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Electrochemistry

Electrochemistry. Electrochemistry Terminology . Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na + + e - Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e -  2Cl - .

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Electrochemistry

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  1. Electrochemistry

  2. Electrochemistry Terminology • Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e- • Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e- 2Cl- • Oxidizing agent -The substance that is reduced is the oxidizing agent • Reducing agent - The substance that is oxidized is the reducing agent • LEO says GER

  3. Electrochemistry Terminology • Anode -The electrode where oxidation occurs • Cathode - The electrode where reduction occurs Reduction at the Cathode Leo is a

  4. Balancing Equations Consider the reduction of Ag+ ions with copper metal. Cu + Ag+----> Cu2+ + Ag

  5. Balancing Equations Cu + Ag+----> Cu2+ + Ag Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. Ox Cu ---> Cu2+ Red Ag+ ---> Ag Step 2: Balance each element for mass. Already done in this case. Step 3:Balance each half-reaction for charge by adding electrons. Ox Cu ---> Cu2+ + 2e- Red Ag+ + e----> Ag

  6. Balancing Equations Step 4: Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu ---> Cu2+ + 2e- Oxidizing agent 2 Ag+ + 2 e- ---> 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ ---> Cu2+ + 2Ag The equation is now balanced for both charge and mass.

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

  8. Balancing Equations Step 3: Balance half-reactions for charge. Ox Zn ---> Zn2+ +2e- Red e-+ 2 H++ VO2+--> VO2++ H2O Step 4: Multiply by an appropriate factor. Ox Zn ---> Zn2+ +2e- Red 2e-+ 4 H+ + 2 VO2+---> 2 VO2+ + 2 H2O Step 5: Add balanced half-reactions Zn + 4 H++ 2 VO2+---> Zn2++ 2 VO2++ 2 H2O

  9. Balancing Equations for RedoxReactions A great example of a thermodynamically spontaneous reaction is the thermite reaction. Here, iron oxide (Fe2O3 = rust) and aluminum metal powder undergo a redox (reduction-oxidation) reaction to form iron metal and aluminum oxide (Al2O3 = alumina): Fe2O3(s) + Al(s) ↔ Al2O3(s) + Fe(l) Fe = +3 Al = 0 Al = +3 Fe = 0

  10. Tips on Balancing Equations • 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.

  11. How many electrons are transferred in the following reaction?2ClO3– +12H+ + 10I– → 5I2 + Cl2 + 6H2O • 12 • 5 • 2 • 30 • 10 0 of 30 9

  12. Which of the following reactions is possible at the anode of a galvanic cell? 10 Seconds Remaining • Zn → Zn2+ + 2e– • Zn2+ + 2e– → Zn • Zn2+ + Cu → Zn + Cu2+ • Zn + Cu2+ →Zn2+ + Cu • two of these 0 of 30

  13. Which of the following species cannot function as an oxidizing agent? • S(s) • NO3–(ag) • Cr2O72–(aq) • I– (aq) • MnO4– (aq) 0 of 30 15

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

  15. Basic Concepts of Electrochemical Cells Anode Cathode

  16. Terms Used for Voltaic Cells

  17. CELL POTENTIAL, E • For Zn/Cu cell, potentialis+1.10 Vat 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.

  18. 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.

  19. Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

  20. Table of Reduction Potentials Measured against the StandardHydrogenElectrode

  21. oxidizing o ability of ion E (V) 2+ Cu + 2e- Cu +0.34 + 2 H + 2e- H 0.00 2+ Zn + 2e- Zn -0.76 reducing ability of element TABLE OF STANDARD REDUCTION POTENTIALS 2 To determine an oxidation from a reduction table, just take the opposite sign of the reduction!

  22. + Zn/Cu Electrochemical Cell 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) Eo = +1.10 V Anode, negative, source of electrons Cathode, positive, sink for electrons

  23. break

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

  25. Eo for a Voltaic Cell From the table, you see • Fe is a better reducing agent than Cd • Cd2+ is a better oxidizing agent than Fe2+

  26. Eo for a Voltaic Cell e- Cathode Anode Cd2+Fe --> Fe2+Cd E=+0.10V All ingredients are present. Which way does reaction proceed?

  27. More About Calculating Cell Voltage Assume I- ion can reduce water. 2 H2O + 2e- ---> H2 + 2 OH- 2 I----> I2 + 2e- ------------------------------------------------- 2 I- + 2 H2O --> I2 + 2 OH- + H2 Cathode Anode Assuming reaction occurs as written, E˚ = E˚cat+ E˚an= (-0.828 V) + (- 0.535 V) = -1.363 V Negative E˚ means rxn. occurs in opposite direction (the connection is backwards or you are recharging the battery)

  28. Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E0 A negative free energy change, (-G)

  29. Zn - Cu Galvanic Cell From a table of reduction potentials: Zn2+ + 2e- Zn E = -0.76V Cu2+ + 2e- Cu E = +0.34V

  30. Zn - Cu Galvanic Cell The less positive, or more negative reduction potential becomes the oxidation… Cu2+ + 2e- Cu E = +0.34V Zn  Zn2+ + 2e- E = +0.76V Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

  31. Line Notation An abbreviated representation of an electrochemical cell • Zn(s) | Zn2+(aq) (1.0M) || Cu2+(aq) (1.0M) | Cu(s) Anode material Cathode material Anode solution Cathode solution | || | • Line notation is cool, just like AC

  32. Zn(s) | || H+(aq) (1.0M) | H2(g)(1.00 atm) | Pt(s) Zn2+(aq) (1.0M)

  33. Calculating G0 for a Cell G0 = -nFE0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e- Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

  34. Try this one Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Fe+2(aq)+ 2e-® Fe(s) Eº = 0.44 V Cu+2(aq) +2e- ® Cu(s) Eº = 0.34 V

  35. Day 3 (dahditdadahditahh…Charge!)

  36. The Nernst Equation Standard potentials assume a concentration of 1.0 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(molK) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e-

  37. Nernst Equation Simplified At 25 C (298 K) the Nernst Equation is simplified this way:

  38. Equilibrium Constants and Cell Potential At equilibrium, forward and reverse reactions occur at equal rates, therefore: • The battery is “dead” • The cell potential, E, is zero volts Modifying the Nernst Equation (at 25 C):

  39. Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

  40. Apply to Problem 70

  41. ??? Concentration Cell Both sides have the same components but at different concentrations. Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

  42. ??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration Zn2+ (1.0M) + 2e- Zn (reduction) Zn  Zn2+ (0.10M) + 2e- (oxidation) Zn2+ (1.0M)  Zn2+ (0.10M)

  43. Concentration Cell ??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C). Zn2+ (1.0M)  Zn2+ (0.10M)

  44. Nernst Calculations Zn2+ (1.0M)  Zn2+ (0.10M)

  45. Try Problem 62

  46. Alkaline Battery Nearly same reactions as in common dry cell, but under basic conditions. Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2e- Cathode (+): 2 MnO2 + H2O + 2e- ---> Mn2O3+ 2 OH-

  47. Mercury Battery Anode: Zn is reducing agent under basic conditions Cathode: HgO + H2O + 2e- ---> Hg + 2 OH-

  48. Lead Storage Battery Anode (-) Eo = +0.36 V Pb + HSO4- ---> PbSO4 + H+ + 2e- Cathode (+) Eo = +1.68 V PbO2 + HSO4- + 3 H+ + 2e- ---> PbSO4 + 2 H2O

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