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Chapter 20 Electrochemistry

Chemistry, The Central Science , 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 20 Electrochemistry. John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. APRIL 8. REDOX REACTIONS –

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Chapter 20 Electrochemistry

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  1. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 20Electrochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

  2. APRIL 8 • REDOX REACTIONS – • HOW TO ASSIGN OXIDATION NUMBERS • HOW TO BALANCE REACTIONS IN ACID AND BASIC MEDIUM USING THE HALF REACTION METHOD • HW : 13,15, 17,19

  3. Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. Metals tend to lose electrons and are oxidized, non metals tend to gain electrons and are reduced.

  4. LEOGER Losing Electrons is Oxidation. Gaining Electrons is Reduction

  5. OIL RIG Oxidation Is Loss. Reduction Is Gain.

  6. REDOX REACTIONS Are reduction – oxidation reactions. Electrons are transferred. When an atom is losing electrons its O.N. increases. It is being oxidized. When an atom gains electrons its O.N. decreases. It is being reduced

  7. Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

  8. Assigning Oxidation Numbers • Elements in their elemental form have an ON= 0. • The oxidation number of a monatomic ion is its charge. • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. • Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. • Hydrogen is +1 except in metal hydrides when is −1 . • Fluorine always has an oxidation number of −1.

  9. Assigning Oxidation Numbers • The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions • The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. • The sum of the oxidation numbers in a neutral compound is 0.

  10. Oxidation and Reduction • A species is oxidized when it loses electrons. • Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

  11. Oxidation and Reduction • A species is reduced when it gains electrons. • Here, each of the H+ gains an electron and they combine to form H2.

  12. Oxidation and Reduction • What is reduced is the oxidizing agent. • H+ oxidizes Zn by taking electrons from it. • What is oxidized is the reducing agent. • Zn reduces H+ by giving it electrons.

  13. Oxidation-Reduction Reactions • Zn added to HCl yields the spontaneous reaction • Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g). • The oxidation number of Zn has increased from 0 to 2+. • The oxidation number of H has reduced from 1+ to 0. • Zn is oxidized to Zn2+ while H+ is reduced to H2. • H+ causes Zn to be oxidized and is the oxidizing agent. • Zn causes H+ to be reduced and is the reducing agent. • Note that the reducing agent is oxidized and the oxidizing agent is reduced.

  14. Balancing Oxidation-Reduction Reactions • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. • Conservation of charge: electrons are not lost in a chemical reaction. • Half Reactions • Half-reactions are a convenient way of separating oxidation and reduction reactions.

  15. Half Reactions • The half-reactions for • Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq) • are • Sn2+(aq)  Sn4+(aq) +2e- • 2Fe3+(aq) + 2e- 2Fe2+(aq) • Oxidation: electrons are products. • Reduction: electrons are reactants. • Loss of Gain of • Electrons is Electrons is • Oxidation Reduction

  16. Balancing Equations by the Method of Half Reactions • Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). • MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. • The equivalence point is given by the presence of a pale pink color. • If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

  17. What is the balanced chemical equation? • 1. Write down the two half reactions. • 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H+. d. If it is in basic solution, remove H+ by adding OH- e. Finish by balancing charge by adding electrons. • Multiply each half reaction to make the number of electrons equal. • Add the reactions and simplify. • Check!

  18. Half-Reaction Method Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq) Mn2+(aq) + CO2(aq)

  19. Balancing Equations by the Method of Half Reactions • Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 (deep purple). • MnO4- is reduced to Mn2+ (pale pink) while the C2O42- is oxidized to CO2. • The equivalence point is given by the presence of a pale pink color. • If more KMnO4 is added, the solution turns purple due to the excess KMnO4.

  20. +7 +3 +2 +4 MnO4− + C2O42- Mn2+ + CO2 Half-Reaction Method First, we assign oxidation numbers. Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

  21. For KMnO4 + Na2C2O4: • The two incomplete half reactions are • MnO4-(aq)  Mn2+(aq) • C2O42-(aq)  2CO2(g) 2. Adding water and H+ yields • 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O • There is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left: • 5e- + 8H+ + MnO4-(aq)  Mn2+(aq) + 4H2O

  22. In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons: • C2O42-(aq)  2CO2(g) + 2e- 3. To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying gives: • 10e- + 16H+ + 2MnO4-(aq)  2Mn2+(aq) + 8H2O • 5C2O42-(aq)  10CO2(g) + 10e-

  23. 4. Adding gives: 16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq)  2Mn2+(aq) + 8H2O(l) + 10CO2(g) 5. Which is balanced!

  24. Balancing in Basic Solution • If a reaction occurs in basic solution, one can balance it as if it occurred in acid. • Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place. • If this produces water on both sides, you might have to subtract water from each side.

  25. Examples – Balance the following oxidation-reduction reactions: • Cr (s) + NO3- (aq)  Cr3+ (aq) + NO (g) (acidic) • Al (s) + MnO4- (aq)  Al3+ (aq) + Mn2+ (aq) (acidic) • PO33- (aq) + Mn4- (aq)  PO43- (aq) + MnO2 (s) (basic) • H2CO (aq) + Ag(NH3)2+ (aq)  HCO3- (aq) + Ag (s) + NH3 (aq) (basic)

  26. April 9Section 20-3 20-4 • Voltaic Cells – Spontaneous reactions • ELECTRODES – POLARITIES • SALT BRIDGE • DRIVING FORCE- EMF • STANDARD REDUCTION POTENTIALS • Hw: • 23, 25, 27, 29, 31, 33 (a,b), 35

  27. Voltaic Cells In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

  28. Voltaic Cells • We can use that energy to do work if we make the electrons flow through an external device. • We call such a setup a voltaic cell.

  29. Voltaic Cells • A typical cell looks like this. • The oxidation occurs at the anode. • The reduction occurs at the cathode.

  30. Voltaic Cells Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

  31. Voltaic Cells • Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. • Cations move toward the cathode. • Anions move toward the anode.

  32. Voltaic Cells • In the cell, then, electrons leave the anode and flow through the wire to the cathode. • As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

  33. Voltaic Cells • As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode. • The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

  34. April 10 • Section 4.5 – 4.6 • Free energy and redox rx • Cell EMF under nonstandard conditions • Nerst Equation- Concentration cells

  35. Electromotive Force (emf) • Water only spontaneously flows one way in a waterfall. • Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

  36. Electromotive Force (emf) • The potential difference between the anode and cathode in a cell is called the electromotive force (emf). • It is also called the cell potential, and is designated Ecell.

  37. Cell Potential Cell potential is measured in volts (V). One volt is the potential difference required to impart 1J of energy to a charge of 1 coulomb. (1 electron has a charge of 1.6 x 10-19 Coulombs). The potential difference between the 2 electrode provides the driving force that pushes the electron through the external circuit.

  38. J C 1 V = 1 Cell Potential or Electromotive Force (emf) • The “pull” or driving force on the electrons. • ELECTROMOTIVE FORCE EMF • CAUSES THE ELECTRON MOTION!

  39. Standard Reduction Potentials Reduction potentials for many electrodes have been measured and tabulated.

  40. Standard Hydrogen Electrode • Their values are referenced to a standard hydrogen electrode (SHE). • By definition, the reduction potential for hydrogen is 0 V: 2 H+(aq, 1M) + 2 e− H2(g, 1 atm)

  41. = Ered (cathode) −Ered (anode)   Ecell  Standard Cell Potentials The cell potential at standard conditions can be found through this equation: Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

  42. For the reaction to be SPONTANEOUS E has to be positive

  43. Ered = −0.76 V  Ered = +0.34 V  Cell Potentials • For the oxidation in this cell, • For the reduction,

  44. = (anode) (cathode) − Ered Ered   Ecell  Cell Potentials = +0.34 V − (−0.76 V) = +1.10 V

  45. Since Ered = -0.76 V we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. • The oxidation of Zn with the SHE is spontaneous. • Changing the stoichiometric coefficient does not affect Ered. • Therefore, • 2Zn2+(aq) + 4e- 2Zn(s), Ered = -0.76 V. • Reactions with Ered > 0 are spontaneous reductions relative to the SHE.

  46. Reactions with Ered < 0 are spontaneous oxidations relative to the SHE. • The larger the difference between Ered values, the larger Ecell. • In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode). • Recall

  47. PNEUMONICS • LEO GER • OIL RIG • RED CAT • AND ALWAYS THE SOURCE OF ELECTRONS IS THE NEGATIVE ELECTRODE!!!

  48. Spontaneity of Redox Reactions • In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode) since • Or • More generally, for any electrochemical process • A positive E indicates a spontaneous process (galvanic cell). • A negative E indicates a nonspontaneous process.

  49. Cell Potential Calculations • To Calculate cell potential using Standard Reduction Potentials: • 1. One reaction and its cell potential’s sign must be reversed--it must be chosen such that the overall cell potential is positive. • 2. The half-reactions must often be multiplied by an integer to balance electrons--this is notdone for the cell potentials.

  50. Cell Potential Calculations Continued • Fe3+(aq) + Cu(s) ----> Cu2+(aq) + Fe2+(aq) • Fe3+(aq) + e- ----> Fe2+(aq) Eo= 0.77 V • Cu2+(aq) + 2 e- ----> Cu(s)Eo = 0.34 V • Reaction # 2 must be reversed.

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