1 / 118

Chapter 18 Electrochemistry

Chapter 18 Electrochemistry. Oxidation States: Review. An atom’s oxidation state is the imaginary charge that the atom would have if the shared electrons in the bond(s) were divided equally between the atoms in the bond. Oxidation state rule:

cale
Télécharger la présentation

Chapter 18 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. Chapter 18Electrochemistry

  2. Oxidation States: Review • An atom’s oxidation state is the imaginary charge that the atom would have if the shared electrons in the bond(s) were divided equally between the atoms in the bond. • Oxidation state rule: • The overall sum of all oxidation states in a species must be equal to the overall charge of the species

  3. Oxidation States: Review • The oxidation state (OS) of … • An atom in any element is 0 • Ne, Na, Fe • A monatomic ion is the same as its charge • Na+ has an OS of +1 • Fluorine is -1 in its compounds • The F’s in PF3 are all -1.

  4. Oxidation States: Review • The oxidation state (OS) of … • Oxygen is -2 (except in peroxides, where it is -1) • H2O and CO2 both have oxygens with an OS of -2, H2O2 has two oxygens with a -1 each. • Hydrogen is +1 in its compounds • in NH3, hydrogen has a +1 OS, nitrogen has a -3 OS.

  5. Oxidation States: Review • What is the oxidation state of the bolded atom in each of the following molecules? • SF6 • NO3- • HClO4

  6. Oxidation States: Review • LEO the lions says GER • Loss of electrons is oxidation • Gain of electrons is reduction

  7. Oxidation States: Review • If one species is oxidized, another must be reduced. • The species that is oxidized is called an reducing agent. • The species that is reduced is called the oxidizing agent.

  8. Oxidation States: Review • Example,

  9. Oxidation States: Review • Example, Carbon goes from -4 in CH4 to a +2 in CO. CH4 is oxidized, therefore it is the reducing agent

  10. Oxidation States: Review • Example, +2 -4 RA Carbon goes from -4 in CH4 to a +2 in CO. CH4 is oxidized, therefore it is the reducing agent

  11. Oxidation States: Review • Example, +1 +2 0 -4 RA OA The H’s in water goes from +1 in CH4 to a 0 in H2(g). Water is reduced, therefore it is the oxidizing agent

  12. Oxidation States: Review • Example, Silver’s oxidation state goes from +1 to 0. Silver is reduced, therefore it is the oxidizing agent

  13. Oxidation States: Review • Example, 0 +1 OA Silver’s oxidation state goes from +1 to 0. Silver is reduced, therefore it is the oxidizing agent

  14. Oxidation States: Review • Example, 0 +1 +0 +2 RA OA Copper’s oxidation state goes from +0 to +2. It is oxidized, therefore it is the reducing agent.

  15. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method • Used to balance redox reactions. • The method depends on whether the reaction occurs in acidic or basic solution. We’ll do acidic first (it’s easier)

  16. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 1: • Split your redox reaction into two half reactions that show just the oxidation and another that shows just the reduction. • Example, Cu(s) + NO3-(aq) Cu2+(aq) + NO(g) Cu(s)  Cu2+(aq) NO3-(aq)  NO(g)

  17. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 2 – In both half reactions: • Balance all elements except hydrogen and water. • Balance the oxygens using H2O. • Balance the hydrogens using H+ (since it is in acidic solution) • Balance the charge using electrons Cu(s)  Cu2+(aq) Cu(s)  Cu2+(aq) + 2e-

  18. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 2 – In both half-reactions: • Balance all elements except hydrogen and water. • Balance the oxygens using H2O. • Balance the hydrogens using H+ (since it is in acidic solution) • Balance the charge using electrons NO3-(aq)  NO(g) NO3-(aq)  NO(g) + 2H2O(l) NO3-(aq)+ 4 H+(aq) NO(g) + 2H2O(l) NO3-(aq) + 4 H+(aq) + 3e-  NO(g) + 2H2O(l)

  19. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 3: • Multiply each balanced half-reaction by a coefficient to ensure that the number of electrons in each rxn is equal. Cu(s)  Cu2+(aq) + 2e- NO3-(aq) + 4 H+(aq) + 3e-  NO(g) + 2H2O(l) 3Cu(s)  3Cu2+(aq) + 6e- 2NO3-(aq) + 8 H+(aq) + 6e-  2NO(g) + 4H2O(l)

  20. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 4: • Add the half-reactions and cancel identical species (the electrons should cancel out, if not go back): 3Cu(s)  3Cu2+(aq) + 6e- 2NO3-(aq) + 8 H+(aq) + 6e-  2NO(g) + 4H2O(l) ----------------------------------------------------------------- 3Cu(s) + 2NO3-(aq) + 8H+(aq) 3Cu2+(aq) + 2NO(g) + 4H2O(l)

  21. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Step 5: • Check that each element (and the overall charge) is balanced 3Cu(s)  3Cu2+(aq) + 6e- 2NO3-(aq) + 8 H+(aq) + 6e-  2NO(g) + 4H2O(l) ----------------------------------------------------------------- 3Cu(s) + 2NO3-(aq) + 8H+(aq) 3Cu2+(aq) + 2NO(g) + 4H2O(l)

  22. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (acidic) • Example, H3AsO4(aq) + Zn(s) AsH3(aq) + Zn2+(aq)

  23. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 1: • Follow the same steps as if it took place in an acidic environment • Example, CN-(aq) + MnO4-(aq) CNO-(aq) + MnO2(s) CN-(aq)  CNO-(aq) MnO4-(aq) MnO2(s)

  24. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 1: • Follow the same steps as if it took place in an acidic environment Example, CN-(aq)  CNO-(aq) CN-(aq) + H2O(l)  CNO-(aq) CN-(aq)+ H2O(l)  CNO-(aq) + 2H+(aq) CN-(aq) + H2O(l)  CNO-(aq) + 2H+(aq) + 2e-

  25. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 1: • Follow the same steps as if it took place in an acidic environment Example, MnO4-(aq)  MnO2(s) MnO4-(aq)  MnO2(s) + 2H2O(l) MnO4-(aq)+ 4H+  MnO2(s) + 2H2O(l) MnO4-(aq) + 4H++ 3e-  MnO2(s) + 2H2O(l)

  26. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 2: • Then, in each half-reaction, add a number of OH- ions equivalent to the number of H+’s that you added. Add them to the reactants’ side and the products’ side. Example, CN-(aq) + H2O(l) + 2OH- CNO-(aq) + 2H+(aq) + 2e- + 2OH- CN-(aq) + H2O(l) + 2OH- CNO-(aq)+ 2e- + 2H2O(l) CN-(aq) + 2OH- CNO-(aq)+ 2e- + H2O(l)

  27. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 2: • Then, in each half-reaction, add a number of OH- ions equivalent to the number of H+’s that you added. Add them to the reactants’ side and the products’ side. Example, MnO4-(aq) + 4H+ + 3e- + 4OH- MnO2(s) + 2H2O(l) + 4OH- MnO4-(aq) + 4H2O + 3e-  MnO2(s) + 2H2O(l) + 4OH- MnO4-(aq) + 2H2O + 3e-  MnO2(s) + 4OH-

  28. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 3: • Multiply the half-reactions in order to make the number of electrons equal in both reactions. CN-(aq) + 2OH- CNO-(aq) + 2e- + H2O(l) 3CN-(aq) + 6OH- 3CNO-(aq) + 6e- + 3H2O(l) MnO4-(aq) + 2H2O + 3e-  MnO2(s) + 4OH- 2MnO4-(aq) + 4H2O + 6e-  2MnO2(s) + 8OH-

  29. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 4: • Add the two half-reactions, cancelling out all species that appear on both sides. Example, 3CN-(aq) + 6OH- 3CNO-(aq) + 6e- + 3H2O(l) 2MnO4-(aq) +4H2O+ 6e- 2MnO2(s) + 8OH- -------------------------------------------------------- 3CN- + 2MnO4- + H2O  3CNO- + 2MnO2 + 2OH-

  30. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Step 5: • Check that each element (and the overall charge) is balanced Example, 3CN- + 2MnO4- + H2O  3CNO- + 2MnO2 + 2OH-

  31. 18.1 – Balancing Oxidation- Reduction Equations • Half-Reaction Method (basic) • Example: MnO4-(aq) + S2-(aq)  MnS(s) + S(s)

  32. 18.2 – Galvanic Cells • In the previous systems, electrons were flowing spontaneously from one substance to another. The energy that is formed is released as heat. • If we were to separate the two substances into separate beakers, and connect them with a wire, the electrons will travel through it. If we hook up a device to the wire (i.e., a motor), we could produce useful work via the galvanic (or voltaic) cell.

  33. 18.2 – Galvanic Cells • Drawing of a generic galvanic cell: voltmeter cathode (reduced) anode (oxidized) salt bridge M(s) N(s) M+(aq) N+(aq)

  34. 18.2 – Galvanic Cells • Example, voltmeter cathode (reduced) anode (oxidized) KCl salt bridge Zn(s) Cu(s) Cu2+(aq) SO42-(aq) Zn2+(aq) SO42-(aq) Cu(s) dipped into a CuSO4 solution Zn(s) dipped into a ZnSO4 solution

  35. 18.2 – Galvanic Cells • The potential (i.e., the pull) for electrons to travel from one electrode to another is called the cell potential Ecell, or electromotive force (emf). • The unit of electrical potential is the volt (V). • 1V = 1 joule of work done per coulomb of charge transferred

  36. 18.2 – Galvanic Cells • Line notation for galvanic cells • Shorthand way of drawing a galvanic cell • The metal and metal ion on the anode side is written first with a single vertical line between them, | • A set of double vertical lines separates the anode from the cathode, || • The metal ion and metal on the cathode side is then written with a single vertical line between them, | The previous galvanic cell would be shortened to: Zn(s) | Zn2+(aq) || Cu2+(aq)| Cu(s)

  37. 18.3 – Standard Reduction Potentials • In order to determine in which direction the electrons will flow, we must consider the standard reduction potentials, Eo of the half-reactions • The reduction of hydrogen (i.e., 2H+ + 2e- H2(g)) is assigned a standard reduction potential (Eo) of 0, everything else is measured in relation to it

  38. 18.3 – Standard Reduction Potentials • The cell’s potential (Ecell) is given by: Eocell= Eocathode – Eoanode • Reduction potential is an intensive property. Therefore you can multiply the reaction by whole number and the E will remain unchanged. • When a reduction reaction is reversed, the sign of E is reversed

  39. 18.3 – Standard Reduction Potentials • Example, What is the overall cell potential in the following galvanic cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) From table A5.5, Zn2+(aq) + 2e- Zn(s) Eo = -0.76V Cu2+(aq) + 2e- Cu(s) Eo = +0.34V

  40. 18.3 – Standard Reduction Potentials • Example, Step 1: The half-reaction with the largest potential will run as written (as a reduction). The other half-reaction will be forced to run in reverse (as an oxidation). So, according to our values, Zn will lose electrons at the anode, Cu2+ will pick them up at the cathode to become Cu(s) : Zn(s) Zn2+(aq) + 2e-Eo = +0.76V (an, ox) Cu2+(aq) + 2e- Cu(s) Eo = +0.34V (red, cat)

  41. 18.3 – Standard Reduction Potentials • Example, Step 2: From this, we can come up with the overall cell potential by adding the equations (and Eo’s): Zn(s) Zn2+(aq) + 2e-Eo = +0.76V (an, ox) Cu2+(aq) + 2e- Cu(s) Eo = +0.34V (red, cat) --------------------------------------------------------------------- Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Ecell = 0.76 + 0.34V = 1.10V

  42. 18.3 – Standard Reduction Potentials • Example, Alternatively, we could recognize that a more negative Eo means that the Zn has a lower reduction potential (i.e., is oxidized). This would occur at the anode. The Eo for Zn stays as -0.76V as its reduction potential. Therefore we could use: Eocell= Eocathode – Eoanode Eocell= 0.34V – (-0.76V) Eocell = 1.10V

  43. 18.3 – Standard Reduction Potentials • Salt bridge Ensures that the overall charge on each half-cell is balanced. voltmeter cathode (reduced) anode (oxidized) salt bridge M(s) N(s) M+(aq) N+(aq)

  44. 18.3 – Standard Reduction Potentials • Standard hydrogen electrode • Defined as having a standard reduction potential of 0.

  45. 18.3 – Standard Reduction Potentials • If you work out a electrochemistry problem and get a … • A positive Ecell, it means the electrons flow spontaneously from the anode to the cathode, as you wrote it. It can act as a galvanic cell • A negative Ecell, it means a battery is required to make the electrons flow in the direction you have written. It can act as an electrolytic cell

  46. 18.3 – Standard Reduction Potentials • Electrochemical series • Looking at the electrochemical series in table A5.5, the more negative the Eo value… • The better the species will behave as a reducing agent. • The easier the species will be oxidized (lose electrons) • The more positive the Eo value… • The better the species will behave as an oxidizing agent. • The easier the species will be reduced (gain electrons)

  47. 18.3 – Standard Reduction Potentials • Electrochemistry of the electron transport chain • Recall that the ETC is a series of proteins embedded in the inner membrane of mitochondria and the thylakoid membrane of chloroplast. The purpose of the ETC is to receive high energy electrons from NADH (i.e., be reduced by NADH, a strong reducing agent) and transfer them to proteins within the chain, while at the same time, using this energy to pump protons against its gradient, creating an electrochemical potential difference. The final electron acceptor is oxygen (which becomes water)

  48. 18.3 – Standard Reduction Potentials • Electrochemistry of the electron transport chain • The first protein is reduced the easiest (has a high Eo). Each successive protein in the ETC must be lower in Eo, as it gets passed down the line. • Sometimes an electron “leaks” by reducing oxygen in the middle of the chain (instead of at the end), forming superoxide (O2-) something which may cause oxidative stress in the cell.

  49. 18.4 – Cell Potential, Electrical Work, and Free Energy • We can relate the spontaneity (ΔGo) of a reaction to the potential of a cell Ecell. n is the number of moles of electrons F is the Faraday constant, 96 485 C/mol

  50. 18.4 – Cell Potential, Electrical Work, and Free Energy • Example, • For the copper/zinc cell that we studied earlier, the Eocellwas 1.10V (or 1.10 J/C),the number of moles of electrons in that reaction were 2

More Related