Review Unit 8 (Chp 20): Electrochemistry Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Review Unit 8 (Chp 20): Electrochemistry John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

LEO says GER A species is oxidized when it loses e– . A species is reduced when it gains e– . LEO says GER oxidized reduced

Balance RedOx by Half-Rxns 5 Steps: Balance RedOx by Half-Rxns +3 +2 1: Ox #’s Zn + Fe(NO3)3  Zn(NO3)2 + Fe (spectator ion: NO3–) comp–diss–cross –net– balanced? 2: Half rxns RED: Fe+3  Fe 3 e− + OX: Zn  Zn+2 + 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)

(maintain charge balance) Voltaic Cells oxidation occurs at the anode. reduction occurs at the cathode. E = + DG = – Galvanic Electrochemical RED CAT OX AN salt bridge (maintain charge balance)

Voltaic Cells +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

Standard Cell Potential (Eo)  = Ered + Eox CAT AN  (–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

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)

Potential (Eo) & Free Energy (DGo) Go for a redox reaction can be found by using the equation: Go = −nFEo n : moles of e– transferred F : Faraday’s constant (96,485) 1 F = 96,485 C/mol e– = 96,485 J/V∙mol e– (in J) on equation sheet on equation sheet 1 V = 1 J C

(Eo) & (DGo) & ______________ (K) Equilibrium G = −nFE G = −RT ln K   on equation sheet  on equation sheet –∆Go RT = ln K UNITS!!! ∆Go & R both in J or kJ Solved for K : –∆Go RT K = e^ NOT on equation sheet

– + – + Eo, ΔGo , & K Go = −RT ln K Go = −nFEo K ∆Go = –RT(ln K) Fav or UNfav therm. fav. – –RT ( + ) –nF( ) + > 1 = – = –nF( ) = –RT ( 0 ) 0 = = 1 neither therm. UNfav. – + –RT ( – ) –nF( ) < 1 = + =

Ecell (nonstandard) (not Eo) [P]y [R]x Q = If Eo, then Q = __ x R  y P Q = 1 [1.0]y [1.0]x NON-standard conditions : [R] or [P] ≠ 1.0 M so Q ≠ 1 (if Q < 1 then E > Eo) [R] & [P] = 1.0 M so Q = 1, & Eo= +(fav) but… …as rxn proceeds  , [R] & [P], and… Q > 1 so E < Eo until Q = K and E = 0 Q ≠ 1 or (if Q > 1 then E < Eo) (Eo unchanged) (equilibrium) (“dead” cell, 0.00 V)

Ecell (not Eo) (not 1.0 M) As Q inc↑, E ________. dec to 0 Q = ? 2.00 V As Q inc↑, E ________. dec to 0 Al Q = ? ___ + ___  ___ + ___ 2 Al 3 Cu2+ 2 Al3+ 3 Cu [Al3+]2 [Cu2+]3 Q = Al3+

recall Faraday’s constant, F = 96,485 C/mol e– Electrolysis electrical energy input used to cause an UNfavorable (E = –) REDOX rxn to plate out a solid mass of neutral metal from aq. ions. on equation sheet current: charge (C) per second (s). q t I = I = current [Amperes (A)] q = charge [Coulombs (C)] t = time [seconds (s)] recall Faraday’s constant, F = 96,485 C/mol e–

Electrolysis Examples: (longer) t = ? s (c) How long will it take to plate out 0.61 g Cu(s) from a solution of Cu2+ ions with 2.5 amps? t = ? s mass current 1 mol Cu 63.55 g 2 mol e– 1 mol Cu2+ 96,485 C 1 mol e– 1 s 2.5 C 0.61 g x x x x = 741 s (d) How many grams of Ag(s) will be produced when 21.0 A flows through a solution of Ag+ ions for 45.0 min? current time 60 s 1 min 21.0 C 1 s 1 mol e– 96,485 C 1 mol Ag+ 1 mol e– 107.87 g 1 mol Ag 63.4 g 45.0 min x x x x x =

Electrolysis = 25.9 g 1 F = 96,485 C 1 mol e– Example: (MC no CALCULATOR) (e) What mass of Pb(s) will be deposited by passing 0.50 F through a solution of Pb4+ ? 1 mol Pb4+ 4 mol e– 207.2 g Pb 1 mol Pb 0.50 mol e– x x = 25.9 g 1 F = 96,485 C 1 mol e– 96,485 C 1 F 1 mol e– 96,485 C 1 mol Pb4+ 4 mol e– 207.2 g Pb 1 mol Pb 0.50 F x x x x =

Electrolysis of Molten (l) Salts P  RED: Na+ + e–  Na Ered = –2.71 V  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 

Electrolysis Aqueous (aq) Salts  RED: K+ + e–  K Ered = –2.92 V P  Ered = –0.83 V 2 H2O + 2 e–  H2 + 2 OH– H2O  Eox = OX: 2 I–  I2 + 2 e– –0.53 V K+ I– KI(aq)  Eox = –1.23 V  RED: highest Ered OX: highest Eox 2 H2O  O2 + 4 H+ + 4 e– 

Redox Titration The analytical technique used to calculate the moles in an unknown soln. mass % (g /gtotal) molarity (mol/L) molar mass (g/mol) buret titrant known vol. (V) known conc. (M) Stoich: L X  mol X  mol Y analyte M X known vol. (V) unknown conc. (M) (or moles)

Redox Titration The analytical technique used to calculate the moles in an unknown soln. mass % (g /gtotal) molarity (mol/L) molar mass (g/mol) buret titrant known vol. (V) known conc. (M) Stoich: L X x mol X x mol Y = mol Y 1 L X mol X analyte known vol. (V) unknown conc. (M) (or moles)

K = e^ E  = Ered + Eox  (–Ered) Go = −nFEo –∆Go RT Go = −RT ln K CAT AN “standard” cell potential (Eo)  (–Ered) free energy change (∆Go) Go = −nFEo UNITS! ∆Go & R J or kJ equilibrium constant (K) –∆Go RT Go = −RT ln K K = e^ (Q = K , E = 0) (Q = 1 , E = Eo) non-“standard” cell potential (E) (≠ 1.0 M) from Eo & Q [P]y [R]x Q = (Q > 1 , E < Eo) (Q < 1 , E > Eo) q (C) t (s) I (A) = electrolysis (E = –) (calculate: g, s, etc.) 96,485 C 1 mol e– 1 F =