1. Electrochemistry Applications of Redox

2. Review l Oxidation reduction reactions involve a transfer of electrons. l OIL- RIG l Oxidation Involves Loss l Reduction Involves Gain

3. Solid lead(II) sulfide reacts with oxygen in the air at high temperatures to form lead(II) oxide and sulfur dioxide. Which substance is a reductant (reducing agent) and which is an oxidant (oxidizing agent)? A.PbS, reductant; O 2, oxidant B.PbS, reductant; SO 2, oxidant C.Pb 2+, reductant; S 2- oxidant D.PbS, reductant; no oxidant E.PbS, oxidant; SO 2, reductant

4. Applications l Moving electrons is electric current. 8H + +MnO 4 - + 5Fe +2 +5e -  Mn +2 + 5Fe +3 +4H 2 O l Helps to break the reactions into half reactions. 8H + +MnO 4 - +5e -  Mn +2 +4H 2 O 5(Fe +2  Fe +3 + e - ) l In the same mixture it happens without doing useful work, but if separate

5. H + MnO 4 - Fe +2 l Connected this way the reaction starts l Stops immediately because charge builds up. e-e- e-e- e-e- e-e- e-e-

6. H + MnO 4 - Fe +2 Galvanic Cell Salt Bridge allows current to flow

7. H + MnO 4 - Fe +2 e-e- l Electricity travels in a complete circuit

8. H + MnO 4 - Fe +2 Porous Disk l Instead of a salt bridge

9. Reducing Agent Oxidizing Agent e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode

10. Cell Potential l Oxidizing agent pulls the electron. l Reducing agent pushes the electron. The push or pull (“driving force”) is called the cell potential E cell l Also called the electromotive force (emf) l Unit is the volt(V) l = 1 joule of work/coulomb of charge l Measured with a voltmeter

11. Zn +2 SO 4 -2 1 M HCl Anode 0.76 1 M ZnSO 4 H + Cl - H 2 in Cathode

12. 1 M HCl H + Cl - H 2 in Standard Hydrogen Electrode l This is the reference all other oxidations are compared to E º = 0 l º indicates standard states of 25ºC, 1 atm, 1 M solutions.

13. Cell Potential Zn(s) + Cu +2 (aq)  Zn +2 (aq) + Cu(s) l The total cell potential is the sum of the potential at each electrode. E º cell = E º Zn  Zn +2 + E º Cu +2  Cu l We can look up reduction potentials in a table. l One of the reactions must be reversed, so change it sign.

14. Cell Potential l Determine the cell potential for a galvanic cell based on the redox reaction. Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq) Fe +3 (aq) + e -  Fe +2 (aq) E º = 0.77 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V Cu(s)  Cu +2 (aq)+2e - E º = -0.34 V 2Fe +3 (aq) + 2e -  2Fe +2 (aq) E º = 0.77 V

15. Reduction potential More negative E º – more easily electron is added – More easily reduced – Better oxidizing agent More positive E º – more easily electron is lost – More easily oxidized – Better reducing agent

16. Line Notation solid  Aqueous  Aqueous  solid Anode on the left  Cathode on the right l Single line different phases. l Double line porous disk or salt bridge. l If all the substances on one side are aqueous, a platinum electrode is indicated.

17. Cu 2+ Fe +2 l For the last reaction Cu(s)  Cu +2 (aq)  Fe +2 (aq),Fe +3 (aq)  Pt(s)

18. Under standard conditions, which of the following is the net reaction that occurs in the cell? Cd|Cd 2+ || Cu 2+ |Cu a. Cu 2+ + Cd → Cu + Cd 2+ b. Cu + Cd → Cu 2+ + Cd 2+ c. Cu 2+ + Cd 2+ → Cu + Cd d. Cu + Cd 2+ → Cd + Cu 2+

19. Galvanic Cell l The reaction always runs spontaneously in the direction that produced a positive cell potential. l Four things for a complete description. 1)Cell Potential 2)Direction of flow 3)Designation of anode and cathode 4)Nature of all the components- electrodes and ions

20. Potential, Work and  G l emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q l Charge is measured in coulombs. -w = q E l Faraday = 96,485 C/mol e - l q = nF = moles of e - x charge/mole e - w = -q E = -nF E =  G

21. Potential, Work and  G  Gº = -nF E º if E º > 0, then  Gº < 0 spontaneous if E º 0 nonspontaneous l In fact, reverse is spontaneous. Calculate  Gº for the following reaction: Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq) Fe +2 (aq) + e -  Fe(s) E º = 0.44 V Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V

22. Cell Potential and Concentration Qualitatively - Can predict direction of change in E from LeChâtelier. 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) Predict if E cell will be greater or less than E º cell if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M l if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M l if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M

23. The Nernst Equation  G =  Gº +RTln(Q) -nF E = -nF E º + RTln(Q) E = E º - RT ln(Q) nF 2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E º = 0.48 V l Always have to figure out n by balancing. l If concentration can gives voltage, then from voltage we can tell concentration.

24. The Nernst Equation l As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and E cell = 0 0 = E º - RTln(K) nF E º = RT ln(K) nF nF Eº = ln(K) RT

25. Batteries are Galvanic Cells l Car batteries are lead storage batteries. Pb +PbO 2 +H 2 SO 4  PbSO 4 (s) +H 2 O

26. Batteries are Galvanic Cells Dry Cell Zn + NH 4 + +MnO 2  Zn +2 + NH 3 + H 2 O + Mn 2 O 3

27. Batteries are Galvanic Cells Alkaline Zn +MnO 2  ZnO+ Mn 2 O 3 (in base)

28. Batteries are Galvanic Cells l NiCad NiO 2 + Cd + 2H 2 O  Cd(OH) 2 +Ni(OH) 2

29. Corrosion l Rusting - spontaneous oxidation. l Most structural metals have reduction potentials that are less positive than O 2. Fe  Fe +2 +2e - E º= 0.44 V O 2 + 2H 2 O + 4e -  4OH - E º= 0.40 V Fe +2 + O 2 + H 2 O  Fe 2 O 3 + H + l Reactions happens in two places.

30. Water Rust Iron Dissolves- Fe  Fe +2 e-e- Salt speeds up process by increasing conductivity O 2 + 2H 2 O +4e -  4OH - Fe 2+ + O 2 + 2H 2 O  Fe 2 O 3 + 8 H + Fe 2+

31. Preventing Corrosion l Coating to keep out air and water. l Galvanizing - Putting on a zinc coat l Has a lower reduction potential, so it is more easily oxidized. l Alloying with metals that form oxide coats. l Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

32. l Running a galvanic cell backwards. l Put a voltage bigger than the potential and reverse the direction of the redox reaction. l Used for electroplating. Electrolysis

33. 1.0 M Zn +2 e-e- e-e- Anode Cathode 1.10 Zn Cu 1.0 M Cu +2

34. 1.0 M Zn +2 e-e- e-e- Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2

35. Calculating plating l Have to count charge. Measure current I (in amperes) l 1 amp = 1 coulomb of charge per second q = I x t l q/nF = moles of metal l Mass of plated metal l How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag +

36. Calculating plating 1.Current x time = charge 2.Charge ∕Faraday = mole of e - 3.Mol of e - to mole of element or compound 4.Mole to grams of compound Or the reverse if you want time to plate

37. Calculate the mass of copper which can be deposited by the passage of 12.0 A for 25.0 min through a solution of copper(II) sulfate.

38. How long would it take to plate 5.00 g Fe from an aqueous solution of Fe(NO 3 ) 3 at a current of 2.00 A?

39. Other uses l Electrolysis of water. l Separating mixtures of ions. l More positive reduction potential means the reaction proceeds forward. l We want the reverse. l Most negative reduction potential is easiest to plate out of solution.

40. Redox Know the table 2. Recognized by change in oxidation state. 3. “Added acid” 4. Use the reduction potential table on the front cover. 5. Redox can replace. (single replacement)

41. 6. Combination Oxidizing agent of one element will react with the reducing agent of the same element to produce the free element. I - + IO 3 - + H +  I 2 + H 2 O 7. Decomposition. a) peroxides to oxides b) Chlorates to chlorides c) Electrolysis into elements. d) carbonates to oxides

42. A way to remember l An Ox – anode is where oxidation occurs l Red Cat – Reduction occurs at cathode l Galvanic cell- spontaneous- anode is negative l Electrolytic cell- voltage applied to make anode positive