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Electrochemistry

Chapter Eighteen. Electrochemistry. Today…. Turn in: Nothing Our Plan: Test Results Notes – Redox Equations Worksheet #1 Homework (Write in Planner): Redox Equation WS due Friday. Oxidation–Reduction: The Transfer of Electrons.

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Electrochemistry

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  1. Chapter Eighteen Electrochemistry

  2. Today… • Turn in: • Nothing • Our Plan: • Test Results • Notes – Redox Equations • Worksheet #1 • Homework (Write in Planner): • Redox Equation WS due Friday

  3. Oxidation–Reduction:The Transfer of Electrons Silver metal is formed, and the solution turns blue from copper(II) ions formed. Electrons from copper metal are transferred to silver ions.

  4. Half-Reactions (Review) • In any oxidation–reduction reaction, there are two half-reactions: • Oxidation: a species loses electrons to another species. (LEO) • Reduction: a species gains electrons from another species. (GER) • Both oxidation and reduction must occur simultaneously. • A species that loses electrons must lose them to something else (something that gains them). • A species that gains electrons must gain them from something else (something that loses them).

  5. Oxidation Numbers • An oxidation number is the charge on an ion, or a hypothetical charge assigned to an atom in a molecule or polyatomic ion. • Examples: in NaCl, the oxidation number of Na is +1, that of Cl is –1 (the actual charge). • In CO2 (a molecular compound, no ions) the oxidation number of oxygen is –2, because oxygen as an ion would be expected to have a 2– charge. • What is the oxidation number on carbon in CO2?

  6. Rules for Assigning Oxidation Numbers • For the atoms in a neutral species—an isolated atom, a molecule, or a formula unit—the sum of all the oxidation numbers is 0. • This includes elements in their standard state (Cu (s), Cl2 (g), etc.) • For the atoms in an ion, the sum of the oxidation numbers is equal to the charge on the ion. • In compounds, the group 1A metals all have an oxidation number of +1and the group 2A metals all have an oxidation number of +2. • In compounds, the oxidation number of fluorine is –1. • In compounds, hydrogen has an oxidation number of +1. • In most compounds, oxygen has an oxidation number of –2. • In binary compounds with metals, group 7A elements have an oxidation number of –1, group 6A elements have an oxidation number of –2, and group 5A elements have an oxidation number of –3.

  7. Let’s Try It… • Assign oxidation numbers to the elements in each compound: • NO3-1 • SO2 • Fe2O3 • Cu (s)

  8. Identifying Oxidation–Reduction Reactions • In a redox reaction, the oxidation number of a species changes during the reaction. • Oxidation occurs when the oxidation number increases (species loses electrons). LEO • Reduction occurs when the oxidation number decreases (species gains electrons). GER • Another mnemonic is OIL RIG! • If any species is oxidized or reduced in a reaction, that reaction is a redox reaction. • Examples of redox reactions: displacement of an element by another element; combustion; incorporation of an element into a compound, etc.

  9. Review • In the reactions below, write oxidation numbers for each substance and identify which substance is oxidized and which is reduced? • N2O + H2 --> H2O + NH3 • K + KNO3 --> N2 + K2O • Fe2O3+ S --> Fe + SO2

  10. The Half-Reaction Method ofBalancing Redox Equations • Separate a redox equation into two half-equations, one for oxidation and one for reduction. • Balance the number of atoms of each element in each half-equation. Usually we balance O and H atoms last. • Balance each half-reaction for charge by adding electrons to the left in the reduction half-equation and to the right in the oxidation half-equation. • Adjust the coefficients in the half-equations so that the same number of electrons appears in each half-equation. • Add together the two adjusted half-equations to obtain an overall redox equation. • Simplify the overall redox equation as necessary.

  11. Redox Reactions in Acidic and in Basic Solution • Redox reactions in acidic solution and in basic solution may be very different from one another. • If acidic solution is specified, we must add H2O and/or H+ as needed when we balance the number of atoms. • If basic solution is specified, the final equation may have OH– and/or water molecules in it. • A simple way to balance an equation in basic solution: • Balance the equation as though it were in acidic solution. • Add as many OH– ions to each side as there are H+ ions in the equation. • Combine the H+ and OH– ions to give water molecules on one side, and simplify the equation as necessary.

  12. Example 18.1 • Balance the equation in acidic solution: MnO4-1 + S2O3-2--> Mn+2 + SO4-2

  13. Example 18.1 B • Write a balanced equation for the oxidation of phosphorus by nitric acid, which is described by P4 (s) + H+1 (aq) + NO3-1 (aq) --> H2PO4-1 (aq) + NO (g)

  14. Example 18.2 • Balance the following in basic solution: Br2 (l) --> Br-1 (aq) + BrO3-1 (aq)

  15. Example 30a Balance the following in basic solution: CrO4-2 (aq) + AsH3 (g) --> Cr(OH)3 (s) + As (s)

  16. WS Example 1 HCl + K2Cr2O7--> KCl + CrCl3 + H2O + Cl2

  17. WS Example 2 FeCl2 + KMnO4 + HCl--> FeCl3 + KCl + MnCl2 + H2O

  18. WS Example 3 S-2 + MnO4-1 --> S + MnO2 (basic solution)

  19. WS Example 4 CuS + NO3-1 --> Cu+2 + S + NO (acidic)

  20. Stop! • Do the Redox Equations Worksheet.

  21. Today… • Turn in: • Get out Redox WS • Our Plan: • Questions on Redox WS • Quiz • Notes • Galvanic Cells WS • Homework (Write in Planner): • Worksheet Due Wednesday

  22. A Qualitative Description of Voltaic Cells • A voltaic cell uses a spontaneous redox reaction to produce electricity. • A half-cell consists of an electrode (strip of metal or other conductor) immersed in a solution of ions. This Zn2+ becomes a Zn atom. Both oxidation and reduction occur at the electrode surface, and equilibrium is reached. This Zn atom leaves the surface to become a Zn2+ ion.

  23. Important Electrochemical Terms • An electrochemical cell consists of two half-cells with the appropriate connections between electrodes and solutions. • Two half-cells may be joined by a salt bridge(U shaped tube suspended in gel) that permits migration of ions, without completely mixing the solutions. • The anodeis the electrode at which oxidation occurs. • The cathode is the electrode at which reduction occurs.

  24. Helpful Mnemonic • AUTO (anode = oxidation) • CAR (cathode = reduction) OR: • To remember the charge: Ca+ions are attracted to the Ca+hode (the t is a plus sign) • To remember which reaction occurs at which terminal: An Ox and Red Cat - Anode Oxidation, Reduction Cathode

  25. Important Electrochemical Terms • In a voltaic cell, a spontaneous redox reaction occurs and current (electricity) is generated. • Cell potential (Ecell) is the potential difference in volts between anode and cathode. • Ecell is the driving force that moves electrons and ions.

  26. A Zinc–Copper Voltaic Cell Positive and negative ions move through the salt bridge to equalize the charge. … the electrons produced move through the wire … … to the Cu(s) electrode, where they are accepted by Cu2+ ions to form more Cu(s). Zn(s) is oxidized to Zn2+ ions, and … Reaction:Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)

  27. Cell Diagrams • A cell diagram is “shorthand” for an electrochemical cell. • The anode is placed on the left side of the diagram. • The cathode is placed on the right side. • A single vertical line ( | ) represents a boundary between phases, such as between an electrode and a solution. • A double vertical line ( || ) represents a salt bridge or porous barrier separating two half-cells.

  28. Reaction:Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)

  29. Another look at Cell Diagrams

  30. Example 18.3 Describe the half-reactions and the overall reaction that occur in the voltaic cell represented by the cell diagram: Pt(s) |Fe2+(aq), Fe3+(aq) ||Cl–(aq) |Cl2(g) |Pt(s)

  31. Standard Electrode Potentials • Since an electrode represents only a half-reaction, it is not possible to measure the absolute potential of an electrode. • The standard hydrogen electrode(SHE) provides a reference for measurement of other electrode potentials. • The SHE is arbitrarily assigned a potential of 0.000 V.

  32. Standard Electrode Potentials • The standard electrode potential, E°, is based on the tendency for reduction to occur at an electrode. (Sometimes it is referred to as standard reduction potential) • All standard reduction potentials are for 25 ᵒC and 1 M solutions. • E°for the standard hydrogen electrode is arbitrarily assigned a value of 0.000 V. • All other values of E° are determined relative to the standard hydrogen electrode.

  33. Standard Electrode Potentials • The standard cell potential (E°cell) is the difference between E° of the cathode and E°of the anode. • E°cell = E°(cathode) – E°(anode) • Remember AUTO CAR!

  34. Measuring the Standard Potentialof the Cu2+/Cu Electrode The voltmeter reading and the direction of electron flow tell us that … … Cu2+ is more easily reduced than H+, by 0.340 volts. Standard hydrogen electrode Cu2+ + 2e --> Cu E° = +0.340 V

  35. Measuring the Standard Potentialof the Zn2+/Zn Electrode The voltmeter reading and the direction of electron flow tell us that … … Zn2+ is harder to reduce than H+, by 0.763 volts. Standard hydrogen electrode Zn2+ + 2e--> ZnE° = – 0.763 V

  36. F2 is the strongest oxidizing agent F– is the weakest reducing agent Li is the strongest reducing agent

  37. Important Note • The formula E°cell= E°(cathode) – E°(anode) allows us to simply use the numbers in the table, but if you were asked for the oxidation value of a reaction, you would have to reverse the sign since the values are all reduction potentials.

  38. Important Points about Electrode and Cell Potentials • Standard electrode potentials and cell voltages are intensive properties; they do not depend on the total amounts of the species present. • A “flashlight battery” (D-cell) and a “penlight battery” (AA cell) produce the same potential—1.5 volts. • E° does depend on the particular species in the reaction (or half-reaction). • As we shall learn later, cell and electrode potentials can depend on concentration of the species present.

  39. Example 18.4 Determine E° for the reduction half-reaction Ce4+(aq) + e–-->Ce3+(aq), given that the cell voltage for the voltaic cell Co(s) | Co2+(1 M) || Ce4+(1 M), Ce3+(1 M) | Pt(s) is E°cell = 1.887 V.

  40. Example 18.5 Balance the following oxidation–reduction equation, and determine E°cell for the reaction. O2(g) + H+(aq) + I–(aq) --> H2O(l) + I2(s)

  41. Electrode Potentials, Spontaneous Change, and Equilibrium • An electrochemical cell does work. • welec = nFEcell • n = number of electrons in the balanced equation • F = 96,485 coulombs per mole. • The amount of electrical work is also equal to –DG: • DG = –nFEcell • Under standard conditions: • DG° = –nFE°cell

  42. Criteria for SpontaneousChange in Redox Reactions • If Ecell is positive, the forward reaction is spontaneous. • If Ecell is negative, the forward reaction is nonspontaneous (the reverse reaction is spontaneous). • If Ecell = 0, the system is at equilibrium. • When a cell reaction is reversed, Ecell and DG change signs.

  43. Example 18.6 Will copper metal displace silver ion from aqueous solution? That is, does the reaction Cu(s) + 2 Ag+(1 M) --> Cu2+(1 M) + 2 Ag(s) occur spontaneously from left to right?

  44. The Activity Series Revisited • In the activity series of metals (Section 4.4), any metal in the series will displace a metal below it from a solution of that metal’s ions. • Theoretical basis: The activity series lists metals in order of their standard potentials. • Displacement of a metal from a solution of its ions by a metal higher in the series corresponds to a positive value of Ecell and a spontaneous reaction.

  45. Visual Example • http://intro.chem.okstate.edu/1515F01/Laboratory/ActivityofMetals/home.html

  46. Equilibrium Constants in Redox Reactions • Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another. • DG° = –nFE°cell and DG° = –RTlnKeq therefore –RTlnKeq = –nFEocell RTlnKeqRT E°cell = ––––––––– = –––– lnKeq nFnF R and F are constant, therefore at 298 K: 0.025693 V E°cell = –––––––– lnKeq n

  47. Example 18.8 Calculate the values of ΔG° and Keq at 25 °C for the reaction Cu(s) + 2 Ag+(1 M) --> Cu2+(1 M) + 2 Ag(s)

  48. Thermodynamics, Equilibrium, and Electrochemistry: A Summary From any one of the three quantities Keq, ΔG°, E°cell, we can determine the others.

  49. Stop! • Begin working on Worksheet #2!

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