottom of the table. It has a greater tendency to be reduced. Gold also has a lower tendency to lose electrons (to be oxidized).

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38 out of 49 In the cell diagram: Zinc replaces copper Zn + CuSO 4 → ZnSO 4 + Cu

39 out of 49 Calculating EMF Try sample exercise 20.5

40 out of 49 E ˚ red and spontaneity 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. Try sample exercise 20.8

41 out of 49 EMF and Free Energy Change ΔG = - nFE “n” is a positive # that represents the # of electrons transferred in the reaction “F” is Faraday’s constant (this constant is the quantity of electrical charge on 1 mole of electrons) 1F = 96,500 C/mol = 96,500 J / V mole The units for ΔG are J/mole

42 out of 49 EMF and Free Energy Change ΔG = - nFE If “n” and “F” are positive values, a positive value of E leads to a negative ΔG. Remember that a negative ΔG indicates a spontaneous reaction. The equation can be altered slightly if the reactants and products are in their standard states: ΔG˚ = - nFE˚

43 out of 49 Cell EMF Under Nonstandard Conditions The next set of slides explains how to calculate the cell emf while working under non-standard conditions, specifically concentrations other than 1M.

44 out of 49 Concentration and Cell EMF Remember ΔG = ΔG˚ + RT ln Q Combining ΔG = ΔG˚ + RT ln Q and ΔG = - nFE You have the Nernst equation: E = E˚ – RT ln Q nF

45 out of 49 Nernst Equation The Nernst equation can be expressed two ways: E = E˚ – RT ln Q nF E = E˚ – 2.303RT log Q nF Base 10 logs are related to natural logarithms by a factor of

46 out of 49 Nernst Equation Variation The equation can be simplified if the cell is run under standard conditions: E = E˚ – V log Q n * Temp. is 298 K

47 out of 49 Remember when we talked about equilbrium situations, E = E˚ – V log [products ] n[reactants ] Remember: this form of the Nernst equation can only be used is the temp. is 298K.

48 out of 49 Use the Nernst Equation to work through the example on the bottom of page 773 Try the sample exercise and practice exercise

49 out of 49 Electrolysis Electrolysis reactions are driven by an outside source of electrical energy and take place in electrolytic cells. This process is the opposite of a voltaic cell. Voltaic cells are spontaneous processes.

50 out of 49 Electrolysis An electrolytic cell consists of two electrodes in a molten salt or a solution.

51 out of 49 Ch 20 Problems 6, 7, 9, 14, 24, 25 a,b, 27, 29, 31, 36, 43, 44, 46, 49, 51, 52, 56, 59