1. Warmup Assign oxidation numbers to each element: NaNaClCl 2 H 2 O Ca(OH) 2 NO 3 -

2. Electrochemistry is the basis for: Batteries and CorrosionBatteries and Corrosion Industrial production of chemicalsIndustrial production of chemicals Biological redox reactionsBiological redox reactions

3. The oxidation numbers of elements in their compounds 4.4

4. Reviewing Oxidation Numbers 2(+1) + (-2) = 0 H O H O (+2) + 2(-2) + 2(+1) = 0 Ca O H Ca O H X + 3(-2) = -1 N O  X = +5

5. LEO says GER: LEO says GER: Lose Electrons = Oxidation Sodium is oxidized Gain Electrons = Reduction Chlorine is reduced

6. Sr + 2H 2 O Sr(OH) 2 + H 2 TiCl 4 + 2Mg Ti + 2MgCl 2 0 +1 +2 0 0 +40 +2 Assign oxidation numbers to each species. Then identify the oxidizer and reducer.

7. But how do we know who will give electrons to who and if the reaction will be spontaneous? Some reactions are spontaneous and can generate an electric current Redox (oxidation-reduction) reactions involve the transfer of electrons Some are nonspontaneous and can be caused by imposing an electric current (lab)

8. The Activity Series for Metals Lithium Lithium Potassium Potassium Calcium Calcium Sodium Sodium Magnesium Magnesium Aluminum Aluminum Zinc Zinc Chromium Chromium Iron Iron Nickel Nickel Lead Lead Hydrogen Bismuth Copper Copper Mercury Mercury Silver Silver Platinum Platinum Gold Gold Some standard Reduction Potentials at 25  C E  (volts) F 2 (g) + 2e -  2F - (aq)+2.87 Au 3+ + 3e -  Au(s) +1.50 Cl 2 (g) + 2e -  2Cl - (aq)+1.36 O 2 (g) + 4H 3 O + (aq) + 4e -  6H 2 O(l)+1.23 Br 2 (l) + 2e -  2Br - (aq)+1.08 Ag + (aq) + e -  Ag(s) +0.80 Hg 2 2+ (aq) + 2e -  2Hg(l) +0.79 I 2 (s) + 2e -  2I - (aq)+0.535 Cu 2+ (aq) + 2e -  Cu(s) +0.337 Sn 4+ (aq) + 2e -  Sn 2+ (aq)+0.15 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Cd 2+ (aq) + 2e -  Cd(s) -0.40 Zn 2+ (aq) + 2e -  Zn(s) -0.763 2H 2 O(l) + 2e -  H 2 (g) + 2OH - (aq)-0.828 Al 3+ (aq) + 3e -  Al(s) -1.66 K + (aq) + e -  K(s) -2.93 Li + (aq) + e -  Li(s) -3.045

9. Using data from the previous reduction potential table, predict which of the following is the best oxidizer Hint: good oxidizing agents are really good at stealing electrons and “getting reduced” a)F 2 b) Ag + c) Cd 2+ d) Al 3+ Using data from the previous reduction potential table, predict which of the following is the best reducing agent. Hint: good reducers are really good at donating electrons……and really bad at STEALING them. a) I 2 b) Au 3+ c) Br 2 e) Sn 2+

10. The Activity Series for Metals Lithium Lithium Potassium Potassium Calcium Calcium Sodium Sodium Magnesium Magnesium Aluminum Aluminum Zinc Zinc Chromium Chromium Iron Iron Nickel Nickel Lead Lead Hydrogen Bismuth Copper Copper Mercury Mercury Silver Silver Platinum Platinum Gold Gold Some standard Reduction Potentials at 25  C E  (volts) F 2 (g) + 2e -  2F - (aq)+2.87 Au 3+ + 3e -  Au(s) +1.50 Cl 2 (g) + 2e -  2Cl - (aq)+1.36 O 2 (g) + 4H 3 O + (aq) + 4e -  6H 2 O(l)+1.23 Br 2 (l) + 2e -  2Br - (aq)+1.08 Ag + (aq) + e -  Ag(s) +0.80 Hg 2 2+ (aq) + 2e -  2Hg(l) +0.79 I 2 (s) + 2e -  2I - (aq)+0.535 Cu 2+ (aq) + 2e -  Cu(s) +0.337 Sn 4+ (aq) + 2e -  Sn 2+ (aq)+0.15 Sn 2+ (aq) + 2e -  Sn(s) -0.14 Cd 2+ (aq) + 2e -  Cd(s) -0.40 Zn 2+ (aq) + 2e -  Zn(s) -0.763 2H 2 O(l) + 2e -  H 2 (g) + 2OH - (aq)-0.828 Al 3+ (aq) + 3e -  Al(s) -1.66 K + (aq) + e -  K(s) -2.93 Li + (aq) + e -  Li(s) -3.045

11. Calculate E° for a voltaic cell in which the reaction is: 2Ag + (aq) + Cd(s)  2Ag(s) +Cd 2+ (aq) a) split into 2 half reactions 2Ag + (aq) + 2e-  2Ag(s) Cd 2+ (aq) + 2e-  Cd(s) Cd(s)  Cd 2+ (aq) + 2e- b) Find E° values on table Don’t multiply potential by # moles: electric potentials are ”intensive properties” and do not depend on the amount of the substance E t ° +0.80 + +0.40 = 1.20 V The reaction will always happen in a way where the E t is + (spontaneous) E° = -0.40V reversed, = + 0.40 V E° = + 0.80V

12. Calculate E T ° for the following reaction: 2Al 3+ (aq) + 3Cd(s)  2Al(s) + 3Cd 2+ (aq) a)-2.06 V b) -4.52 Vc) +4.52 V d) -1.26 Ve) +2.06 V Al 3+ (aq) + 3e -  Al(s) E° = -1.66 Cd 2+ (aq) + 2e -  Cd(s E° = -0.40 E° = -1.66 + 0.40

13. Write the cell notation: Zn|Zn 2+ ||Cu 2+ |Cu Which reaction occurs at the anode? Oxidation Zn(s)  Zn 2+ (aq) + 2e- Which reaction occurs at the cathode? Reduction Cu 2+ (aq) + 2e-  Cu(s) Voltaic Cells e- are produced at the anode and travel to the cathode. This causes ions to migrate to the electrodes and undergo redox reactions. After some time, the [ ] change. salt bridge maintains electroneutrality and allows a current to flow. Otherwise, there would be a pile-up of electrons, and the current would stop. If there were no salt bridge, the reactants would simply mix together and react directly rather than sending electrons through a wire. Consequently, no current would flow, and there would not be a cell.

14. Reduction Potentials Pb 2+ + 2e¯  Pb -0.13 V Al 3+ + 3e¯  Al -1.68 V Label the cathode and anode. What voltage will be produced by the voltaic cell? a)2.97V b)1.55V c)-1.81V d)-2.97V Cathode (reduction, e- accepted here) anode (oxidation, e- given away here)