1 / 77

Chapter 19: Electrochemistry

Vanessa N. Prasad-Permaul Valencia College CHM 1046. Chapter 19: Electrochemistry. Oxidation–Reduction Reactions. Redox reactions are those involving the oxidation and reduction of species. Oxidation and reduction must occur together. They cannot exist alone.

bobbieboyd
Download Presentation

Chapter 19: Electrochemistry

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. Vanessa N. Prasad-Permaul Valencia College CHM 1046 Chapter 19: Electrochemistry

  2. Oxidation–Reduction Reactions • Redox reactions are those involving the oxidation and reduction of species. • Oxidation and reductionmustoccur together. They cannot exist alone. Fe2+ + Cu0 Fe0 + Cu2+ Reduced: Iron gained 2 electrons Fe2+ + 2 e  Fe0 Oxidized: Copper lost 2 electrons Cu0 Cu2+ + 2 e • Remember that electrons are negative so if you gain electrons your oxidation # decreases and if you lose electrons your oxidation # increases

  3. Oxidation–Reduction Reactions Fe2+ + Cu0 Fe0 + Cu2+ • Fe2+ gains electrons, is reduced, and we call it an oxidizing agent • Oxidizing agent is a species that can gain electrons and this facilitates in the oxidation of another species. (electron deficient) • Cu0 loses electrons, is oxidized, and we call it a reducing agent • Reducing agent is a species that can lose electrons and this facilitates in the reduction of another species. (electron rich)

  4. Example 1: Oxidation–Reduction Reactions Which is a reduction half reaction? • Fe  Fe2+ + 2e- • Fe2+ Fe3+ + 1e- • Fe  Fe3+ + 3e- • Fe3+ + 1e-  Fe2+

  5. Example 2: Oxidation–Reduction Reactions For each of the following, identify which species is the reducing agent and which is the oxidizing agent. • Ca(s) + 2 H+(aq) Ca2+(aq) + H2(g) • 2 Fe2+(aq) + Cl2(aq) 2 Fe3+(aq) + 2 Cl–(aq) C) SnO2(s) + 2 C(s) Sn(s) + 2 CO(g)

  6. Oxidation–Reduction Reactions • Assigning Oxidation Numbers:All atoms have an “oxidation number” regardless of whether it carries an ionic charge. 1. An atom in its elemental state has an oxidation number of zero. Elemental state as indicated by single elements with no charge. Exception: diatomics H2 N2 O2 F2 Cl2 Br2 and I2

  7. Oxidation–Reduction Reactions 2.An atom in a monatomic ion has an oxidation number identical to its charge.

  8. Oxidation–Reduction Reactions 3.An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. A. Hydrogen can be either +1 or –1. B. Oxygen usually has an oxidation number of –2. In peroxides, oxygen is –1. C. Halogens usually have an oxidation number of –1. • When bonded to oxygen, chlorine, bromine, and iodine have positive oxidation numbers.

  9. Oxidation–Reduction Reactions 4. The sum of the oxidation numbers must be zero for a neutral compound and must be equal to the net charge for a polyatomic ion. A. H2SO4 neutral atom, no net charge SO42- sulfate polyatomic ion [SO4]2- [Sx O42-] = -2 X + -8 = -2 X = 6 so sulfur has an oxidation # of +6

  10. Oxidation–Reduction Reactions B. ClO4– , net charge of -1 [ClO4]-1 [Clx O42-] = -1 X + -8 = -1 X = 7 so the oxidation number of chloride is +7

  11. Example 3: Oxidation–Reduction Reactions Assign oxidation numbers to each atom in the following substances: CdS V2O3 MnO4– H2PO4- VOCl3 S2O32- AlH3 HNO3 Na2Cr2O7 FeSO4 SnCl4 Fe2O3

  12. Example 4: What is the oxidation number of arsenic in AsO43- ? • 8 • 1 • 5 • -1

  13. Activity Series of Elements Lithium, Li can reduce anything under it. Lithium is a very strong reducing agent and this is in part why lithium batteries work so well. Light weight and produce a high voltage Anode: Li  Li+ + e Cathode: MnO2 + Li+ + e  LiMnO2

  14. Activity Series of Elements Activity series looks at the relative reactivity of a free metal with an aqueous cation. Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Cu(s) + 2 Ag+(aq)  2 Ag(s) + Cu2+(aq) Mg(s) + 2 H+(aq) Mg2+(aq) + H2(g)

  15. Example 5: Activity Series of Elements • Given the following three reactions, determine the activity series for Cu, Zn, & Fe. 1. Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) 2. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 3. Fe(s) + Zn2+ (aq)NR

  16. Balancing Redox Reactions • Half-Reaction Method:Allows you to focus on the transfer of electrons. This is important when considering batteries and other aspects of electrochemistry. • The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. (oxidation half and reduction half)

  17. Balancing Redox Reactions Balance for an acidic solution: MnO4–(aq) + Br –(aq) Mn2+(aq) + Br2(aq) 1.Determine oxidation and reduction half-reactions: Oxidation half-reaction: Br – (aq)  Br2 (aq) Reduction half-reaction: MnO4– (aq)  Mn2+ (aq) 2.Balance for atoms other than H and O: Oxidation:2Br – (aq)  Br2 (aq) Reduction: MnO4– (aq)  Mn2+ (aq)

  18. Balancing Redox Reactions 3.Balance for oxygen by adding H2O to the side with less oxygen Oxidation:2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq)  Mn2+(aq) + 4 H2O(l) 4.Balance for hydrogen by adding H+ to the side with less hydrogens Oxidation:2 Br–(aq)  Br2(aq) Reduction: MnO4–(aq) + 8 H+(aq)  Mn2+(aq) + 4 H2O(l)

  19. Balancing Redox Reactions 5.Balance for charge by adding electrons (e–): Oxidation: 2 Br–(aq)  Br2(aq) + 2 e– Reduction: MnO4–(aq) + 8 H+(aq) +5 e– Mn2+(aq) + 4 H2O(l) 6.Balance for numbers of electrons by multiplying: Oxidation: 5[2 Br–(aq)  Br2(aq) + 2 e–] Reduction: 2[MnO4–(aq) + 8 H+(aq) +5 e– Mn2+(aq) + 4 H2O(l)]

  20. Balancing Redox Reactions 7.Combine and cancel to form one equation: Oxidation: 10 Br – (aq)  5 Br2(aq) + 10 e– Reduction: 2 MnO4–(aq) + 16 H+(aq) + 10 e–2 Mn2+(aq) + 8H2O(l) 2 MnO4–(aq) + 10 Br–(aq) + 16 H+(aq) 2 Mn2+(aq) + 5 Br2(aq) + 8 H2O(l)

  21. Example 6: Balancing Redox Reactions Balance the following in an acidic sol’n NO3–(aq) + Cu(s) NO(g) + Cu2+ (aq)

  22. Balancing Redox Reactions Balance for a basic solution: MnO4-(aq) + SO32-(aq) MnO2(s) + SO42-(aq) 1.Determine oxidation and reduction half-reactions: Oxidation half-reaction: SO32–(aq) SO42-(aq) Reduction half-reaction: MnO4–(aq) MnO2(s) 2.Balance for atoms other than H and O: Oxidation:SO32– (aq) SO42- (aq) Reduction: MnO4– (aq) MnO2(aq)

  23. Balancing Redox Reactions: Basic Sol’n 3.Balance for oxygen by adding H2O to the side with less oxygen Oxidation:H2O(l) +SO32– SO42- Reduction: MnO4– MnO2 + 2 H2O(l) 4.Balance for hydrogen by adding H+ to the side with less hydrogens Oxidation:H2O(l) +SO32– SO42- + 2H+(aq) Reduction: 4H+(aq)+MnO4– MnO2 + 2 H2O(l)

  24. Balancing Redox Reactions: Basic Sol’n 5.Balance for charge by adding electrons (e–): Oxidation:H2O +SO32– SO42- + 2H++2 e- Reduction: 3 e- +4H++MnO4– MnO2 + 2 H2O 6.Balance for numbers of electrons by multiplying: Oxidation:3[H2O +SO32– SO42- + 2H++2 e-] Reduction: 2[3 e- +4H++MnO4– MnO2 + 2 H2O]

  25. Balancing Redox Reactions: Basic Sol’n 7.Combine and cancel to form one equation: Ox:3H2O(l) +3SO32–(aq) 3SO42-(aq) + 6H+(aq) +6 e- Red: 6 e- +8H+(aq)+2MnO4–(aq) 2MnO2(s) + 4 H2O(l) 2 1 2H+(aq) + 3SO32-(aq) + 2MnO4-(aq) 3SO42-(aq) + 2MnO2(s) + H2O(l) So far same as acidic sol’n, 2 more steps for basic sol’n

  26. Balancing Redox Reactions: Basic Sol’n • Note # of H+ ions in the equation. Add this # of OH- to both sides of equation 2OH- + 2H+ + 3SO32- + 2MnO4- 3SO42- + 2MnO2 + H2O + 2OH- • Simplify equation by combining H+ and OH- to make H2O, reduce to lowest terms 2H2O + 3SO32- + 2MnO4- 3SO42- + 2MnO2 + H2O + 2OH- 1 Write balanced equation

  27. Balancing Redox Reactions: Basic Sol’n H2O(l) + 3SO32-(aq) + 2MnO4-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq)

  28. Electrochemistry 2 types of electrochemical cells 1.  Galvanic cells (voltaic cells) Spontaneous chemical rxn generates an electric current 2.  Electrolytic cells An electric current drives a non-spontaneous rxn

  29. Review of Redox Overall rxn Zn + Cu2+ Zn2+ + Cu ½ ox. rxn Zn  Zn2+ + 2e- ½ red. rxn Cu2+ + 2e-  Cu • Cu2+ is reduced and it is the oxidizing agent • Zn is oxidized and it is the reducing agent • Electrons are transferred directly from Zn to Cu2+ 4. Enthalpy of rxn is lost to the surroundings as heat

  30. Same rxn with electrochemical cell (galvanic)Some of the chemical energy released by the rxn is converted to electrical energy which can be used to light a light bulb Apparatus

  31. Apparatus Electrodes- strips of zinc and copper connected by an electrically conductive wire (electrons transferred through wire make current) Anode electrode is negative, oxidation takes place Zn strip Cathode electrode reduction takes place Cu strip Salt bridge- U-shaped tube that contains a gel permeated with a sol’n of inert electrolyte (will not react) Anode: Ox ½ Zn  Zn2+ + 2e- Cathode: Red ½ Cu2+ + 2e-  Cu Overall Rxn Zn + Cu2+ Zn2+ + Cu

  32. Apparatus Why is the salt bridge necessary? 1. Completes the electrical circuit Without it the anode cell would become positively charged as Zn2+ ions appear the sol’n in the cathode beaker would become negatively charged as Cu2+ ions are removed 2. Because of the charge imbalance the elctrode rxns would quickly stop and electron flow through the wire stops 3. With the salt bridge the electrical neutrality is maintained in both beakers by a flow of ions Anions- SO42- flow through salt bridge from cathode to anode Cations- from anode to cathode

  33. Overview Anode: Oxidation occurs, electrons are produced, anions move toward, neg. sign Cathode: Reduction occurs, Electrons are consumed, cations move towards, positive sign Remember: An ox ate a red cat

  34. Shorthand Notation for Galvanic Cells Example: Zn(s) + Cu2+ Zn2+ + Cu(s) Shorthand notation: Zn(s) Zn2+(aq) Cu2+(aq) Cu(s) 1. , single vertical line represents a phase boundary (between a solid electrode and an aqueous solution 2. , double vertical line denotes a salt bridge 3. The anode half-cell is always on the left of the salt bridge, with the solid electrode to the far left 4. The cathode half-cell is always on the right of the salt bridge, with the solid electrode on the far right 5. The reactants in each half cell are always written first, followed by the products 6. Electrons move through the external circuit from left to right (from anode to cathode)

  35. Example 7: Write the shorthand notation for a galvanic cell that uses the reaction Fe(s) + Sn2+(aq) Fe2+(aq) + Sns)

  36. Cell Potentials and Free-Energy Changes for Cell Reactions Electrons move through the external circuit from the zinc anode to the copper cathode because they have lower energy when on copper than on zinc. The driving force that pushes the negatively charges electrons away from the anode (- electrode) and pulls them toward the cathode (+ electrode) is an electrical potential called the electromotive force (emf) also known as cell potential (E) or the cell voltage. SI units in volts (V)

  37. Cell Potentials and Free-Energy Changes for Cell Reactions 1 J = 1C x 1V G = -nFE n = number of moles of electrons transferred in the reaction F = faraday constant 9.648534 x 104 C/mol G = gibbs free energy G = -nFE Standard free-energy change and standard cell potential Because both are directly proportional a voltmeter can be regarded as a “free-energy meter” When a voltmeter measures E, it is also indirectly measuring G

  38. Example 8: The standard cell potential at 25C is 0.92 V for the reaction Al(s) + Cr3+(aq) Al3+(aq) + Cr(s) What is the standard free-energy change for this reaction at 25C in kJ?

  39. Standard Reduction Potentials The standard potential of any galvanic cell is the sum of the standard half-cell potentials for oxidation at the anode and reduction at the cathode: Ecell = Eox + Ered H2 gas is oxidized to H+ ions at the anode and Cu2+ ions are reduced to copper metal at the cathode

  40. Standard Reduction Potentials Anode (oxidation): H2(g) 2 H+(aq) + 2 e Cathode (reduction): Cu2+(aq) + 2 e  Cu(s) Overall: H2(g) + Cu2+(aq) 2 H+(aq) + Cu(s) Ecell = Eox + Ered = EH2H+ + ECu2+  Cu = .34 V To come up with a table of standard potentials we must first chose a reference half-cell to measure all others with.

  41. Standard Reduction Potentials Standard Hydrogen Electrode (SHE) (last cell we considered!!) E = 0 V .34 V = 0 V + .34 V Because the Cu2+/Cu half-reaction is a reduction the half-cell potential is called a standard reduction potential Cu2+ + 2 e  Cu E = .34 V Cu  Cu2+ + 2 e E = -.34 V

  42. Standard Reduction Potential Table 1. The half-reactions are all written as reductions. Oxidizing agents and electrons are on the left side of each half-reaction and reducing agents are on the right side 2. The listed half-cell potentials are standard reduction potentials, also known as standard electrode potentials 3. The half-reactions are listed in order of decreasing standard reduction potential(decreasing tendency to occur in the forward direction; increasing tendency to occur in the reverse direction) The strongest oxidizing agents are located in the upper left of the table (F2, H2O2, MNO4-) the strongest reducing agents are found in the lower right of the table (Li, Na, Mg)

  43. Using Standard Reduction Potentials Table arranges oxidizing or reducing agents in order of increasing strength, this allows us to predict the spontaneity or nonspontaneity of thousands of redox rxns. Let’s calculate E for the oxidation of Zn(s) by Ag+(aq): 2 Ag+(aq) + Zn(s) 2 Ag(s) + Zn2+(aq)

  44. Using Standard Reduction Potentials Step 1: Find half-rxns in Table and write them in the appropriate direction Step 2: Multiply the Ag+/Ag half-reaction by a factor of 2 so that the electrons cancel. Do not multiply the E by 2 because electric potential is an intensive property, which does not depend on the amount of substance. E = -G / nF G will double and n will double so that E will remain constant Step 3: Change E for oxidations since the table is based on reductions Step 4: Add the half-reactions to get the overall reaction.

  45. Using Standard Reduction Potentials Reduction: 2 x [Ag+ + e-  Ag] E = 0.80 V Oxidation: Zn  Zn2+ + 2e- E = -(-0.76 V) Overall reaction: 2 Ag+(aq) + Zn(s) 2 Ag(s) + Zn2+(aq) E= 1.56 V E = -G / nF G =   -E(nF) = -3.01 x 105 Because E is positive and G is negative, oxidation of zinc by Ag+ is a spontaneous reaction under SS conditions. Note: Ag+ can oxidize any reducing agent that lies below it in the table. Can’t oxidize a reducing agent that appears above it on the table. Because E will have a negative value.

  46. Example 9: Arrange the following oxidizing agents in order of increasing strength under SS conditions: Br2(l) , Fe3+(aq) , Cr2O72-(aq)

  47. Example 10: Arrange the following reducing agents in order of increasing strength under SS conditions: Al(s) , Na(s) , Zn(s)

  48. Example 11: Predict from the Reduction table whether each of the following reactions can occur spontaneously under SS conditions: a)      2 Fe3+(aq) + 2 I-(aq) 2 Fe2+(aq) + I2(s) b)      3 Ni(s) + 2 Al3+(aq) 3 ni2+(aq) + 2 Al(s)

  49. Cell potentials and composition of the rxn mixture: The Nernst Equation Cell pot. and gibbs free energy depend on temp. and composition of rxn mixture -Concentration of solutes -partial pressures of gases G = G + RT ln Q Q = Rxn quotient nFE = -nFE + RT ln Q

More Related