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ELECTROCHEMISTRY Chapter 21

ELECTROCHEMISTRY Chapter 21. redox reactions electrochemical cells electrode processes construction notation cell potential and G o standard reduction potentials (E o ) non-equilibrium conditions (Q) batteries corrosion. Electric automobile. - 2 e-. 2 x +1 e-.

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ELECTROCHEMISTRY Chapter 21

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  1. ELECTROCHEMISTRYChapter 21 • redox reactions • electrochemical cells • electrode processes • construction • notation • cell potential and Go • standard reduction potentials (Eo) • non-equilibrium conditions (Q) • batteries • corrosion Electric automobile Electrochemistry (Ch. 21)

  2. - 2 e- 2 x +1 e- TRANSFER REACTIONS Atom transfer HCl (g) + H2O (l)  Cl- (aq) + H3O+ (aq) Electron transfer Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) Loss of Electrons = OXIDATION (LEO) Gain of Electrons = REDUCTION (GER) Electrochemistry (Ch. 21)

  3. Electron Transfer Reactions • Electron transfer reactions are oxidation-reduction or redox reactions. • Redox reactions can result in : • generation of an electric current, or • be caused by imposing an electric current. • When external electric current is involved, this field of chemistry is called ELECTROCHEMISTRY. Electrochemistry (Ch. 21)

  4. Terminology for Redox Reactions • OXIDATION—loss of electron(s) by a species; increase in oxidation number. • REDUCTION—gain of electron(s); decrease in oxidation number. • OXIDIZING AGENT—electron acceptor; species is reduced. • REDUCING AGENT—electron donor; species is oxidized. Electrochemistry (Ch. 21)

  5. Direct Redox Reactions Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s) 2Al (s) + 3Cu2+ 2 Al3+ + 3 Cu (s) 11_CuAg.mov 21mo2an1.mov Electrochemistry (Ch. 21)

  6. Electron transfer Reduction Oxidation Ions Indirect Redox Reactions A battery functions by transferring electrons through an external wire from the reducing agent to the oxidizing agent. 11_battry.mov 21mo2an2.mov Electrochemistry (Ch. 21)

  7. Balancing Equations How to balance for both charge and mass ? Cu (s) + Ag+(aq) Cu2+(aq) + Ag (s) Step 1: Identify the oxidation and reduction HALF-REACTIONS: OX: Cu  Cu2+ + 2e- RED: Ag+ + e-  Ag Step 2: Balance each HALF-REACTION for charge and mass (done) Step 3: Multiply each half-reaction by a factor that makes the reducing agent supply as many electrons as the oxidizing agent requires - ELECTRON TRANSFER NUMBER (2 here) Oxidation (Reducing agent) Cu  Cu2+ + 2e- Reduction (Oxidizing agent) 2 Ag+ + 2 e-  2 Ag Step 4: Add half-reactions to give the overall equation. Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2Ag (s) Electrochemistry (Ch. 21)

  8. Balancing Equations (2) Balance the following in acid solution— VO2+ + Zn  VO2+ + Zn2+ Step 1: Write the half-reactions Ox Zn  Zn2+ Red VO2+ VO2+ Step 2: Balance each half-reaction for mass. Ox Zn  Zn2+ Red 2 H++ VO2+ VO2+ + H2O Add H2O on O-deficient side and add H+ on other side for H-balance. Electrochemistry (Ch. 21)

  9. Balancing Equations (3) Step 3: Balance half-reactions for charge. Ox Zn  Zn2+ + 2e- Red e- + 2 H+ + VO2+ VO2+ + H2O Step 4: Multiply by an appropriate factor to balance the electron transfer in OX. and RED. Ox Zn  Zn2+ +2e- Red 2e-+ 4 H+ + 2 VO2+ 2 VO2+ + 2 H2O Step 5: Add half-reactions Zn + 4 H+ + 2 VO2+ Zn2+ + 2 VO2+ + 2 H2O Electrochemistry (Ch. 21)

  10. Tips on Balancing Equations • Never add O2, O atoms, or O2- to balance oxygen. • Never add H2 or H atoms to balance hydrogen. • Be sure to write the correct charges on all the ions. • Check your work at the end to make sure mass and charge are balanced. Electrochemistry (Ch. 21)

  11. The heme group Why Study Electrochemistry? • Batteries • Industrial production of chemicalssuch as Cl2, NaOH, F2 and l • Biological redox reactions • Corrosion Electrochemistry (Ch. 21)

  12. Electrochemical Cells • An apparatus in which a redox reaction occurs by transferring electrons through an external connector. • VOLTAIC CELL • Product favored reaction • chemical reaction  electric current Batteries are voltaic cells • ELECTROLYTIC CELL • Reactant favored reaction • electric current  chemical reaction Electrochemistry (Ch. 21)

  13. CHEMICAL CHANGE  ELECTRIC CURRENT • Zn is oxidizedand is the reducing agent Zn(s)  Zn2+(aq) + 2e- • Cu2+ is reduced and is the oxidizing agent Cu2+(aq) + 2e-  Cu(s) With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Electrochemistry (Ch. 21)

  14. Electrons are transferred from Zn to Cu2+, but there is no useful electric current. CHEMICAL CHANGE  ELECTRIC CURRENT (2) Oxidation: Zn(s)  Zn2+(aq) + 2e- Reduction: Cu2+(aq) + 2e-  Cu(s) -------------------------------------------------------- Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s) Electrochemistry (Ch. 21)

  15. CHEMICAL CHANGE  ELECTRIC CURRENT (2) • To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. • This is accomplished in a VOLTAIC cell. • (also called GALVANIC cell) • A group of such cells is called a battery. Electrochemistry (Ch. 21)

  16. ANODE OXIDATION CATHODE REDUCTION • Electrons travel thru external wire. • Salt bridge allows anions and cations to move between electrode compartments. • This maintains electrical neutrality. Electrochemistry (Ch. 21)

  17. Electrochemical Cell Electrons move from anode to cathode in the wire. Anions & cations move through the salt bridge. 1!_cell.mov 21mo4an1.mov Electrochemistry (Ch. 21)

  18. Phase boundary Salt bridge Phase boundary Anode electrode Cathode electrode Active electrolyte in oxidation half-reaction Active electrolyte in reduction half-reaction Standard Notation for Electrochemical Cells ANODE Zn / Zn2+ // Cu2+ / Cu CATHODE REDUCTION OXIDATION Electrochemistry (Ch. 21)

  19. CELL POTENTIAL, E • Electrons are “driven” from anode to cathode by an electromotive force or emf. • For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25°C and when [Zn2+] and [Cu2+] = 1.0 M. Zn  Zn2+ + 2e- ANODE 2e- + Cu2+ Cu CATHODE Electrochemistry (Ch. 21)

  20. CELL POTENTIAL, Eo For Zn/Cu, voltage is 1.10 V at 25°C and when [Zn2+] and [Cu2+] = 1.0 M. • This is the STANDARD CELL POTENTIAL, Eo • Eo is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25 °C. Electrochemistry (Ch. 21)

  21. Michael Faraday 1791-1867 Eo and DGo Eo is related to DGo, the free energy change for the reaction. DGo = - n F Eo • F = Faraday constant = 9.6485 x 104 J/V•mol • n = the number of moles of electrons transferred. Discoverer of • electrolysis • magnetic props. of matter • electromagnetic induction • benzene and other organic chemicals Zn / Zn2+ // Cu2+ / Cu n = 2 n for Zn/Cu cell ? Electrochemistry (Ch. 21)

  22. DGo = - n F Eo Eo and DGo (2) • For a product-favored reaction • battery or voltaic cell: Chemistry  electric current Reactants  Products DGo < 0 and so Eo > 0 (Eo is positive) • For a reactant-favored reaction • - electrolysis cell: Electric current  chemistry • Reactants  Products • DGo > 0 and so Eo < 0 (Eo is negative) Electrochemistry (Ch. 21)

  23. Calculating Cell Voltage • Balanced half-reactions can be added together • to get the overall, balanced equation. • If we know Eo for each half-reaction, we can calculate Eo for the net reaction. Anode: 2 I- I2 + 2e- Cathode: 2 H2O + 2e-  2 OH- + H2 Net rxn: 2 I- + 2 H2O  I2 + 2 OH- + H2 Electrochemistry (Ch. 21)

  24. 2 H+(aq, 1 M) + 2e- H2(g, 1 atm) Eo = 0.0 V STANDARD CELL POTENTIALS, Eo • Can’t measure half- reaction Eo directly. Therefore, measure it relative to a standard HALF CELL: the Standard Hydrogen Electrode (SHE). Electrochemistry (Ch. 21)

  25. Zn/Zn2+ versus H+/H2 Zn/Zn2+ half-cell combined with a SHE. Eo for the cell is +0.76 V Electrochemistry (Ch. 21)

  26. Eo for Zn/Zn2+ half-cell Overall reaction is reduction of H+ by Zn metal. Zn(s) + 2 H+ (aq)  Zn2+ + H2(g) Eo = +0.76 V Therefore, Eo for Zn  Zn2+ (aq) + 2e- is ?? +0.76 V. Electrochemistry (Ch. 21)

  27. Standard REDUCTION potentials Zn  Zn2+ (aq) + 2e- Eo = +0.76 V Q. Relative to H2 is Zn a (better/worse) reducing agent ? A. Zn is a better reducing agent than H2. What is Eo for the reverse reaction ? Zn2+ + 2e-  Zn The value for the REDUCTION 1/2-cell is the negative of that for the OXIDATION 1/2-cell: Zn  Zn2+ (aq) + 2e- Eo = +0.76 V THUS Zn2+ + 2e-  Zn Eo = -0.76 V Electrochemistry (Ch. 21)

  28. Cu/Cu2+ and H2/H+ Cell Eo = +0.34 V Electrochemistry (Ch. 21)

  29. Cu/Cu2+ half cell Eo • Cu2+ (aq) + H2(g)  Cu(s) + 2 H+(aq) • Measured Eo = +0.34 V • Therefore, Eo for Cu2+ + 2e-  Cu is ?? Overall reaction is reduction of Cu2+ by H2 gas. +0.34 V Electrochemistry (Ch. 21)

  30. Zn/Cu Electrochemical Cell What is Eo for the Zn/Cu cell (Daniel’s cell) ?? Anode: Zn(s)  Zn2+(aq) + 2e- Eo = +0.76 V Anode, negative, source of electrons Cathode, positive, sink for electrons Cathode: Cu2+(aq) + 2e-  Cu(s) Eo = +0.34 V Net: Cu2+(aq) + Zn(s)  Zn2+(aq) + Cu(s) Eo = Eo(anode) + Eo(cathode) = 0.76 + 0.34 = +1.10 V Electrochemistry (Ch. 21)

  31. Uses of Eo Values This shows we can a) decide on relative ability of elements to act as reducing agents (or oxidizing agents) b) assign a voltage to a half-reaction that reflects this ability. Electrochemistry (Ch. 21)

  32. Oxidizing ability of ion Reducing ability of element STANDARD REDUCTION POTENTIALS Half-Reaction Eo (Volts) Cu2+ + 2e-  Cu + 0.34 2 H+ + 2e-  H2 0.00 Zn2+ + 2e-  Zn -0.76 BEST Oxidizing agent ? ? Cu2+ BEST Reducing agent ? ? Zn Electrochemistry (Ch. 21)

  33. Standard Redox Potentials, Eo • Any substance on the right will reduce any substance higher than it on the left. • Zn can reduce H+ and Cu2+. • H2 can reduce Cu2+ but not Zn2+ • Cu cannot reduce H+ or Zn2+. Use tabulated reduction potentials to analyse spontaneity of ANY REDOX REACTION Electrochemistry (Ch. 21)

  34. Determining Eo for a Voltaic Cell Cd  Cd2+ + 2e- or Cd2+ + 2e-  Cd Fe  Fe2+ + 2e- or Fe2+ + 2e-  Fe Electrochemistry (Ch. 21)

  35. LHS species is better oxidizing agent RHS species is better reducing agent Eo for Fe/Cd Cell • Fe is a better reducing agent than Cd Cd 2+ + 2e-  Cd -0.40 Fe 2+ + 2e-  Fe -0.44 • Cd 2+ is a better oxidizing agent than Fe 2+ • Overall reaction as written is spontaneous: • Fe + Cd 2+ Cd + Fe 2+ Eo = +0.04 V • The reverse reaction is not spontaneous: • Cd + Fe 2+ Fe + Cd 2+ Eo = -0.04 V Electrochemistry (Ch. 21)

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