1. 1 2-5-2011

2. 2 Galvanic Cell: Electrochemical cell in which chemical reactions are used to create spontaneous current (electron) flow

3. 3 Galvanic Cells spontaneous redox reaction anode oxidation cathode reduction

4. 4 Galvanic Cells The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anodecathode Cell Representation

5. 5 For example, look at the following cell representation: Cu(s) | CuSO 4 (0.100 M) || ZnCl 2 (0.200 M) | Zn(s) This cell is read as follows: a copper electrode is immersed in a 0.100 M CuSO 4 solution (this is the first half-cell, anode), 0.200 M ZnCl 2 solution in which a Zn(s) electrode is immersed (this is the second half-cell, cathode). If the E cell is a positive value, the reaction is spontaneous (galvanic cell) and if the value is negative, the reaction is nonspontaneous (electrolytic cell) in the direction written and will be spontaneous in the reverse direction.

6. 6 Standard Reduction Potentials and the Standard hydrogen electrode (SHE) Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm (standard state). E 0 = 0.00 V 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

7. 7 Calculating the Cell Potential The process of calculating the cell potential is simple and involves calculation of the potential of each electrode separately, then the overall cell potential can be determined from the simple equation: E cell = E cathode - E anode Sometimes, this relation is written as: E cell = E right – E left E cell = E reduction half cell - E oxidation half cell

8. 8 Standard Reduction Potentials Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm)

9. 9 E 0 = 0.76 V cell Standard emf (E 0 ) cell 0.76 V = 0 - E Zn /Zn 0 2+ E Zn /Zn = -0.76 V 0 2+ Zn 2+ (1 M) + 2e - Zn E 0 = -0.76 V E 0 = E H /H - E Zn /Zn cell 00 + 2+ 2 Standard Reduction Potentials E 0 = E cathode - E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s)

10. 10 Standard Reduction Potentials Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = E cathode - E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu – E H /H 2++ 2 000 0.34 = E Cu /Cu - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

11. 11 These electrodes are Active (they take part in the reaction) Combining the two half reactions:

12. 12 Electrodes are passive (not involved in the reaction)

13. 13 Standard Electrode Potential

14. 14

15. 15 E 0 is for the reaction as written The more positive E 0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

16. 16 Diagonal Rule It is possible to determine whether an electrochemical reaction would occur spontaneously or not by considering the values of the standard electrode potentials of the two half reactions. Steps: 1.Write down the two half reactions placing the one with the larger E o up. 2.Draw a diagonl from up left pointing to down right. 3.The species in the up left will react with the species in the products of the down right.

17. 17 Predict what will happen if Br 2 is added to a solution containing NaCl and NaI. Assume all species are in their standard states. Solution: From E o Table we have: Cl 2 + 2e  2Cl - E o = 1.36V Br 2 (l) + 2e  2Br - E o = 1.07V I 2(s) + 2e  2I - E o = 0.53V Applying the diagonal rule suggests that Br 2 will not be able to react with Cl -, but will react with I -.

18. 18 Br 2 (l) + 2e  2Br - E o = 1.07V 2I -  I 2(s) + 2e E o = 0.53V Br 2 (l) + 2I -  I 2(s) + 2Br - E o rxn = E o cathode - E o anode E o rxn = 1.07 – 0.53 E o rxn = 0.54V

19. 19 Can Sn(s) reduce Zn 2+ under standard state conditions? Sn(s) + Zn 2+  Sn 2+ + Zn(s) ? To answer this question, we should write the two half reactions starting with the one with larger E o Sn 2+ + 2e  Sn(s)E o = -0.14V Zn 2+ + 2e  Zn(s)E o = -0.76V Applying the diagonal rule, we conclude that Sn(s) will not be able to reduce Zn 2+. The spontaneous reaction will be: Sn 2+ + Zn(s)  Zn 2+ + Sn(s) This means that Zn(s) will reduce Sn 2+.

20. 20 Calculation of the standard emf of an electrochemical cell The procedure is simple: 1.Arrange the two half reactions placing the one with the larger E o up (the cathode). 2.The half reaction with the lower E o is placed down (the anode). 3.E o cell = E o cathode - E o anode