1. 17-Nov-97Electrochemistry (Ch. 21)1 ELECTROCHEMISTRY Chapter 21 Electric automobile redox reactions electrochemical cells electrode processes construction notation cell potential and  G o standard reduction potentials (E o ) non-equilibrium conditions (Q) batteries corrosion

2. 17-Nov-97Electrochemistry (Ch. 21)2 TRANSFER REACTIONS Atom transfer HCl (g) + H 2 O (l)  Cl - (aq) + H 3 O + (aq) Electron transfer Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) L oss of E lectrons = O XIDATION ( LEO ) G ain of E lectrons = R EDUCTION ( GER ) - 2 e- 2 x +1 e-

3. 17-Nov-97Electrochemistry (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.

4. 17-Nov-97Electrochemistry (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.

5. 17-Nov-97Electrochemistry (Ch. 21)5 Direct Redox Reactions Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) 11_CuAg.mov 21mo2an1.mov 2Al (s) + 3Cu 2+  2 Al 3+ + 3 Cu (s)

6. 17-Nov-97Electrochemistry (Ch. 21)6 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 Oxidation Reduction Electron transfer Ions

7. 17-Nov-97Electrochemistry (Ch. 21)7 Balancing Equations Cu (s) + Ag + (aq)  Cu 2+ (aq) + Ag (s) 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  Cu 2+ + 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)  Cu 2+ (aq) + 2Ag (s) How to balance for both charge and mass ? Step 1: Identify the oxidation and reduction HALF-REACTIONS: OX: Cu  Cu 2+ + 2e- RED: Ag + + e-  Ag Step 2: Balance each HALF-REACTION for charge and mass (done)

8. 17-Nov-97Electrochemistry (Ch. 21)8 Balance the following in acid solution— VO 2 + + Zn  VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn  Zn 2+ RedVO 2 +  VO 2+ Step 2:Balance each half-reaction for mass. OxZn  Zn 2+ Red2 H + + VO 2 +  VO 2+ + H 2 O Add H 2 O on O-deficient side and add H + on other side for H-balance. Balancing Equations (2)

9. 17-Nov-97Electrochemistry (Ch. 21)9 Step 3:Balance half-reactions for charge. OxZn  Zn 2+ + 2e- Rede- + 2 H + + VO 2 +  VO 2+ + H 2 O Step 4:Multiply by an appropriate factor to balance the electron transfer in OX. and RED. OxZn  Zn 2+ + 2e- Red 2e- + 4 H + + 2 VO 2 +  2 VO 2+ + 2 H 2 O Step 5:Add half-reactions Zn + 4 H + + 2 VO 2 +  Zn 2+ + 2 VO 2+ + 2 H 2 O Balancing Equations (3)

10. 17-Nov-97Electrochemistry (Ch. 21)10 Tips on Balancing Equations Never add O 2, O atoms, or O 2- to balance oxygen. Never add H 2 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.

11. 17-Nov-97Electrochemistry (Ch. 21)11 Why Study Electrochemistry? Industrial production of chemicals such as Cl 2, NaOH, F 2 and l Biological redox reactions The heme group Batteries Corrosion

12. 17-Nov-97Electrochemistry (Ch. 21)12 Electrochemical Cells An apparatus in which a redox reaction occurs by transferring electrons through an external connector. Batteries are voltaic cells ELECTROLYTIC CELL Reactant favored reaction electric current  chemical reaction VOLTAIC CELL Product favored reaction chemical reaction  electric current

13. 17-Nov-97Electrochemistry (Ch. 21)13 CHEMICAL CHANGE  ELECTRIC CURRENT With time, Cu plates out onto Zn metal strip, and Zn strip “disappears.” Zn is oxidized and is the reducing agent Zn(s)  Zn 2+ (aq) + 2e- Cu 2+ is reduced and is the oxidizing agent Cu 2+ (aq) + 2e-  Cu(s)

14. 17-Nov-97Electrochemistry (Ch. 21)14 Oxidation: Zn(s)  Zn 2+ (aq) + 2e- Reduction: Cu 2+ (aq) + 2e-  Cu(s) -------------------------------------------------------- Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Electrons are transferred from Zn to Cu 2+, but there is no useful electric current. CHEMICAL CHANGE  ELECTRIC CURRENT (2)

15. 17-Nov-97Electrochemistry (Ch. 21)15 To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. CHEMICAL CHANGE  ELECTRIC CURRENT (2) This is accomplished in a VOLTAIC cell. (also called GALVANIC cell) A group of such cells is called a battery.

16. 17-Nov-97Electrochemistry (Ch. 21)16 Electrons travel thru external wire. Salt bridge allows anions and cations to move between electrode compartments. This maintains electrical neutrality. ANODE OXIDATION CATHODE REDUCTION

17. 17-Nov-97Electrochemistry (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

18. 17-Nov-97Electrochemistry (Ch. 21)18 Standard Notation for Electrochemical Cells ANODE Zn / Zn 2+ // Cu 2+ / Cu CATHODE OXIDATION Anode electrode Active electrolyte in oxidation half-reaction Cathode electrode Active electrolyte in reduction half-reaction Salt bridge Phase boundary REDUCTION

19. 17-Nov-97Electrochemistry (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 [Zn 2+ ] and [Cu 2+ ] = 1.0 M. Zn  Zn 2+ + 2e- ANODE 2e- + Cu 2+  Cu CATHODE

20. 17-Nov-97Electrochemistry (Ch. 21)20 This is the STANDARD CELL POTENTIAL, E o E o is a quantitative measure of the tendency of reactants to proceed to products when all are in their standard states at 25  C. CELL POTENTIAL, E o For Zn/Cu, voltage is 1.10 V at 25  C and when [Zn 2+ ] and [Cu 2+ ] = 1.0 M.

21. 17-Nov-97Electrochemistry (Ch. 21)21 E o and  G o E o is related to  G o, the free energy change for the reaction.  G o = - n F E o F = Faraday constant = 9.6485 x 10 4 J/Vmol n = the number of moles of electrons transferred. Michael Faraday 1791-1867 Discoverer of electrolysis magnetic props. of matter electromagnetic induction benzene and other organic chemicals n for Zn/Cu cell ? n = 2 Zn / Zn 2+ // Cu 2+ / Cu

22. 17-Nov-97Electrochemistry (Ch. 21)22 For a reactant-favored reaction - electrolysis cell: Electric current  chemistry Reactants  Products  G o > 0 and so E o < 0 (E o is negative) For a product-favored reaction – battery or voltaic cell: Chemistry  electric current Reactants  Products  G o 0 (E o is positive) E o and  G o (2)  G o = - n F E o

23. 17-Nov-97Electrochemistry (Ch. 21)23 Calculating Cell Voltage If we know E o for each half-reaction, we can calculate E o for the net reaction. Balanced half-reactions can be added together to get the overall, balanced equation. Anode:2 I -  I 2 + 2e- Cathode:2 H 2 O + 2e-  2 OH - + H 2 Net rxn: 2 I - + 2 H 2 O  I 2 + 2 OH - + H 2

24. 17-Nov-97Electrochemistry (Ch. 21)24 STANDARD CELL POTENTIALS, E o Can’t measure half- reaction E o directly. Therefore, measure it relative to a standard HALF CELL: the S tandard H ydrogen E lectrode ( SHE ). 2 H + (aq, 1 M) + 2e- H 2 (g, 1 atm) E o = 0.0 V

25. 17-Nov-97Electrochemistry (Ch. 21)25 Zn/Zn 2+ versus H + /H 2 Zn/Zn 2+ half-cell combined with a SHE. E o for the cell is +0.76 V

26. 17-Nov-97Electrochemistry (Ch. 21)26 E o for Zn/Zn 2+ half-cell Zn(s) + 2 H + (aq)  Zn 2+ + H 2 (g) E o = +0.76 V Therefore, E o for Zn  Zn 2+ (aq) + 2e- is ?? +0.76 V. Overall reaction is reduction of H + by Zn metal.

27. 17-Nov-97Electrochemistry (Ch. 21)27 Standard REDUCTION potentials What is E o for the reverse reaction ? Zn 2+ + 2e-  Zn The value for the REDUCTION 1/2-cell is the negative of that for the OXIDATION 1/2-cell: Zn  Zn 2+ (aq) + 2e- E o = +0.76 V THUS Zn 2+ + 2e-  Zn E o = -0.76 V Zn  Zn 2+ (aq) + 2e- E o = +0.76 V Q. Relative to H 2 is Zn a (better/worse) reducing agent ? A. Zn is a better reducing agent than H 2.

28. 17-Nov-97Electrochemistry (Ch. 21)28 E o = +0.34 V Cu/Cu 2+ and H 2 /H + Cell

29. 17-Nov-97Electrochemistry (Ch. 21)29 Cu 2+ (aq) + H 2 (g)  Cu(s) + 2 H + (aq) Measured E o = +0.34 V Therefore, E o for Cu 2+ + 2e-  Cu is ?? Cu/Cu 2+ half cell E o Overall reaction is reduction of Cu 2+ by H 2 gas. +0.34 V

30. 17-Nov-97Electrochemistry (Ch. 21)30 Zn/Cu Electrochemical Cell Anode: Zn(s)  Zn 2+ (aq) + 2e- E o = +0.76 V Cathode, positive, sink for electrons Anode, negative, source of electrons What is E o for the Zn/Cu cell (Daniel’s cell) ?? E o = E o ( anode ) + E o ( cathode ) = 0.76 + 0.34 = +1.10 V Net: Cu 2+ (aq) + Zn(s)  Zn 2+ (aq) + Cu(s) Cathode: Cu 2+ (aq) + 2e-  Cu(s) E o = +0.34 V

31. 17-Nov-97Electrochemistry (Ch. 21)31 Uses of E o 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.

32. 17-Nov-97Electrochemistry (Ch. 21)32 BEST Reducing agent ? ? STANDARD REDUCTION POTENTIALS Half-Reaction E o (Volts) Cu 2+ + 2e-  Cu+ 0.34 Oxidizing ability of ion Reducing ability of element 2 H + + 2e-  H 2 0.00 Zn 2+ + 2e-  Zn -0.76 BEST Oxidizing agent ? ? Cu 2+ Zn

33. 17-Nov-97Electrochemistry (Ch. 21)33 Standard Redox Potentials, E o Any substance on the right will reduce any substance higher than it on the left. Zn can reduce H + and Cu 2+. H 2 can reduce Cu 2+ but not Zn 2+ Cu cannot reduce H + or Zn 2+. Use tabulated reduction potentials to analyse spontaneity of ANY REDOX REACTION

34. 17-Nov-97Electrochemistry (Ch. 21)34 Determining E o for a Voltaic Cell Cd  Cd 2+ + 2e- or Cd 2+ + 2e-  Cd Fe  Fe 2+ + 2e- or Fe 2+ + 2e-  Fe

35. 17-Nov-97Electrochemistry (Ch. 21)35 Fe is a better reducing agent than Cd E o for Fe/Cd Cell Cd 2+ + 2e-  Cd -0.40 Fe 2+ + 2e-  Fe -0.44 RHS species is better reducing agent LHS species is better oxidizing agent Cd 2+ is a better oxidizing agent than Fe 2+ Overall reaction as written is spontaneous: Fe + Cd 2+  Cd + Fe 2+ E o = +0.04 V The reverse reaction is not spontaneous: Cd + Fe 2+  Fe + Cd 2+ E o = -0.04 V