1. 1 UNIT 7 Reduction / Oxidation Reactions “Redox” and Electrochemistry

2. 2 Reduction / Oxidation -- Reactions where electrons are transferred from one atom to another loss of electrons = oxidation gain of electrons = reduction -- Shown by a change in the oxidation state of an atom

3. 3 Rules for Oxidation States 1. Pure elements:oxidation state = 0 2. Compounds:total of ox. states = 0 example: Na 2 SO 4 Na = +1 O = -2 S = ? (+1)+(+1)+S+(-2)+(-2)+(-2)+(-2) = 0 S = +6 3. Ions:ox. state = charge 4. Polyatomic ions:total ox. states = charge example: PO 4 -3 O = -2 P = ? P+(-2)+(-2)+(-2)+(-2) = -3 P = +5 (+2) + S + (-8) = 0 P + (-8) = -3

4. 4 Example of a Redox reaction: 2 Al + 3 CuCl 2  2 AlCl 3 + 3 Cu 0+2+30 Cl - : -1  -1 no change ; spectator ion Al 0 : 0  +3 Ox. state increases electrons lost = oxidation Cu +2 : +2  0 Ox. state decreases electrons gained = reduction Oxidation and reductions always happen together -- one can’t happen without the other

5. 5 More examples: 2 HCl + Mg  MgCl 2 + H 2 +1 –1 0 +2 -1 0 2H 2 + O 2  2 H 2 O 0 0 +1 -2 2 NaCl + CaBr 2  2 NaBr + CaCl 2 +1 –1 +2 –1 +1 –1 +2 -1 not a redox reaction

6. 6 Half - Reactions 1. Write equation, and balance 2 Al + 3 CuCl 2  2 AlCl 3 + 3 Cu 2. Find oxidation states 3. Write oxidation and reduction half-reactions separately ox. 2Al 0  2Al +3 red. 3Cu +2  3Cu 0 4. Show electrons to balance charges ox. 2Al 0  2Al +3 + 6e - red. 3Cu +2 + 6e -  3Cu 0 Half-reactions must be balanced with respect to mass and charge 0 +2 -1 +3 -1 0

7. 7 Each half-reaction has a tendency to occur or not Elements can be ranked in an activity series – see ref. table J The most easily oxidized metals are at the top  active metals The most easily reduced non-metals are at the top  active non-metals A redox reaction will occur spontaneously if the pure element reacting is the more active one (higher on table J) Example: Pb + 2NaCl  PbCl 2 + 2Na -- Pb, the pure metal, is below Na  not spontaneous Example: CaI 2 + Cl 2  CaCl 2 + I 2 -- Cl 2, the pure non-metal, is above I 2  spontaneous

8. 8 Electrochemistry Spontaneous redox reactions can be used to produce electricity in a voltaic cell Half-reactions are separated, creating a flow of electrons Alessandro Volta 2Al 0  2Al +3 + 6e - 3Cu +2 + 6e -  3Cu 0 Al Al +3 Cl - Cu Cu +2 Cl - oxidation reduction e-e- ---- ++++ Na + Cl - mass of bar= corrosion mass of bar= plating anode cathode The metal bars are called electrodes: oxidation happens at the anode reduction happens at the cathode

9. 9 2Al 0  2Al +3 + 6e - 3Cu +2 + 6e -  3Cu 0 Al Al +3 Cl - Cu Cu +2 Cl - oxidation reduction ---- ++++ Na + Cl - mass of bar= corrosion mass of bar= plating anode cathode The cell consists of 2 half-cells, connected by: 1. a wire, for electron flow 2. a salt bridge, for ion flow The overall redox reaction is: 2Al 0 + 3Cu +2  2Al +3 + 3Cu 0 (balanced for mass & charge)

10. 10 There are 2 possible reactions: 1. Ag 0 ox.; Pb +2 red. Ag below Pb, not spontaneous 2. Ag + red.; Pb 0 ox. Pb above Ag, spontaneous The more active metal is always oxidized

11. 11 Sample Questions 1. What gets oxidized? reduced? ox.red. 2. Write ½ reactions under the appropriate beaker Pb 0  Pb +2 + 2e - Ag + + e -  Ag 0 2 2 2 3. Balance ½ reactions 4. Label anode & cathode anode cathode 5. Which metal gets corroded? Plated? corrodedplated 6. Indicate the direction of electron flow though the wire e-e- 7. Indicate the direction of ion movement in the salt bridge + _ 8. Write & balance the overall redox equation Pb 0 + 2Ag +  Pb +2 + 2Ag 0 Balance these redox equations: a. ____Na 0 + ____Mg +2  ____Na + + ____Mg 0 b. ____Cr +3 + ____Ba 0  ____Cr 0 + ____Ba +2 2 2 2 3 3 2

12. 12 If a reaction is non-spontaneous, we can make it happen -- by adding electrical energy Electrolytic Cell -- Electrolysis Example: 2NaCl  2Na + Cl 2 Battery + - Na + Cl - e-e- e-e- ---- ++++ gain electrons reduction Na + + e -  Na 0 loses electrons oxidation 2Cl -  Cl 2 0 + 2e - 2 2 2 Overall redox reaction = 2NaCl  2Na + Cl 2 anode cathode (molten)

13. 13 Electrolytic cells can be used to electroplate metals Battery Sn +2 Cl - e-e- ++++ ---- e-e- - + steel can Sn bar gain e - reduction lose e - oxidation Sn +2 + 2e -  Sn 0 Sn 0  Sn +2 + 2e - cathode object to be plated anode metal used for plating

14. 14 Real-Life Redox Reactions I. Lead-Acid Storage Battery(car battery) H 2 SO 4 PbO 2 Pb Half-reactions: ox. Pb 0 (s) + SO 4 -2 (aq)  Pb +2 SO 4 (s) + 2e - red. Pb +4 O 2 (s) + SO 4 -2 (aq) + 4H + (aq) + 2e -  Pb +2 SO 4 (s) + 2H 2 O Overall reaction: Pb In charging, a power source is applied, and both half-reactions are reversed + PbO 2 + 2H 2 SO 4  2PbSO 4 + 2H 2 Ovoltage = 2.0V

15. 15 II. Corrosion of Iron a 2-step process; requires Fe, O 2, H 2 O Step one: ox. 2Fe 0  2Fe +2 +4e - red. O 2 0 + 4H + + 4e -  2H 2 O -2 overall: 2Fe 0 + O 2 0 + 4H +  2Fe +2 + 2H 2 O Step two: ox. 4Fe +2  4Fe +3 + 4e - red. O 2 0 + 4H 2 O + 4e -  6O -2 + 8H + overall: 4Fe +2 + O 2 0 + 4H 2 O  2Fe 2 O 3 + 8H + rust – helps step 1;prevents step 2Acid (H + ) Base (OH - ) -- prevents step 1