1. CHEM 163 Chapter 21 Spring 2009 1

2. 3-minute review What is a redox reaction? 2

3. Half-Reactions Split overall reaction into two reactions 3 OxidationReduction Step 1. Divide reaction into half reactions. Step 2. Balance atoms in each half reaction. Do O and H last! Step 3. Balance charges in each half reaction. Add e- Step 4. Make # e- gained equal # e- lost. Multiply by integer! Step 5. Add reactions together. Step 6. Check that atoms and charges are balanced. Need O?Add H 2 O Need H?Add H +

4. Electrochemical Cells 4 Voltaic (Galvanic) CellElectrolytic Cell ∆G < 0∆G > 0 Sys does work on surrSurr do work on sys E rxt > E prod E lost  electricityE rxt < E prod Electricity  rxn Electrodes: Conduct electricity between cell and surroundings Anode (oxidation) Cathode (reduction) Electrolyte: contains ions

5. 5 Fig. 21.3

6. Voltaic Cells Half-cells: to complete the circuit, electrons must flow externally Oxidation half-cell: Anode (Zn) “reactant” Electrolyte Reduction half-cell: Cathode (Cu) “product” Electrolyte Fig. 21.5 6 Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s)

7. Voltaic Cells Electrode charges: e - flow left to right e - created at anode, used up at cathode Anode has excess e - Salt bridge: Completes circuit Keeps each cell neutral Direction of ions 7 Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) Anode - Cathode + AnodeCathode

8. Electrodes Conduct electricity between cell and surroundings Active Electrodes: electrodes are components of half-reactions Inactive Electrodes: conduct electrons but are not reactants or products Ex. Graphite, Pt 8 2I - (aq)  I 2 (s) + 2e - MnO 4 - (aq) + 8H + (aq) + 5e -  Mn 2+ (aq) + 4H 2 O (l) Anode Cathode

9. How much electricity? Zn gives up electrons more easily Zn is a stronger reducing agent Potential difference between two electrodes –Cell potential (E cell ) –Cell voltage –Electromotive force (emf) 9 Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) Zn (s) Zn 2+ (aq) + 2 e - Cu (s) Cu 2+ (aq) + 2 e - E cell > 0 (spontaneous process)

10. Standard Cell Potentials E cell at standard conditions –Specific T (usually 298 K) –All components in standard states 1 M (aq) 1 atm (g) Pure solid Standard Electrode Potential –Half-cell potential –Always shown as a reduction 10 reductionoxidation

11. How can you measure a half-cell? Half-cell potentials are relative to a standard Standard Hydrogen Electrode (SHE) 11 2H + (aq; 1 M) + 2e -  H 2 (g; 1 atm) Stronger oxidizing agents… are easily reduced themselves Reduction reaction occurs more easily have more positive E o are weaker reduction agents M + (aq) + e -  M (s)

12. Writing spontaneous redox reactions 1.Which is the oxidizing agent? 2.Write reduction rxn for oxidizing agent (incl. E o ) 3.Flip oxidation rxn for reducing agent (incl. -E o ) 4.Multiply to make # e - lost = # e - gained 5.Add together 12 2 Ag + (aq) + Sn (s) 2 Ag (s) + Sn 2+ (aq)? Ag + (aq) + e -  Ag (s) Sn 2+ (aq) + 2e -  Sn (s) Ag E o value does not change!

13. Activity Series of Metals 1.Metals that can displace H 2 from acid –E cell is positive for reaction with H + –Any negative E half-cell (reduction potential) 2.Metals that cannot displace H 2 from acid –E cell is negative for reaction with H + –Any positive E half-cell (reduction potential) 3.Metals that can displace H 2 from water –E cell is positive for reaction with water 13

14. How much Work? 14 Volt electrical potential Joule energy Coulomb electrical charge Max work: How much charge flows? Faraday constant Charge of 1 mol of e - = 96,485 C / mol e - # mols of e - transferred = 96,485 J/V mol e - (standard state)

15. 15 Spontaneous At equilibrium Nonspontaneous

16. Effect of Concentration on E cell 16 Nernst Equation (at 298 K)

17. Concentration Cells Cells with different concentrations of same half-reaction 17 Not standard conditions! ?

18. Primary Batteries Nonrechargable Alkaline Zn (s) + MnO 2 (s) + H 2 O (l)  ZnO (s) + Mn(OH) 2 (s) Mercury and Silver –Zn anode; Hg or Ag cathode –Steady output Primary Lithium Batteries –High energy/mass ratio –Lithium metal anode –Implanted medical devices, watches 18 E = 1.5 V

19. Secondary Batteries Rechargeable Reverse reaction using electricity Lead-Acid PbO 2 (s) + Pb (s) + 2H 2 SO 4 (aq)  2 PbSO 4 (s) + 2 H 2 O (l) E cell = 2.1 V Nickel-Metal Hydride (Ni-MH) Lithium-Ion –Anode: Li atoms between graphite sheets –Cathode: Lithium metal oxide 19

20. Corrosion Natural redox: metal  metal oxides and metal sulfides Anodic regions: –Dents, ridges –Iron loss Cathodic regions: –Surface –Forms water Fe 2+ reacts with O 2 : –Rust deposits 20

21. Electrolytic Cells electrical energy  nonspontaneous reaction 21 E cell < 0 oxidation at anode reduction at cathode anode is positive cathode is negative

22. Electrolysis Splitting a substance using electrical energy Way to harvest elements (for industrial use) from substances What types of substances? Pure molten salts –Isolate metal or nonmetal Mixed molten salts –Isolate more easily reduced metal (based on EA) 22

23. Electrolysis of Water Anode Cathode Net 23 Not at standard state: E cell determined using Nernst equation: [H + ] = [OH - ] = 10 -7 M

24. Electrolysis of Aqueous Salts Which is going to react: water or salt? –Reduction with less negative E electrode occurs –Oxidation with less positive E electrode occurs 24 Example: KI (aq) Reduction: Oxidation: H 2 forms at cathode I 2 forms at anode

25. Electrolysis of Aqueous Salts (con’t) Overvoltage: Additional voltage used to produce gases (including H 2 and O 2 ) at electrodes –Usually 0.4 – 0.6 V So what forms? 1.Cations of less active metals are reduced 2.Cations of more active metals are not reduced; Water is reduced instead 3.Anions that are oxidized are typically halides 4.F -, common oxoanions are not oxidized; water is oxidized instead 25

26. How much product forms? The amount of product is directly proportional to quantity of charge that flows 26 How long does it take to produce 0.0423 mol of Cl 2 (g) by electrolysis of NaCl (aq) with power supply current of 12 A?

27. Homework due TUESDAY, May 19 th Chap 21: #16, 21, 30, 33, 38, 42, 56, 60, 70, 89, 94, 105 27