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Electrochemistry

Electrochemistry. Chapter 3. 2Mg ( s ) + O 2 ( g ) 2MgO ( s ). 2Mg 2Mg 2+ + 4e -. O 2 + 4e - 2O 2-. Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or

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Electrochemistry

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  1. Electrochemistry Chapter 3

  2. 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- • Electrochemical processes are oxidation-reduction reactions in which: • the energy released by a spontaneous reaction is converted to electricity or • electrical energy is used to cause a nonspontaneous reaction to occur 0 0 2+ 2- Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-) 19.1

  3. - Oxidation is the loss of electrons and reduction is the gain of electrons. One can’t occur without the other, but we frequently break them up into the two half reactions. Example: Zno + Cu+2 -----) Zn+2 + Cuo Oxidn: Zno -----) Zn+2 + 2 e- Redn: Cu+2 + 2 e------) Cuo a) Oxidizing Agent causes oxidn. & is reduced (Cu2+) b) Reducing Agentcauses redn. &is oxidized (Zno) (Know Terms & Definitions) - Classes of Redox Reactions: Combination, Decomposition, Displacement, Combustion

  4. A Summary of an Oxidation-Reduction Process

  5. 3CH3CH2OH + 2K2Cr2O7 + 8H2SO4 Chemistry in Action: Breath Analyzer +6 +3 3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O 4.4

  6. Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. In H2O2 and O22- it is –1. 4.4

  7. Oxidation numbers of all the atoms in HCO3- ? • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1 (H -1 when combined with Ia, IIa, IIIa). • Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO3- O = -2 H = +1 3x(-2) + 1 + ? = -1 C = +4 4.4

  8. Examples: Determine the ON of singleN in each: N2 N2O NO NF3 N2O4 NO3-1

  9. - In any redox reaction you should be able to determine: The ON for each element Which reagent has been oxidized Which reagent has been reduced Which reagent is the oxidizing agent Which reagent is the reducing agent What are the two half reactions for each How many moles of e- are transferred in each half rxn

  10. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ Balancing Redox Equations The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? • Write the unbalanced equation for the reaction ion ionic form. • Separate the equation into two half-reactions. Oxidation: Reduction: • Balance the atoms other than O and H in each half-reaction. 19.1

  11. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. • Add electrons to one side of each half-reaction to balance the charges on the half-reaction. • If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 19.1

  12. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O Balancing Redox Equations • Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: Reduction: • Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 19.1

  13. For reactions in basic solutions: • Do all previous steps from 1 to 6 and: *Add OH- equal to # H+to both sides. ** Convert H+ & OH-on same side to H2O& simplify by canceling H2O if needed. *** Check result by noting if following are both balanced: elements & charge.

  14. Examples (Acidic)Zn + NO3- + H+ -----) Zn+2 + NH4+ + H2O 1)Zn -----) Zn2+0 ---) +2 on Zn (Oxidn.) NO3- -----) NH4++5 ---) -3 on N (Redn.) 2) Zn -----) Zn2+ + 2 e-(balance all but H&O) NO3- -----) NH4+ + 3 H2O (balance O with H2O) 10 H+ + NO3- -----) NH4+ + 3 H2O (balance H with H+) 10 H+ + NO3- + 8 e------) NH4+ + 3 H2O (bal. Charge w e-) 3) 4[Zn -----) Zn2+ + 2 e-](8e- lost) 1 [10 H+ + NO3- + 8 e------) NH4+ + 3 H2O](8e- gained) 4 Zn + 10 H+ + NO3- + 8 e----) 4 Zn2+ + 8 e- + NH4+ + 3 H2O 4 Zn + 10 H+ + NO3- -----) 4 Zn2+ + NH4+ + 3 H2O

  15. Example MnO4- + S2O32- -----) Mn2+ + SO42- (acidic) 1)a)MnO4- ----) Mn2+ [+7 ---) +2 Redn] b)S2O32- ----) SO42- [+2 ---) +6 per S Oxn] 2)a)5e- + 8H+ + MnO4- ----) Mn2+ + 4H2O (x 8) b)1S2O32- -----) 2SO42- 5H2O + S2O32- -----) 2SO42- + 10H+ 5H2O + S2O32- -----) 2SO42- + 10H+ + 8e-(x 5) 3) 40e- + 64H+ + 8MnO4- + 25H2O + 5S2O32- ---) 8Mn2+ + 32H2O +10SO42- + 50H+ + 40e- 14 H+ + 8 MnO4- + 5 S2O32-----) 8 Mn2+ + 7 H2O + 10 SO42-

  16. Example (Basic)MnO4- + I- -----) MnO2 + IO3- 1) MnO4- -----) MnO2 +7 ---) +4 Mn I- -----) IO3- -1 ----) +5 I 2) 4H+ + MnO4- + 3e- -----) MnO2 + 2H2O 3H2O + I- -----) IO3- + 6H+ + 6e- 3) 2 [4H+ + MnO4- + 3e- -----) MnO2 + 2H2O] 1 [3H2O + I- -----) IO3- + 6H+ + 6e-] 8H+ + 2MnO4- + 6e- + 3H2O + I- ----) 2MnO2 + 4H2O + IO3- + 6H+ + 6e- 2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3- 4) 2OH- + 2H+ + 2MnO4- + I- ----) 2MnO2 + H2O + IO3-+ 2OH- 5) 2H2O + 2MnO4- + I- ----) 2MnO2 + H2O + IO3-+ 2OH- H2O + 2MnO4- + I- ----) 2MnO2 + IO3-+ 2OH- 6) Double check balancing of atoms & charge.

  17. Voltaic Cells A. Introduction - A voltaic cell (battery) is an electrochemical cell in which a spontaneous redox reaction generates an electric current. - It consists of two half cells which are connected: Oxidation half cell consists of electrode (anode = - ) & electrolyte. Reduction half cell consists of electrode (cathode = + ) & electrolyte. External connection of anode to the cathode. Internal connection of a salt bridge that allows the flow of ions but prevents mixing of electrolytes from two half cells. Electrons travel from anode to cathode.

  18. Galvanic Cells anode oxidation cathode reduction spontaneous redox reaction 19.2

  19. Note the following Conventions: Zn + Cu2+ -----) Zn2+ + Cu Cell Diagram Zn | Zn2+ (1.0 M) || Cu2+(1.0 M) | Cu Anode (-) (oxidation) Zn ---) Zn2+ + 2e- Cathode(+) (reduction) Cu2+ + 2e- ---) Cu Anode written on left Cathode written on right | = phase boundary || = salt bridge Zn = anode electrode Cu = cathode electrode , = separates ions in the same solution M = Molar concentration of ions

  20. Zn (s) + Cu2+(aq) Cu (s) + Zn2+(aq) Galvanic Cells • The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential Cell Diagram [Cu2+] = 1 M & [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) anode cathode 19.2

  21. 2e- + 2H+ (1 M) H2 (1 atm) Standard Reduction Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction EH +/H2 E0= 0 V Standard hydrogen electrode (SHE) 19.3

  22. 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) Zn2+ (1 M) + 2e- Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Cathode (reduction): 19.3

  23. E0 = EH /H - EZn /Zn E0 = 0.76 V E0 = Ecathode - Eanode cell cell cell Standard emf (E0 ) cell 0 0 0 0 2+ + 2 0.76 V = 0 - EZn /Zn 0 2+ EZn /Zn = -0.76 V 0 2+ Zn2+ (1 M) + 2e- ZnE0 = -0.76 V Standard Reduction Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) 19.3

  24. 0 0 0 Ecell = ECu /Cu – EH /H 2+ + 2 0.34 = ECu /Cu - 0 0 2+ 0 ECu /Cu = 0.34 V 2+ E0 = 0.34 V E0 = Ecathode - Eanode cell cell 0 0 H2 (1 atm) 2H+ (1 M) + 2e- 2e- + Cu2+ (1 M) Cu (s) H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M) Standard Reduction Potentials Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): Cathode (reduction): 19.3

  25. E0 is for the reaction as written • The more positive E0 the greater the tendency for the substance to be reduced • The half-cell reactions are reversible • The sign of E0changes when the reaction is reversed • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3

  26. Cd2+(aq) + 2e- Cd (s)E0 = -0.40 V Cr3+(aq) + 3e- Cr (s)E0 = -0.74 V Cr (s) Cr3+ (1 M) + 3e- E0 = 0.34 V E0 = -0.40 – (-0.74) E0 = Ecathode - Eanode cell cell cell 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) 0 0 2e- + Cd2+ (1 M) Cd (s) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd is the stronger oxidizer Cd will oxidize Cr x 2 Anode (oxidation): Cathode (reduction): x 3 19.3

  27. DG0 = -nFEcell 0 0 0 0 0 = -nFEcell Ecell Ecell Ecell F = 96,500 J RT V • mol ln K nF (8.314 J/K•mol)(298 K) ln K = 0.0257 V 0.0592 V ln K log K n (96,500 J/V•mol) = n n = = Spontaneity of Redox Reactions DG = -nFEcell n = number of moles of electrons in reaction = 96,500 C/mol DG0 = -RT ln K 19.4

  28. 0 = -nFEcell Spontaneity of Redox Reactions DG0 = -RT ln K 19.4

  29. 2Ag 2Ag+ + 2e- 0 Ecell 2e- + Fe2+ Fe 0 0 2+ + E0 = EFe /Fe – EAg /Ag E0 = -0.44 – (0.80) What is the equilibrium constant for the following reaction at 250C? Fe2+(aq) + 2Ag (s) Fe (s) + 2Ag+(aq) E0 = -1.24 V E0 cell x n = exp K = 0.0257 V 0.0257 V -1.24 V x2 0.0257 V ln K exp n = Oxidation: n = 2 Reduction: K = 1.23 x 10-42 19.4

  30. DG0 = -nFE 0 RT nF E = E0 - ln Q 0 0 E = E = E E 0.0257 V 0.0592 V log Q ln Q n n - - The Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFE -nFE = -nFE0+ RT ln Q Nernst equation At 298 19.5

  31. Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+(aq) + Cd (s) Fe (s) + Cd2+(aq) Cd Cd2+ + 2e- 2e- + Fe2+ 2Fe 0 0 2+ 2+ E0 = -0.44 – (-0.40) E0 = EFe /Fe – ECd /Cd E = -0.04 V E0 = -0.04 V 0.010 0.60 0 E = E 0.0257 V 0.0257 V ln ln Q 2 n - - Oxidation: n = 2 Reduction: E = 0.013 E > 0 Spontaneous 19.5

  32. Zn (s) Zn2+ (aq) + 2e- + 2NH4(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O (l) Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s) Batteries Dry cell Leclanché cell Anode: Cathode: 19.6

  33. Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e- HgO (s) + H2O (l) + 2e- Hg (l) + 2OH-(aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Batteries Mercury Battery Anode: Cathode: 19.6

  34. Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4 PbO2(s) + 4H+(aq) + SO2-(aq) + 2e- PbSO4(s) + 2H2O (l) 4 Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l) 4 Batteries Lead storage battery Anode: Cathode: 19.6

  35. Batteries Solid State Lithium Battery 19.6

  36. 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e- O2(g) + 2H2O (l) + 4e- 4OH-(aq) 2H2 (g) + O2 (g) 2H2O (l) Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: 19.6

  37. CH3COO- + 2O2 + H+ 2CO2 + 2H2O Chemistry In Action: Bacteria Power

  38. Corrosion 19.7

  39. Cathodic Protection of an Iron Storage Tank 19.7

  40. Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. 19.8

  41. Electrolysis of Water 19.8

  42. Electroplating Plating means depositing the neutral metal on the electrode by reducing the metal ions in solution Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e- = 96,500 C The quantity of electricity carried by 1 mole of electrons is called a faraday, which is equal to 96,500 C 19.8

  43. How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? 2 mole e- = 1 mole Ca mol Ca = 0.452 x 1.5 hr x 3600 C s 2Cl- (l) Cl2 (g) + 2e- hr s Ca2+(l) + 2e- Ca (s) 1 mol Ca 1 mol e- x x 96,500 C 2 mol e- Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) Anode: Cathode: = 0.0126 mol Ca = 0.50 g Ca 19.8

  44. 2+ 2+ 2+ Hg2 /Ag2Hg3 0.85 V Sn /Ag3Sn -0.05 V Sn /Ag3Sn -0.05 V Chemistry In Action: Dental Filling Discomfort

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