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Chapter 20 - Electron Transfer Reactions

Chapter 20 - Electron Transfer Reactions

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Chapter 20 - Electron Transfer Reactions

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  1. Chapter 20 - Electron Transfer Reactions Objectives: 1. Carry out balancing of redox reactions in acidic or basic solutions; 2. Recall the parts of a basic and commercial voltaic cells; 3. Perform cell potential calculations from standard reduction potentials; 4. Classify oxidizing and reducing agents; 5. Apply the Nerst equation to redox problems; 6. Determine K from Ecell; 7. Perform electrolysis calculations.

  2. Introduction • NADH + (1/2)O2 + H+ -----> NAD+ + H2O • Medicinal Biochemistry: • http://web.indstate.edu/thcme/mwking/home.html

  3. Introduction • Pyruvate + CoA + NAD+ ------> CO2 + acetyl-CoA + NADH + H+

  4. Redox Reactions Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) Electron transfer reactions are ________________or redoxreactions. Redox reactions can result in the generation of an _________________ or be caused by imposing ___________________. Therefore, this field of chemistry is often called _____________________.

  5. Why study electrochemistry? • Batteries • Corrosion • Industrial production of chemicalssuch as Cl2, NaOH, F2 and Al • Biological redox reactions The heme group

  6. Review • OXIDATION • _______________________________ • REDUCTION • _______________________________ • OXIDIZING AGENT • _______________________________ • REDUCING AGENT • _______________________________

  7. Redox Reactions Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s)

  8. Redox Reactions Indirect Redox Reaction A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent.

  9. Electrochemical Cells • An apparatus that allows a redox reaction to occur by transferring electrons through an external connector. • Product favored reaction ---> ________________cell ----> electric current. • Reactant favored reaction ---> ________________ cell ---> electric current used to cause chemical change. Batteries are voltaic cells

  10. Electrochemistry Alessandro Volta, 1745-1827, Italian scientist and inventor. Luigi Galvani, 1737-1798, Italian scientist and inventor.

  11. Balancing Redox Equations • Some redox reactions have equations that must be balanced by special techniques. MnO4- + 5 Fe2+ + 8 H+ ---> Mn2+ + 5 Fe3+ + 4 H2O

  12. Balancing Redox Equations Cu + Ag+ --give--> Cu2+ + Ag

  13. Balancing Redox Equations Cu + Ag+ --give--> Cu2+ + Ag • Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. • Ox • Red • Step 2: Balance each for mass. • Step 3: Balance each half-reaction for charge by adding electrons. • Ox • Red

  14. Balancing Redox Equations Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electronsas the oxidizing agent requires. Reducing agent Oxidizing agent Step 5: Add half-reactions to give the overall equation. The equation is now balanced for both charge and mass.

  15. Reduction of VO2+ with Zn

  16. Balance the following in ACID solution: VO2+ + Zn ---> VO2+ + Zn2+ Step 1: Write the half-reactions Ox Red Step 2: Balance each half-reaction for mass. Ox Red Add H2O on O-deficient side and add H+ on other side for H-balance.

  17. Balancing… Step 3: Balance half-reactions for charge. Ox Red Step 4:Multiply by an appropriate factor. Ox Red Step 5: Add balanced half-reactions

  18. Tips on Balancing • Never add O2, O atoms, or O2- to balance oxygen. • Never add H2 or H atoms to balance hydrogen. • Be sure to write the correct charges on all the ions. • Check your work at the • end to make sure mass • and charge are balanced. • PRACTICE!

  19. Balance the following in basic solution: MnO4- + NO2- MnO2 + NO3- Oxidation half reaction:

  20. Balancing… MnO4- + NO2- MnO2 + NO3- Reduction half reaction:

  21. Balancing… MnO4- + NO2- MnO2 + NO3- Oxidation half reaction: Reduction half reaction: Multiply by appropriate factor to cancel e- and add both half-reactions Oxid. X Red. X Sum: Study Exp 11 – Procedure to balance redox reactions – practice and E calculation

  22. Electrons are transferred from Zn to Cu2+, but there is no useful electric current. Chemical Change --->Electric Current With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Oxidation: Zn(s) ---> Zn2+(aq) + 2e- Reduction: Cu2+(aq) + 2e- ---> Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

  23. Chemical Change --->Electric Current • To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. • This is accomplished in a GALVANIC or VOLTAIC cell. • A group of such cells is called a _____________.

  24. Chemical Change --->Electric Current • • Electrons travel thru external wire. • _____________ allows anions and cations to move between electrode compartments. Zn --> Zn2+ + 2e- Cu2+ + 2e- --> Cu Oxidation Anode Negative Reduction Cathode Positive <--Anions Cations-->

  25. The Cu|Cu2+ and Ag|Ag+ Cell

  26. Electrochemical Cell • _________ move from anode to cathode in the wire. • _______ & _________move thru the salt bridge.

  27. Terminology Figure 20.6

  28. 1.10 V 1.0 M 1.0 M What Voltage does a Cell Generate? • Electrons are “driven” from anode to cathode by an _______________________or emf. • For Zn/Cu cell, this is indicated by a voltage of • 1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M. Zn and Zn2+, anode Cu and Cu2+, cathode

  29. Cell Potential, E For Zn/Cu cell, potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M. This is the STANDARD CELL POTENTIAL, Eo —a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 ˚C.

  30. Calculating Cell Voltage • Balanced half-reactions can be added together to get overall, balanced equation. Zn(s) ---> Zn2+(aq) + 2e- Cu2+(aq) + 2e- ---> Cu(s) --------------------------------------------------------- Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s) • If we know Eo for each half-reaction, we could get Eo for net reaction. • But we need a reference!

  31. Cell Potential: SHE • Can’t measure 1/2 reaction Eo directly. Therefore, measure it relative to a ______________________, SHE. 2 H+(aq, 1 M) + 2e- <----> H2(g, 1 atm) Eo = 0.0 V

  32. Negative electrode Positive electrode Zn/Zn2+ half-cell hooked to a SHE.Eo for the cell = +0.76 V Zn(s) ---> Zn2+(aq) + 2e- Supplier of electrons Acceptor of electrons Zn --> Zn2+ + 2e- Oxidation Anode 2 H+ + 2e- --> H2 Reduction Cathode

  33. Reduction of Protons (H+) by Zn

  34. Overall reaction is reduction of H+ by Zn metal. Zn(s) + 2 H+ (aq) --> Zn2+ + H2(g) Eo = +0.76 V Therefore, Eo for Zn ---> Zn2+ (aq) + 2e- is +0.76 V Zn is a (better) (poorer) reducing agent than H2.

  35. Cu/Cu2+ and H2/H+ Cell,E0 for the cell = + 0.34 V Eo = +0.34 V Cu2+(aq) + 2e- ---> Cu(s) e- e- Positive Negative Acceptor of electrons Supplier of electrons Cu2+ + 2e- --> Cu Reduction Cathode H2 --> 2 H+ + 2e- Oxidation Anode

  36. Overall reaction is reduction of Cu2+ by H2 gas. • Cu2+ (aq) + H2(g) ---> Cu(s) + 2 H+(aq) • Measured Eo = +0.34 V • Therefore, Eofor Cu2+ + 2e- ---> Cu is + 0.34 V

  37. + Zn/Cu Electrochemical Cell Anode, negative, source of electrons Cathode, positive, sink for electrons • Zn(s) ---> Zn2+(aq) + 2e- Eo = +0.76 V • Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V • --------------------------------------------------------------- • Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)

  38. Uses of Eo values • Organize half-reactions by relative ability to act as oxidizing agents. Cu2+(aq) + 2e- ---> Cu(s) Eo = +0.34 V Zn2+(aq) + 2e- ---> Zn(s) Eo = –0.76 V Note that when a reaction is reversed the sign of E˚ is reversed! Cu2+ is better oxidazing agent than Zn2+; Cu2+ will be reduced and Zn will be oxidized Cu2+ reaction (reduction) will occur at the cathode Zn reaction (oxidation) will occur at the anode. E˚net = E˚cathode - E˚anode

  39. Std. Reduction Potentials • Organize half-reactions by relative ability to act as oxidizing agents. (All references are written as reduction processes). Table 20.1 • Use this to predict direction of redox reactions and cell potentials.

  40. Potential Ladder for Reduction Half-Reactions Best oxidizing agents Figure 20.14 Best reducing agents

  41. 2+ Cu + 2e- Cu +0.34 + 2 H + 2e- H2 0.00 2+ Zn + 2e- Zn -0.76 reducing ability of agent Using Standard Potentials, Eo Which is the best oxidizing agent: O2, H2O2, or Cl2? Which is the best reducing agent: Hg, Al, or Sn? oxidizing ability of agent Eo (V)

  42. Standard Reduction Potentials Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H+ and Cu+. H2 can reduce Cu2+ but not Zn2+ Cu cannot reduce H+ or Zn2+.

  43. 2+ Cu + 2e- --> Cu +0.34 + 2 H + 2e- --> H2 0.00 2+ Zn + 2e- --> Zn -0.76 Standard Reduction Potentials Ox. agent Red. agent Any substance on the right will reduce any substance higher than it on the left. • Northwest-southeast rule: product-favored reactions occur between • reducing agent at southeast corner • oxidizing agent at northwest corner

  44. Using Standard Reduction Potentials • In which direction do the following reactions go? • Cu(s) + 2 Ag+(aq) ---> Cu2+(aq) + 2 Ag(s) • 2 Fe2+(aq) + Sn2+(aq) ---> 2 Fe3+(aq) + Sn(s) • What is Eonet for the overall reaction?

  45. Calculating Cell Potential E˚net = “distance” from “top” half-reaction (cathode) to “bottom” half-reaction (anode) E˚net = E˚cathode - E˚anode Eonet for Cu/Ag+ reaction = +0.46 V

  46. Eo for a Cell Cd --> Cd2+ + 2e- or Cd2+ + 2e- --> Cd Fe --> Fe2+ + 2e- or Fe2+ + 2e- --> Fe All ingredients are present. Which way does reaction proceed?

  47. Eo for a Cell • From the table, you see • Fe is a better reducing agent than Cd • Cd2+ is a better oxidizing agent than Fe2+ Overall reaction: Fe + Cd2+ ---> Cd + Fe2+ Eo = E˚cathode - E˚anode = = Fe/Fe2+ // Cd2+/Cd

  48. More about Eo for a Cell Assume I- ion can reduce water. 2 H2O + 2e- ---> H2 + 2 OH- Cathode 2 I----> I2 + 2e- Anode ------------------------------------------------- 2 I- + 2 H2O --> I2 + 2 OH- + H2 Assuming reaction occurs as written, E˚net = E˚cathode - E˚anode = _________ E˚ means rxn. occurs in ___________ direction It is ____________ favored.

  49. Eo at non-standard conditions E = Eo - (RT/nF) lnQ E = Eo - 0.0257/n lnQ The NERNST EQUATION E = potential under nonstandard conditions R = gas constant (8.314472 J/Kmol) T = temperature (K) n = no. of electrons exchanged F = Faraday constant (9.6485338 x 104 C/mol) ln = “natural log” Q = reaction quotent (concentration of products/concentration of reactants to the appropriate power) One ___________ is the quantity of electric charge carried by one mole of electrons. If [P] and [R] = 1 mol/L, then E = E˚ If [R] > [P], then E is ______________ than E˚ If [R] < [P], then E is ______________ than E˚

  50. A voltaic cell is set up at 25oC with the following half-cells: Al3+(0.0010 M)/Al and Ni2+(0.50 M)/Ni. Write an equation for the reaction that occurs when the cell generates an electric current. • Determine which substance is oxidized (decide which is the better reducing agent). • Al is best reducing agent. • Then Al is oxidized and Ni2+ is reduced. • Ox (Anode): • Red (Cathode): • b) Add the half-reactions to determine the net ionic equation. • Net eq: • c) Calculate Eo and use Nernst eq. to calculate E. • Eo = Eocathode – Eoanode E = Eo – 0.0257/n ln Q