1. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)1 ELECTROCHEMISTRY Chapter 21 redox reactions electrochemical cells electrode processes construction notation cell potential and  G o standard reduction potentials (E o ) non-equilibrium conditions (Q) batteries corrosion Electric automobile

2. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)2 CHEMICAL CHANGE  ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ is reduced and is the oxidizing agent Cu 2+ (aq) + 2e-  Cu(s)

3. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)3 Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments. This maintains electrical neutrality. ANODE OXIDATION CATHODE REDUCTION

4. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)4 This is the STANDARD CELL POTENTIAL, E o E o is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25  C. CELL POTENTIAL, E o For Zn/Cu, voltage is 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M.

5. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)5 E o and  G o E o is related to  G o, the free energy change for the reaction.  G o = - n F E o F = Faraday constant = 9.6485 x 10 4 J/Vmol n = the number of moles of electrons transferred. Michael Faraday 1791-1867 Discoverer of electrolysis magnetic props. of matter electromagnetic induction benzene and other organic chemicals n for Zn/Cu cell ? n = 2 Zn / Zn 2+ // Cu 2+ / Cu

6. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)6 For a reactant-favored reaction - electrolysis cell: Electric current  chemistry Reactants  Products  G o > 0 and so E o < 0 (E o is negative) For a product-favored reaction – battery or voltaic cell: Chemistry  electric current Reactants  Products  G o 0 (E o is positive) E o and  G o (2)  G o = - n F E o

7. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)7 STANDARD CELL POTENTIALS, E o Can’t measure half- reaction E o directly. Therefore, measure it relative to a standard HALF CELL: the S tandard H ydrogen E lectrode ( SHE ). 2 H + (aq, 1 M) + 2e- H 2 (g, 1 atm) E o = 0.0 V

8. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)8 BEST Reducing agent ? ? STANDARD REDUCTION POTENTIALS Half-Reaction E o (Volts) Cu 2+ + 2e-  Cu+ 0.34 Oxidizing ability of ion Reducing ability of element 2 H + + 2e-  H 2 0.00 Zn 2+ + 2e-  Zn -0.76 BEST Oxidizing agent ? ? Cu 2+ Zn

9. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)9 Using Standard Potentials, E o H 2 O 2 /H 2 O +1.77 Cl 2 /Cl - +1.36 O 2 /H 2 O +1.23 In which direction does the following reaction go? Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Hg 2+ /Hg +0.86 Sn 2+ /Sn -0.14 Al 3+ /Al -1.66 Ag + /Ag +0.80 Cu 2+ / Cu +0.34 See Table 21.1, App. J for E o (red.) Which is the best oxidizing agent: O 2, H 2 O 2, or Cl 2 ? Which is the best reducing agent: Sn, Hg, or Al ? As written: E o = (-0.34) + 0.80 = +0.43 V reverse rxn: E o = +0.34 + (-0.80) = -0.43 V

10. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)10 Cells at Non-standard Conditions For ANY REDOX reaction, Standard Reduction Potentials allow prediction of direction of spontaneous reaction If E o > 0 reaction proceeds to RIGHT (products) If E o < 0 reaction proceeds to LEFT (reactants) E o only applies to [ ] = 1 M for all aqueous species at other concentrations, the cell potential differs E cell can be predicted by Nernst equation

11. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)11 Cells at Non-standard Conditions (2) E o only applies to [ ] = 1 M for all aqueous species at other concentrations, the cell potential differs E cell can be predicted by Nernst equation E = E o - ln (Q) RT nF n = # e- transferred F = Faraday’s constant = 9.6485 x 10 4 J/Vmol Q is the REACTION QUOTIENT (recall ch. 16, 20) At equilibrium  G = 0 E = 0 Q = K  G o, E o refer to ALL REACTANTS relative to ALL PRODUCTS

12. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)12 Example of Nernst Equation Q. Determine the potential of a Daniels cell with [Zn 2+ ] = 0.5 M and [Cu 2+ ] = 2.0 M; E o = 1.10 V A. Zn / Zn 2+ (0.5 M) // Cu 2+ (2.0 M) / Cu E = 1.10 - (0.0257) ln ( [Zn 2+ ]/[Cu 2+ ] ) 2 E = 1.10 - (-0.018) = 1.118 V Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Q = ? [Zn 2+ ] [Cu 2+ ] E = E o - ln (Q) RT nF

13. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)13 Nernst Equation (2) Q. What is the cell potential and the [Zn 2+ ], [Cu 2+ ] when the cell is completely discharged? A. When cell is fully discharged: chemical reaction is at equilibrium E = 0  G = 0 Q = Kand thus 0 = E o - (RT/nF) ln (K) or E o = (RT/nF) ln (K) or ln (K) = nFE o /RT = (n/0.0257) E o at T = 298 K So... K = e = 1.5 x 10 37 (2)(1.10)/(.0257) E = E o - ln (Q) RT nF Determine K c from E o by K c = e (nFE o /RT)

14. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)14 Primary (storage) Batteries Anode (-) Zn  Zn 2+ + 2e- Cathode (+) 2 NH 4 + + 2e-  2 NH 3 + H 2 Common dry cell (LeClanché Cell) Mercury Battery (calculators etc) Anode (-) Zn (s) + 2 OH - (aq)  ZnO (s) + 2H 2 O + 2e- Cathode (+) HgO (s) + H 2 O + 2e-  Hg (l) + 2 OH - (aq)

15. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)15 Secondary (rechargeable) Batteries Nickel-Cadmium 11_NiCd.mov 21m08an5.mov Anode (-) Cd + 2 OH - Cd(OH) 2 + 2e-  Cathode (+) NiO(OH) + H 2 O + e- Ni(OH) 2 + OH -  DISCHARGE RE-CHARGE

16. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)16 Secondary (rechargeable) Batteries (2) Lead Storage Battery 11_Pbacid.mov 21mo8an4.mov Con-proportionation reaction - same species produced at anode and cathode RECHARGEABLE Cathode (+) E o = +1.68 V PbO 2 (s) + HSO 4 - + 3 H + + 2e- PbSO 4 (s) + 2 H 2 O  Overall battery voltage = 6 x (0.36 + 1.68) = 12.24 V Anode (-) E o = +0.36 V Pb (s) + HSO 4 -  PbSO 4 (s) + H + + 2e-

17. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)17 Corrosion - an electrochemical reaction Electrochemical or redox reactions are tremendously damaging to modern society e.g. - rusting of cars, etc: anode: Fe -  Fe 2+ + 2 e- net: 2 Fe(s) + O 2 (g) + 2 H 2 O (l)  2 Fe(OH) 2 (s) Mechanisms for minimizing corrosion sacrificial anodes (cathodic protection) (e.g. Mg) coatings - e.g. galvanized steel - Zn layer forms (Zn(OH) 2.xZnCO 3 ) this is INERT (like Al 2 O 3 ); if breaks, Zn is sacrificial cathode: O 2 + 2 H 2 O + 4 e -  4 OH - E OX = +0.44 E RED = +0.40 E cell = +0.84

18. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)18 Electrolysis of Aqueous NaOH Anode : E o = -0.40 V 4 OH -  O 2 (g) + 2 H 2 O + 2e- Cathode : E o = -0.83 V 4 H 2 O + 4e-  2 H 2 + 4 OH - E o for cell = -1.23 V since E o 0 - not spontaneous ! - ONLY occurs if E external > 1.23 V is applied Electric Energy  Chemical Change 11_electrolysis.mov 21m10vd1.mov

19. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)19 ELECTROCHEMISTRY Chapter 21 Electric automobile redox reactions electrochemical cells construction electrode processes notation cell potential and  G o standard reduction potentials (E o ) non-equilibrium conditions (Q) batteries corrosion

20. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)20 Phosphorus and Sulfur Chemistry Kotz, Ch 22 the elements physical properties chemical reactions redox chemistry acid/base chemistry

21. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)21 Elemental Sulfur - Obtained from: - free element in volcanic vents ‘mined’ by Frasch process - minerals : FeS 2 (pyrite), PbS 2 (galena) Cu 2 S (chalcocite) (S produced as by-product of metal extraction) - natural gas and oil processing desulfurization: 2 H 2 S (g) + SO 2 (g)  3 S (s) + 2 H 2 O (g)

22. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)22 Elemental Phosphorus - not found free in nature - too easily oxidized “phosphate rock” Ca 3 (PO 4 ) 2 calcium phosphate Ca 5 (PO 4 ) 3 Ffluoro apatite Ca 5 (PO 4 ) 3 OHhydroxy apatite (teeth etc) Ca 5 (PO 4 ) 3 Clchloro apatite Isolate phosphorus from these ‘rocks’ by burning with charcoal and sand 2 Ca 3 (PO 4 ) 2 (l) + 6 SiO 2 (s)  P 4 O 10 (g) + 6 CaSiO 3 (l) P 4 O 10 (g) + 10 C (s)  P 4 + 10 CO (g)

23. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)23 Structure of P P 4 - white (or yellow) phosphorus (m.p. 44 o C) Allotropes : - different structural forms of the same element or compound OTHER EXAMPLES ?? C ( diamond, graphite, fullerene) P n - red or black phosphorus m.p. > 400 o C

24. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)24 Structure of S Solid sulfur : various solid state structures orthorhombic monoclinic plastic (amorphous) > 160 o C - very viscous - S n chains Liquid Sulfur: < 160 o C - free flowing - S 8 rings

25. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)25 Bonding in 3rd row versus 2nd row Gp V Gp VI Multiple bonding between two 3rd-row elements is uncommon due to their LARGER SIZE N2N2 O2O2 P4P4 S8S8

26. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)26 Chemistry of Sulfur Compounds SO 2 - STRONG, diprotic acid - 1st H fully ionized H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - - 2nd partially ionized HSO 4 - + H 2 O H 3 O + + SO 4 2-  S can have more than 8 electrons / 4 electron pairs expanded (>4) valence usually occurs with O, F or Cl SO 3 Molecular structure ? Lewis diagram ? angular, bent planar triangular Sulfuric Acid Oxides O=S=O... O=S=O.. O........

27. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)27 Reactions of Sulfuric Acid 1. Strong acid NaNO 3 + H 2 SO 4  HNO 3 + NaHSO 4 2. Dehydrating agent C 11 H 22 O 11 + H 2 SO 4  12 C + 11 H 3 O + 11 HSO 4 - 3. Strong oxidizing agent 2 Br - + 2 H 2 SO 4 (conc.)  2 Br 2 + SO 4 2- + SO 2 + 2H 2 O 4. Useful solvent : m.p. 10 o C b.p. 338 o C

28. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)28 Oxidation States of Sulfur and Phosphorus Both S and P have many oxidation states - and lots of redox chemistry -2H 2 S sulfide 0S 8 +2SCl 2 +4SF 4, H 2 SO 3 sulfurous SO 3 2- sulfite +6SF 6, H 2 SO 4 sulfuric SO 4 2- sulfate Sulfur O.N.e.g. name -3AlP phosphide 0P 4 +3PCl 3, H 3 PO 3 phosphorus PO 3 3- phosphite +5PF 5, H 3 PO 4 phosphoric PO 4 3- phosphate Phosphorus O.N.e.g. name

29. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)29 Redox chemistry of sulfur compounds Compounds in intermediate oxidation states S(2) or S(4) can act as both oxidizing and reducing agents SO 2 SO 2 (g) + Br 2 (aq) + 6 H 2 O  2 Br - (aq) + SO 4 2- (aq) + 4 H 3 O + (aq) can act as a reducing agent... and can act as an oxidizing agent: SO 2 (g) + 2 H 2 S (g)  3 S (s) + 2 H 2 O Water is both CATALYST and product ! - autocatalysis 5 SO 2 (g) + 2MnO 4 - (aq) + 6 H 2 O  5SO 4 2- (aq) + 2Mn 2+ (aq) + 4 H 3 O + (aq)

30. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)30 Chemistry of phosphorus compounds OXIDES P 4 + 3 O 2  P 4 O 6 P 4 + 5 O 2  P 4 O 10

31. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)31 Phosphoric acid P 4 O 10 + 6 H 2 O  4 H 3 PO 4 - phosphoric acid H 3 PO 4 is a weak tri-protic acid - even 1st H + not fully ionized HPO 4 2- (aq) + H 2 O H 3 O + (aq) + PO 4 3- (aq) phosphate   H 2 PO 4 - (aq) + H 2 O H 3 O + (aq) + HPO 4 2- (aq) hydrogen phosphate  H 3 PO 4 (aq) + H 2 O H 3 O + (aq) + H 2 PO 4 - (aq) dihydrogen phosphate 7.5x10 -3 6.2x10 -8 3.6x10 -13 K c (eq)

32. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)32 Phosphorus Chemistry (2) P 4 O 6 + 6 H 2 O  4 H 3 PO 3 - phosphorus acid H 3 PO 3 is a weak di-protic acid WHY ONLY 2 IONIZABLE hydrogens ? P (III) oxide and its acid are easily oxidized to P (V) so they act as REDUCING agents: Cu 2+ (aq) + H 3 PO 3 (aq) + 3 H 2 O  Cu (s) + H 3 PO 4 (aq) + 2H 3 O + - 2 e- The P-H bond is strong and non-polar - not ionizable

33. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)33 Phosphorus Chemistry (3) P 3- Phosphine. PH 3 - like NH 3 but weaker base Phosphide - ionic compounds with some metals 6 Ca + P 4  2 Ca 3 P 2 (Ca 2+ ) 3 ( P 3- ) 2 P 5+ Phosphoric acid, phosphate compounds Polyphosphates - condensation of hydroxy-acids X-O-H + H-O-X  X-O-X + H 2 O O OO O e.g. 2 H 3 PO 4  H-O-P-O-P-O-H + H 2 O di-phosphoric acid

34. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)34 Phosphorus Chemistry (4) Phosphate condensation/hydrolysis important in Biochemistry: + R [R-O-(PO 2 )-O-PO 3 ] 3- (aq) + H 2 O  [R-O-(PO 3 )] 2- (aq) + H 2 PO 4 - (aq) ATP 3- + H 2 O  AMP 2- + H 2 PO 4 - (aq) + H 2 O enzymes  G o = -30.5 kJ/mol Energy from - removal of e - -e - repulsion in reactant (ATP) - P-O bond converted to P=O bond - more resonance stabilization in products

35. 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)35 P and S Chemistry Kotz, Ch 22 Physical properties Chemical reactions redox chemistry acid/base chemistry