1. Unit 16 Electrochemistry Oxidation & Reduction

2. Oxidation verses Reduction Gain oxygen atoms 2 Mg + O 2  2 MgO Lose electrons (e - ) Mg (s)  Mg + 2 + 2 e - Lose hydrogen CH 3 OH  CH 2 O + H 2 Lose oxygen atoms 2 KClO 3  2 KCl + 3 O 2 Gain electrons (e - ) Cu + 2 + 2 e -  Cu (s) Gain hydrogen CO + 2 H 2  CH 3 OH

3. l l l l l l l l l -4 -3 -2 -1 0 1 2 3 4 reduction oxidation When oxidation occurs, there is also reduction. They go “hand in hand”. Not all reactions are oxidation reduction reactions.

4. Using Appendix 6 chart: To determine the strength as an oxidizing agent, look at the left side of appendix 6. The chemical that is higher on the list (has a greater magnitude + voltage) is the better oxidizing agent. Which is better? Br 2 or Sn + 2 To determine the strength as a reducing agent, look at the right side of appendix 6. The chemical that is lower on the list (has the greater magnitude – voltage) is the better reducing agent. Which is better? Pb (s) or H 2 O 2

5. The chemical that is oxidized acts as the reducing agent. The chemical that is reduced acts as the oxidizing agent. In the following, indicate which chemical is being oxidized, reduced, acts as the oxidizing agent, and acts as the reducing agent. Zn (s) + Cu + 2 (aq)  Zn + 2 (aq) + Cu (s)

6. Electricity is the movement of electrons to provide energy. Two factors affect the amount of energy produced. 1. Voltage is the push behind the e -. Voltage is measured in volts 2. Current which is how many e - pass a point each second. Current is measured in Amperes. 1 Amp = 6.0 x 10 18 e - /s An electrochemical Cell (Galvanic or voltaic cell) is a chemical device to make electricity from chemical reactions. It employs oxidation and reduction. It will have a positive (+) voltage while it works.

7. Cell Potential (E o cell ) or Electromotive force (emf) is the pull or “driving force” on the e -. It is measured in volts (E o ) Standard reduction potential is the potential for the half – reaction at 1M the standard being measured off 2 H + + 2 e -  H 2, Pt electrode = 0.00 v Remember that these are listed as Reduction Potentials. When there is reduction, there is also oxidation. One of the two half – reactions will need to be reversed (so that it is written as an oxidation)

8. To calculate the voltage of an electrochemical cell: 1. Obtain the two half – reactions from the reduction potential chart. 2. Determine which half – reaction is the oxidation, reverse the reaction. When the reaction is reversed, the sign on the voltage is changed. 3. Balance the e - in the two half – reactions by multiplying the coefficients by some whole number. Remember that voltage is a property of the kind of chemical, not the amount. Do not multiply the voltage. 4. Add the two half – reactions to determine the net voltage for the cell.

9. A salt bridge or porous disk connects the half cells and allows e - to flow, completing the circuit. We use a porous cup as the barrier between the 2 half cells

10. Calculate the voltage of a Galvanic cell made with Mg and Cu used to run the clock. Mg + 2 + 2e -  Mg (s) volts Cu + 2 + 2 e -  Cu (s) volts

11. Line Notation : A way to denote a galvanic cell Anode: the oxidation is written first Cathode: the reduction is written second One line separates the phases of a half cell Two lines separate the anode and cathode Mg l Mg + 2 ll Cu + 2 l Cu

12. Corrosion is the process of returning metals to their natural state. The oxidation of metals. Example: 4 Fe (s) + 3 O 2 (g)  2 Fe 2 O 3 (s) All metals corrode easily, except Au, Ag, Cu Preventing corrosion: Coating with Cd or Sb Galvanizing (coating) with Zn Stainless steel is an alloy of Fe with Cd + Ni Sacrificial metals which corrode first are Mg or Zn

13. A battery is a group of galvanic cells connected in series. (add potentials E o cell ) Wet cell – contain fluid medium for e- flow Dry cell - contain a paste medium for e- flow Primary cell – one time use, not rechargeable Secondary cell - rechargeable

14. Dry Cell (Batteries) – do not contain a liquid electrolyte –Acid version Anode reaction - oxidation Zn  Zn 2+ + 2e  +0.76 v Cathode reaction – reduction 2NH 4 + + 2MnO 2 + 2e   Mn 2 O 3 + 2NH 3 + 2H 2 O +0.74 v

15. –Other types Alkaline – Zn + 2OH -  ZnO + H 2 O + 2e - anode 2MnO 2 + H 2 O + 2 e -  Mn 2 O 3 + 2OH - cathode Silver cell – Zn anode, Ag 2 O cathode Mercury cell – Zn anode, HgO cathode Cadmium – anode, NiO 2 – cathode (secondary – rechargable) Mercury cell

16. Overall reaction Pb + PbO 2 + 2H 2 SO 4  2PbSO 4 + 2H 2 O +2.02 v Wet cell - Lead storage battery – rechargeable Anode Pb + HSO 4 -  PbSO 4 + H + + 2 e - +0.34 v Cathode PbO 2 + HSO 4 - + 3 H + + 2 e -  PbSO 4 + 2 H 2 O +1.68 v

17. An Electrolytic cell is a cell with a negative voltage, so there is no reaction. An outside source of energy is used as a way to drive the reaction. This type of cell is often used as a controlled way to deposit thin layers of metals, electroplating, on another metal.

18. Cr + 3 + 3 e -  Cr Cu + 2 + 2 e -  Cu An electrolytic cell can be used to electroplate one metal on another metal. Pb Anode Cu Cathode Power source Balance the reaction and determine the minimum voltage that needs to be supplied to drive the reaction.

19. Electrochemical Review 1. a. Zn +2Ag + 1  Zn + 2 + 2 Ag +1.56 v c. Fe + Sn + 2  Fe + 2 + Sn +0.30 v b. Cu + 2 Ag + 1  Cu + 2 + 2 Ag +0.46 v d. Hg (L) + 2 H + 1  Hg + 2 + H 2 -0.78 v e. Cell d does not work 2. oxidation take place at the anode reduction take place at the cathode Cu is at the anode Au is at the cathode +1.16 v 3 Cu + 2 Au + 3  3 Cu + 2 + 2 Au Au continuously get larger 3.Cl 2 (g) +2 Br - 1  2 Cl - 1 + Br 2 (g) +0.30 volts

20. Pollution Crossword Down 3. Potable 21. Photochemical 25. Organic 37. Zeolite Across 6. Inorganic 8. Aeration 18. Precipitation 29. Detergent 40. Sulfate