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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 0

  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. 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?

  24. 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+

  25. 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)

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

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

  28. 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

  29. 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

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

  31. 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

  32. 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

  33. Day 3 (dahditdadahditahh…Charge!)

  34. 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-

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

  36. 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):

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

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

  39. ??? 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)

  40. 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)

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

  42. Charging a Battery When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal. In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.

  43. Dry Cell Battery Anode (-) Zn ---> Zn2+ + 2e- Cathode (+) 2 NH4+ + 2e- ---> 2 NH3 + H2

  44. 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-

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

  46. 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

  47. Ni-Cad Battery Anode (-) Cd + 2 OH- ---> Cd(OH)2 + 2e- Cathode (+) NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-

  48. The positive electrode is made of Lithium cobalt oxide, or LiCoO2 The negative electrode is made of carbon.

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