Unit 12 (Chp 20): Electrochemistry (E, ∆G, K) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 12 (Chp 20): Electrochemistry (E, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Electrochemical Reactions electrons are transferred from one species to another (redox). assign oxidation numbers to identify which loses e– and which gains e– .

Oxidation Numbers .

Oxidation and Reduction A species is oxidized when it loses e– . Zn loses 2 e– to from Zn metal to the Zn2+ ion. A species is reduced when it gains e– . H+ ions gain 1 e– and combine to form H2 . Video Clips: OxidationReduction1 and OxidationReduction2 from Brown text resources Chp 20 emedia_Library LEO says GER 2 video clips

Assigning Oxidation Numbers All elements are 0. (all compounds are 0) Monatomic ion is its charge. (Ex. Na+ ion) Most nonmetals tend to be negative, but some are positive in certain compounds or ions. (Ex. SO3) O is −2 (but in peroxide ion is −1 [ O2–2 ] H is +1 with nonmetals (but −1 with metals) F is −1 (always) other halogens usually −1, but are positive with O Ex. ClO3– or NO3– or SO42– HW p. 890 #11,16ab

Balancing Redox Reactions by… Half-Reactions “O, he balance you.” O H E BALANCE U 5 Steps: 1 2 3 4 5

Balance RedOx by Half-Rxns 5 Steps: Balance RedOx by Half-Rxns 1: Ox #’s +3 +2 Zn + Fe+3  Zn+2 + Fe comp–diss–cross –net– balanced? 2: Half rxns RED: Fe+3  Fe 2 ( ) 3 e− + OX: Zn  Zn+2 3 ( ) + 2 e− 3: Electrons RED: 6 e− + 2 Fe+3  2 Fe 4: Bal. same e–’s OX: 3 Zn  3 Zn+2 + 6 e− 5: Unite 3 Zn + 2 Fe+3  3 Zn+2 + 2 Fe (Balanced Overall)

Readily Oxidized or Reduced? WS Aq. Soln’s & Chm Rxns II Which region and group in the periodic table shown contains elements that are: most readily oxidized? most readily reduced? metals least readily oxidized? A 1 (alkali) D 17 (halogens) C Ag, Au, Pt, Hg

WS Aq. Soln’s & Chm Rxns II #6 5 Steps: WS Aq. Soln’s & Chm Rxns II #6 +2 +2 1: Ox #’s Fe + Pb(NO3)2  Fe(NO3)2 + Pb (NO3– is a spectator ion) 2: Half rxns RED: Pb+2  Pb 2 e− + OX: Fe  Fe+2 + 2 e− 3: Electrons RED: 2 e− + Pb+2  Pb 4: Bal. same e–’s OX: Fe  Fe+2 + 2 e− 5: Unite Fe + Pb+2  Fe+2 + Pb (Balanced Overall)

WS Aq. Soln’s & Chm Rxns II #10 5 Steps: WS Aq. Soln’s & Chm Rxns II #10 1: Ox #’s +1 –1 +1 –1 Cl2 + KI  KCl + I2 (K+ is a spectator ion) 2: Half rxns RED: Cl2  Cl– 2 e− + 2 OX: I–  I2 2 + 2 e− 3: Electrons RED: 2 e− + Cl2  2 Cl– 4: Bal. same e–’s OX: 2 I–  I2 + 2 e− 5: Unite Cl2 + 2 I–  2 Cl– + I2 (Balanced Overall)

WS Aq. Soln’s & Chm Rxns II #13 5 Steps: WS Aq. Soln’s & Chm Rxns II #13 +1 –2 +2 –2 +1 1: Ox #’s Ca + H2O  Ca(OH)2 + H2 2: Half rxns RED: H2O  OH– + H2 2 e− + 2 2 OX: Ca  Ca+2 + 2 e− 3: Electrons RED: 2 e− + 2 H2O  2 OH– + H2 4: Bal. same e–’s OX: Ca  Ca+2 + 2 e− 5: Unite Ca + 2 H2O  Ca+2 + 2 OH– + H2 (Balanced Overall)

WS Aq. Soln’s & Chm Rxns II #23 5 Steps: WS Aq. Soln’s & Chm Rxns II #23 +1 +2 1: Ox #’s Zn + H2SO4  ZnSO4 + H2 (SO4–2 is a spectator ion) 2: Half rxns RED: H+  H2 2 e− + 2 OX: Zn  Zn+2 + 2 e− 3: Electrons RED: 2 e− + 2 H+  2 H2 4: Bal. same e–’s OX: Zn  Zn+2 + 2 e− 5: Unite Zn + 2 H+  Zn+2 + H2 (Balanced Overall)

mass % (gsample /gtotal) Redox Titration The analytical technique used to calculate the moles in an unknown soln. buret molarity (mol/L) molar mass (g/mol) mass % (gsample /gtotal) titrant mol X = mol Y known vol. (V) known conc. (M) MXVX = MYVY x y analyte known vol. (V) unknown conc. (M) xX + yY  X– + Y+ (red) (ox) (or moles)

Redox Titration Consider a titration of FeCl2 with KMnO4 : 2 MnO4− + 5 H2O2 + 6 H+  2 Mn2+ + 5 O2 + 8 H2O +7 –1 +2 colorless purple excess MnO4– titrant limitedMnO4– titrant Why is it colorless? Why is it purple?

Redox Titration MXVX = MYVY x y mol X = mol Y equivalence point,(Veq): equal stoichiometric amounts react completely MXVX = MYVY x y 2 MnO4– + 5 H2O2  MXVX = MYVY 2 5 mol X = mol Y end point: permanently changes color

Redox Titration MXVX = MYVY Titration of H2O2 with MnO4– : 16.8 mL HW p. 163 #103 Titration of H2O2 with MnO4– : 2 MnO4− + 5 H2O2 + 6 H+  2 Mn2+ + 5 O2 + 8 H2O +7 –1 +2 16.8 mL 0.124 M 10.0 mL ? M mol X = mol Y MXVX = MYVY x y (0.0168 L)(0.124 M) = MY(0.0100 L) 2 5 MY = 0.521 M H2O2

Voltaic Cells favorable redox reactions transfer e–’s release free energy (–ΔG) Zn + Cu+2  Zn2+ + Cu Zn2+ 2+ 2+ 2+ 2+ Cu 2+ 2+ Zn Zn 2+ 2+ 2+ 2+ Cu2+ Cu2+

Voltaic Cells the energy can be used to do work if the electrons flow through an external device. We call this a voltaic cell. (or galvanic cell, or electrochemical cell)

_________ at the cathode Oxidation ________ at the anode _________ at the cathode Reduction Voltaic Cells RED CAT AN OX Zn  Zn+2 + 2 e− 2 e− + Cu+2  Cu

salt bridge maintains charge balance Voltaic Cells As 1 e– flows from AN to CAT, the charges in each half-cell would be unbalanced and e– flow stopped. (Anions to anode) (Cations to cathode) RED CAT AN OX salt bridge maintains charge balance

Voltaic Cells +cations form and dissolve at AN HW p. 891 #22, 24 +cations form and dissolve at AN (e– ’s flow from AN to CAT) e– ’s reduce +cations to deposit solid metal on the CAT animation: VoltaicCopperZincCell Zn(s)  Zn+2 + 2e− 2e− + Cu+2  Cu(s) Zn Zn 2+ Zn2+ 2+ Cu2+ Cu2+ 2+ Cu 2+ Cu

Electric Potential Energy Water only flows favorably in one direction. (battery) e–’s only favorably flow from: higher to lower potential energy. e– flow

Standard Reduction Potentials (Ered)  SRP’s measured, tabulated. (likely reduced)  defined as Ered = (SHE) (NOT likely reduced) (oxidized)

Cell Potential (E) OX: RED: Zn  Zn2+ + 2 e– Cu2+ + 2 e–  Cu 1.10 V HOW?

BUT… Cell Potential (E) Ered = ? RED: Cu2+ + 2 e–  Cu OX: Zn  Zn2+ + 2 e– Eox = ? Overall: Zn + Cu+2  Zn2+ + Cu Ecell = 1.10 V E = Ecathode - Eanode E = Ered + Eox or BUT…

E =Ered of cathode – Ered of anode Standard Reduction Potentials Cell Potential (E) o Ered = ? RED: Cu2+ + 2 e–  Cu o OX: Zn  Zn2+ + 2 e– Eox = ? o Overall: Zn + Cu+2  Zn2+ + Cu Ecell = 1.10 V o o o o o o E = Ered + Eox E =Ered of cathode – Ered of anode Standard Cell Potential (E ) from Standard Reduction Potentials (Ered or SRP’s) o … o

Standard Hydrogen Electrode (SHE)  2 H+(aq, 1M) + 2 e−  H2(g, 1 atm) Ered = 0 V (defined) 1 atm H2(g) or Zn Cu SRP’s are measured against a SHE (0.00 V)

Standard Cell Potential (Eo) Ered = ? RED: Cu2+ + 2 e–  Cu OX: Zn  Zn2+ + 2 e– Eox = ? Overall: Zn + Cu+2  Zn2+ + Cu Ecell = 1.10 V Ecell = Ered + Eox Ecell = Ecathode - Eanode  RED: Cu2+ + 2 e–  Cu Ered = +0.34 V  OX: Zn  Zn2+ + 2 e– Eox = ? +0.76 V  Ered = –0.76 V Ecell = Ered(0.34) + Eox(????)    +0.76 Ecell = +1.10 V (–Ered)  HOW? RED OX Cu2+ Zn

Standard Cell Potential (Eo)  Ecell under standard conditions (1.0 M) using Ered’s (SRP’s) is calculated with the equation:  Ecell  = Ered(CAT) + Eox(AN) RED OX NOT on equation sheet  (–Ered)

Likely Oxidized or Reduced? (based on Ered)   reduced easily (highest Ered) F2(g) + 2 e–  2 F–(aq) 2 H+(aq) + 2 e–  H2(g) Li+(aq) + e–  Li(s) +2.87 V … 0 V –3.05 V (most nonmetals) (halogens) Reduced Easily 2 H+(aq) + 2 e–  H2(g) 0 V Oxidized Easily (most metals) (alkali)  oxidized easily (lowest Ered)

Standard Cell Potential (Eo)  = Ered (CAT) + Eox (AN) RED OX  (–Ered) +0.34  greater difference in Ered’s, greater voltage (∆V) (potential difference) RED OX Ecell = (0.34) + (+0.76)   Ecell = +1.10 V –0.76

Electrolysis Note that the electric current here drives the direction of the reaction and not SRP

 E = –4.07 V  E = (–2.71) + (–1.36)   Ered(Cl2) = +1.36 V Ered(Na+) = –2.71 V Electrolytic Cell OX of anions at AN RED of cations at CAT

Electrolysis electrical energy input used to cause an UNfavorable (E = –) REDOX rxn to plate out a solid mass of neutral metal from aq. ions. Electrolytic Cell: NOT electrochemical (voltaic/galvanic) cell b/c current is applied. Can calculate: time(s) to plate given mass mass (g) plated in given time e– (mol) or charge (C) transferred

recall Faraday’s constant, F = 96,485 C/mol e– Electrolysis current: amount of charge passing through an area per second. q t I = on equation sheet I = current [Amperes (A)] q = charge [Coulombs (C)] t = time [seconds (s)] recall Faraday’s constant, F = 96,485 C/mol e–

Electrolysis of Molten (l ) Salts RED: Na+ + e–  Na  Ered = –2.71 V P What drives the reaction is electric current and not SRP  OX: 2 Cl–  Cl2 + 2 e– Eox = –1.36 V Cl2 + 2 e–  2 Cl– Na+ Cl–  Ered = +1.36 V NaCl(l) (molten) Ecell = (–2.71) + (–1.36)  Ecell = –4.07 V 