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24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)1 ELECTROCHEMISTRY Chapter 21 redox reactions electrochemical cells electrode processes.

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Presentation on theme: "24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)1 ELECTROCHEMISTRY Chapter 21 redox reactions electrochemical cells electrode processes."— Presentation transcript:

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 = x 10 4 J/Vmol n = the number of moles of electrons transferred. Michael Faraday 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 e-  Cu Oxidizing ability of ion Reducing ability of element 2 H + + 2e-  H Zn e-  Zn 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 Cl 2 /Cl O 2 /H 2 O In which direction does the following reaction go? Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) Hg 2+ /Hg Sn 2+ /Sn Al 3+ /Al Ag + /Ag Cu 2+ / Cu 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) = V reverse rxn: E o = (-0.80) = 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 = 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 = (0.0257) ln ( [Zn 2+ ]/[Cu 2+ ] ) 2 E = (-0.018) = 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 (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 e- Cathode (+) 2 NH e-  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 = V PbO 2 (s) + HSO H + + 2e- PbSO 4 (s) + 2 H 2 O  Overall battery voltage = 6 x ( ) = V Anode (-) E o = 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 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 H 2 O + 4 e -  4 OH - E OX = E RED = E cell = +0.84

18 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)18 Electrolysis of Aqueous NaOH Anode : E o = V 4 OH -  O 2 (g) + 2 H 2 O + 2e- Cathode : E o = V 4 H 2 O + 4e-  2 H OH - E o for cell = 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 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 nd partially ionized HSO 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 Strong oxidizing agent 2 Br H 2 SO 4 (conc.)  2 Br 2 + SO 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 O 2  P 4 O 6 P O 2  P 4 O 10

31 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)31 Phosphoric acid P 4 O 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 7.5x x x K c (eq)

32 24-Nov-97Electrochemistry (Ch. 21) & Phosphorus and Sulfur (ch 22)32 Phosphorus Chemistry (2) P 4 O 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 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 = 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


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