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© University of South Carolina Board of Trustees Oxidation States Example Find the oxidation state of… Fe: Fe (s) + O 2  Fe 2 O 3 Al: Al (s) + O 2 

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Presentation on theme: "© University of South Carolina Board of Trustees Oxidation States Example Find the oxidation state of… Fe: Fe (s) + O 2  Fe 2 O 3 Al: Al (s) + O 2 "— Presentation transcript:

1 © University of South Carolina Board of Trustees Oxidation States Example Find the oxidation state of… Fe: Fe (s) + O 2  Fe 2 O 3 Al: Al (s) + O 2  AlO 2 Can this compound exist?

2 © University of South Carolina Board of Trustees Chapt 18 Electrochemistry Sec. 3 Voltaic Cells: Experimental

3 © University of South Carolina Board of Trustees Voltaic Cells A cell with a spontaneous flow of current

4 © University of South Carolina Board of Trustees Salt Bridge Required Two Reaction Compartments

5 © University of South Carolina Board of Trustees Direction of electron flow? Cu (s)  Cu 2+ + 2e - (oxid.) Cu 2+ + 2e -  Cu (s) (reduc.) Ag + + e -  Ag (s) (reduc.) Ag (s)  Ag + + e - (oxid.)

6 © University of South Carolina Board of Trustees Cu (s)  Cu 2+ + 2e - (oxid.) Cu 2+ + 2e -  Cu (s) (reduc.) Ag + + e -  Ag (s) (reduc.) Ag (s)  Ag + + e - (oxid.) Positive Voltage, Spontaneous electron flow

7 © University of South Carolina Board of Trustees Half Cell Sn 2+ and Sn 4+ are both in contact with an inert electrode Inert: ●electrical conductor ●no chemical reactions ●e.g., Pt, graphite

8 © University of South Carolina Board of Trustees Hydrogen gas is bubbled over an inert platinum electrode Half Cell: Gas Reactants

9 © University of South Carolina Board of Trustees The Calomel Electrode Hg 2 Cl 2(s) (calomel) is reduced to Hg ( ) in contact with a Pt wire. Common standard for comparison to other electrodes.

10 © University of South Carolina Board of Trustees No Salt Bridge One Reaction Compartment

11 Batteries are Solid Voltaic Cells © University of South Carolina Board of Trustees

12 Chapt 18 Electrochemistry Sec. 4 Voltaic Cells: Determining Cell Voltage

13 © University of South Carolina Board of Trustees Half-Cell Voltages by definition 2H + (aq) + 2e -  H 2(g) E ° = 0.00 V by definition A voltage can be assigned to a ½ reaction, if we establish a reference point Sum of ½-Cell Voltages = Voltage of the Cell Voltage or Potential

14 © University of South Carolina Board of Trustees Standard Potentials ­Value of voltage of ½-reaction is versus H 2 ½- reaction ­Standard States Liquid / Solid = pure Gas = 1 atm Solution = 1 M ­Written as a reduction Reverse sign, if reaction changes direction ­Do not multiply by stoichiometry factor

15 © University of South Carolina Board of Trustees Activity Series +2.87 V +0.77 V +0.34 V +0.00 V -0.25 V -1.66 V -3.05 V High Reduction Potential Easy to Reduce Low Reduction Potential Hard to Reduce

16 © University of South Carolina Board of Trustees Potentials and Spontaneity A positive cell potential is spontaneous ­Half-cell with the larger (positive) potential goes forward (reduction). ­Half-cell with the lower potential (small positive or large negative) goes in reverse (oxidation).

17 © University of South Carolina Board of Trustees Example: Spontaneity Determine the spontaneous direction for the following cell: Pb +2 (aq) + 2e −  Pb (s) E ° = −0.126 V Br 2( ) + 2e −  2Br − (aq) E ° = +1.060 V

18 © University of South Carolina Board of Trustees Putting together a voltaic cell A voltaic cell is made up of an Al electrode in a solution of Al(NO 3 ) 3 and a Cu electrode in a solution of Cu(NO 3 ) 2 with a potassium nitrate salt bridge. Al 3+ (aq) + 3e −  Al (s) E ° = −1.66 V Cu 2+ (aq) + 2e −  Cu (s) E ° = +0.34 V What is the cell potential? Which electrode is positive? Which electrode is negative? What is the oxidizing agent? What is the reducing agent?

19 © University of South Carolina Board of Trustees Student Example Write the cell reaction and calculate E °cell for the voltaic cell made up of a standard Mg/Mg 2+ half-cell and a standard Fe 2+ /Fe 3+ half-cell. Mg 2+ (aq) + 2e −  Mg (s) E ° = -2.37 V Fe 3+ (aq) + e −  Fe +2 (aq) E ° = +0.77 V

20 © University of South Carolina Board of Trustees Activity Series +2.87 V +0.77 V +0.34 V +0.00 V -0.25 V -1.66 V -3.05 V High Reduction Potential Easy to Reduce Low Reduction Potential Hard to Reduce

21 © University of South Carolina Board of Trustees Activity Series +2.87 V +0.77 V +0.34 V +0.00 V -0.25 V -1.66 V -3.05 V High Reduction Potential Easy to Reduce  Low Reduction Potential Hard to Reduce 

22 © University of South Carolina Board of Trustees Activity Series High Reduction Potential Easy to Reduce  Hard to Oxidize  Low Reduction Potential Hard to Reduce  Easy to Oxidize  +2.87 V +0.77 V +0.34 V +0.00 V -0.25 V -1.66 V -3.05 V

23 © University of South Carolina Board of Trustees Example: Activity Series Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Co(NO 3 ) 2 ? Co 2+ (aq) + 2e -  Co (s) E ° = -0.28 V Fe 2+ (aq) + 2e -  Fe (s) E ° = -0.44 V

24 © University of South Carolina Board of Trustees Example: Activity Series Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Co(NO 3 ) 2 ? Co 2+ (aq) Co 2+ (aq) + 2e -  Co (s) E ° = -0.28 V Fe (s) Fe 2+ (aq) + 2e -  Fe (s) E ° = -0.44 V

25 © University of South Carolina Board of Trustees Example: Activity Series Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Zn(NO 3 ) 2 ? Zn 2+ (aq) + 2e -  Zn (s) E ° = -0.76 V Fe 2+ (aq) + 2e -  Fe (s) E ° = -0.44 V

26 © University of South Carolina Board of Trustees Example: Activity Series Which elements are oxidized and which are reduced when Fe is added to a solution of 1 M Zn(NO 3 ) 2 ? Zn 2+ (aq) Zn 2+ (aq) + 2e -  Zn (s) E ° = -0.76 V Fe (s) Fe 2+ (aq) + 2e -  Fe (s) E ° = -0.44 V


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