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Redox Reactions and Electrochemistry

Redox Reactions and Electrochemistry. Redox reactions Galvanic cells Standard reduction potential Cell potential Free energy Effect of concentration Battery Corrosion Electrolysis. 2Fe ( s ) + O 2 ( g ) 2FeO ( s ). 2Fe 2Fe 2+ + 4e -. O 2 + 4e - 2O 2-.

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Redox Reactions and Electrochemistry

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  1. Redox Reactions and Electrochemistry • Redox reactions • Galvanic cells • Standard reduction potential • Cell potential • Free energy • Effect of concentration • Battery • Corrosion • Electrolysis

  2. 2Fe (s) + O2 (g) 2FeO (s) 2Fe 2Fe2+ + 4e- O2 + 4e- 2O2- I. Oxidization–Reduction (Redox) Reaction • Electrochemical processes are redox reactions in which: • energy released by a spontaneous reaction is converted to electricity • electrical energy is used to cause a nonspontaneous reaction 0 0 2+ 2- Oxidation half-reaction (lose e-): Fe is oxidized and Fe is the reducing agent Reduction half-reaction (gain e-): O2 is reduced and O2 is the oxidizing agent

  3. A. Oxidation number/state • Free elements have an oxidation number (ON) =0 • Monatomic ions, the ON is equal to the charge on the ion. • The ON of oxygen isusually –2. In H2O2 and O22-:–1, in O2- : -1/2 • The ON of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Fluorine is always –1, Group IA metals are +1, IIA metals are +2 in compounds. • The sum of the ON of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. Ex. Oxidation numbers of all the atoms in HSO3– 1 +3x (-2)+S= -1 O =-2 H= +1 S=? S=4

  4. Cu+ H+ Cu2+ + H2 Sn2+ (aq) + Fe3+ (aq) Sn4+(aq) + Fe2+(aq) 2 2 Sn2+ (aq) Sn4+(aq) + 2 e– Fe3+ (aq) + e– Fe2+(aq) 2 Fe3+ (aq) + 2e– 2Fe2+(aq) B. Balance redox reaction By inspection 2 • Ion-electron method (half-reaction method) By balancing the number of electrons lost and gained during the reaction Oxidation half reaction Reduction half reaction ( ) x 2

  5. Fe2+ + Cr2O72- Fe3+ + Cr3+ +2 +3 Fe2+ Fe3+ +6 +3 Cr2O72- Cr3+ Cr2O72- 2Cr3+ Ion-electron method 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.

  6. Fe2+ Fe3+ + 1e- 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Cr2O72- 2Cr3+ + 7H2O Ion-electron method • 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.

  7. 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O Ion-electron method • 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 • For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

  8. Summary of balancing redox reaction in acidic solutions 1. Divide reaction into two incomplete half-reactions 2. Balance each half-reaction by doing the following: a. Balance all elements, except O and H b. Balance O by adding H2O c. Balance H by adding H+ d. Balance charge by adding e– as needed 3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to get a common multiple 4. The overall reaction is the sum of the half-reactions 5. The oxidation half-reaction is the one that produces electrons as products, and the reduction half-reaction is the one that uses electrons as reactants

  9. Ex. Balance I– + OCl– I2 + Cl–in basic solution? I– I2 2 I– I2 OCl– Cl– Add up 2 H2O+2 e– + OCl– Cl– + H2O + 2 OH– 2 e– + H2O+ OCl– Cl– + 2 OH– 2 I– + H2O+ OCl– I2 + Cl– + 2 OH– Summary of balancing redox reaction in basic solutions Steps 1-3 are the same as the reaction in acidic solution 4. Since we are under basic conditions, we neutralize any H+ ions in solution by adding an equal number of OH– to both sides 5. The overall reaction is the sum of the half-reactions Oxidation +2 e– Reduction 2 OH– + 2 H+ + 2e– + + H2O + 2 OH–

  10. II. Galvanic Cells e– flow Wire e– Oxidation Zn --> Zn2+ + 2e– Anode Flow of cation Reduction Cu2+ +2 e–--> Cu Cathode Flow of anion Salt bridge • Consider reaction Zn(s) + Cu(NO3)2 --> Cu(s) + Zn(NO3)2 Spontaneous (activity series, chapter 4) half reaction: Zn(s) --> Zn2+ + 2e– Cu2+ +2 e–--> Cu(s) • Galvanic cell: a device uses the spontaneous redox reaction to produce an electric current Electrode: anode (oxidation rxn) and cathode (reduction rxn) Salt bridge: contains conc. soln of strong electrolyte (KCl)

  11. II. Galvanic Cells anode oxidation cathode reduction

  12. II. Galvanic cell anode cathode Zn (s) + Cu2+(aq) Cu (s) + Zn2+(aq) Ex. Draw cell diagram for the following reaction: Al (s) + Ag+ (aq) Al3+ (s) + Ag (s) Cell diagram : a conventional notation to represent galvanic cell [Cu2+] = 1 M and [Zn2+] = 1 M Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) Anode is on the left, cathode on the right | : represent boundary between phases || : represent salt bridge Al (s) | Al3+|| Ag+ | Ag (s)

  13. III. Standard reduction potential 2e- + 2H+ (1 M) H2 (1 atm) Zn (s) Zn2+ (1 M) + 2e- Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm) A. Cell voltage, cell potential, electromotive force (emf) • The potential energy of the e– is higher at the anode than at the cathode • The difference in electrical potential between the anode and cathode is called as emf • Unit: V (volt) = 1 J/C C: coulomb, unit of charge 4. Standard cell potential, Eocell Cell potential under standard condition Zn(s)| Zn2+(1 M) || H+ (1 M)| H2 (1 atm)| Pt (s) Anode (oxidation): Cathode (reduction):

  14. B. Standard reduction potential 2e- + 2H+ (1 M) H2 (1 atm) 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 E0= 0 V Standard hydrogen electrode (SHE) : half-cell contain 1M H+ and 1 atm H2, used as a reference to determine Eo for other species

  15. B. Standard reduction potential E0 = EH /H - EZn /Zn E0 = Ecathode - Eanode E0 = 0.76 V cell cell 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 emf (E0cell ) Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

  16. B. Standard reduction potential 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 = Ecathode - Eanode E0 = 0.34 V 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) Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): Cathode (reduction):

  17. 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

  18. Ex. 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. Ex. 6 H+(aq) + 6e- 3 H2(g) H2O2+2 H+ +2e- 2 H2O Cu(s) Cu2+ + 2e- 2 Al(s) 2 Al3+ + 6e- E0 = E - ECu /Cu E0 = EH /H - EAl /Al E0 = Ecathode - Eanode cell cell cell 0 0 0 0 0 3+ 2+ + 2 H2O2/H2 O Calculate Eocellfor the following reactions: (a) 2 Al (s) + 6 H+-> 2 Al3+ + 3 H2 (g) (b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O (a) 2 Al (s) + 6 H+-> 2 Al3+ + 3 H2 (g) Anode (oxidation): Cathode (reduction): = 1.66 V (b) H2O2 + 2 H+ + Cu (s) -> Cu2+ + 2 H2O Anode (oxidation): Cathode (reduction): = 1.77 V- 0.34 V = 1.43 V

  20. IV. Cell Potential 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 = 0.0592 V 0.0257 V log K ln K n (96,500 J/V•mol) = n n = = • Free energy : spontaneity of redox reactions DG = -nFEcell n = number of moles of electrons in reaction = 96,500 C/mol DG0 = -RT ln K

  21. A. Free energy and spontaneity of redox reaction DG0 = -nFEcell 0 0 0 Ecell Ecell 0.0257 V 0.0592 V ln K log K n n = = DG = -nFEcell

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

  23. B. Concentration effect on cell potential DG0 = -nFE 0 RT nF E = E0 - ln Q 0 0 E = E = E E 0.0257 V 0.0592 V ln Q log Q n n - - DG = DG0 + RT ln Q DG = -nFE -nFE = -nFE0+ RT ln Q Nernst equation At 298 K

  24. Ex. 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+ Fe 0 0 2+ 2+ E0 = EFe /Fe – ECd /Cd = -0.44 – (-0.40) V Q = = -0.04 V E = -0.04 V [Cd2+] 0.010 [Fe2+] 0.60 0 E = E 0.0257 V 0.0257 V ln Q ln n 2 - - Oxidation: n = 2 Reduction: E = 0.013 E > 0 Spontaneous

  25. Ex. Ni Ni2+ + 2e- 2e- +2 Ag+ 2 Ag Ni + 2 Ag+ Ni2+ + 2 Ag 0 0 + 2+ = 0.80 – (-0.25) V E0 = EAg /Ag – E Ni /Ni Q = = 1.05V E = 1.05 V [Ni2+] 0.050 [Ag+]2 5.02 0 E = E 0.0257 V 0.0257 V ln Q ln n 2 - - A redox reaction for the oxidation of nickel by silver ion is set up with an initial concentration of 5.0 M silver ion and 0.050 M nickel ion. What is the cell EMF at 298K? Oxidation: n = 2 Reduction: E = 1.13

  26. 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) V. Battery Dry cell Leclanché cell About 1.5 V Anode: Cathode:

  27. 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) V. Battery Mercury Battery 1.35 V Anode: Cathode:

  28. 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 V. Battery Lead storage battery Total 12V Anode: Cathode:

  29. VI. Corrosion • Deterioration of metal by an electrochemical process

  30. VI. Corrosion • Cathodic Protection of an Iron Storage Tank

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

  32. A. Electrolysis of water

  33. B. Quantitative aspect of electrolysis charge (C) = current (A) x time (s) 1 mole e- = 96,500 C Ex. It will take the greates amount electricity to produce 2 mol of the ____ metal by electrolysis. (1) potassium (2) calcium (3) aluminum (4) silver

  34. Ex. 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) 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? Anode: Cathode: = 0.0126 mol Ca = 0.50 g Ca

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