1. Electrochemistry Chapter 20 Brown-LeMay

2. Review of Redox Reactions Oxidation - refers to the loss of electrons by a molecule, atom or ion - LEO goes Reduction - refers to the gain of electrons by an molecule, atom or ion – GER Chemical reactions in which the oxidation state of one or more substances changes are called oxidation-reduction reactions (or redox reactions)

3. Zn (s) + 2H + (aq)  Zn 2+ (aq) + H 2(g) Zn = 0, Zn 2+ = +2 (LEO) reducing agent H + = +1, H 2 = 0 (GER) oxidizing agent Thus, the oxidation number of both the Zn(s) and H+(aq) change during the course of the reaction, and so, this must be a redox reaction Review balancing redox-equations

4. Balancing Redox by Half Reactions Half reactions are a convenient way of separating oxidation and reduction reactions. Balance the titration of acidic solution of Na 2 C 2 O 4 (colorless) with KMnO 4(deep purple) 1 st Write the incomplete ½ reactions MnO 4(aq)  Mn 2+ (aq) is reduced (pale pink) C 2 O 4(aq)  CO 2(g) is oxidized

5. MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O C 2 O 4 2- (aq)  2CO 2(g) Then bal H by adding H + 8H + + MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O C 2 O 4 2- (aq)  2CO 2(g) finish by balancing the e’s For the permanganate 7 + left and 2 + on the right 5e - + 8H + + MnO 4(aq)  Mn 2+ (aq) + 4H 2 O On the oxalate 2- on the right and o on the left C 2 O 4 2- (aq)  2CO 2(g) + 2e -

6. 2(5e - + 8H + + MnO 4 - (aq)  Mn 2+ (aq) + 4H 2 O) 5(C 2 O 4 2- (aq)  2CO 2(g) + 2e - ) 10e - + 16H + + MnO 4 - (aq)  2Mn 2+ (aq) + 8H 2 O 5C 2 O 4 2- (aq)  10CO 2(g) +10e - 16H + +2 MnO 4 - +5C 2 O 4 2-  2Mn 2+ (aq) +8H 2 O + 10CO 2

7. Balancing Eq in Basic Solution The same method is used but OH - is added to neutralize the H + used. The equation must again be simplified by canceling the terms on both sides of the equation.

8. Voltaic Cells Spontaneous redox reactions may be use to perform electrical work Voltaic or galvanic cells are devices that electron transfer occurs in an external circuit.

9. Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) Zn 0 is spontaneously oxidized to Zn 2+ Cu 2+ is spontaneously reduced to Cu 0 Oxidization half reaction at the anode Zn (s)  Zn 2+ (aq) + 2e - Reduction half reaction at the cathode Cu 2+ (aq) + 2e -  Cu (s)

10. As oxidization occurs, Zn is converted to Zn 2+ and 2e -. The electrons flow toward the cathode, where they are used in the reduction reaction We expect the Zn electrode to lose mass Electrons flow from the anode to cathode so the anode is negative and the cathode is positive

11. Electrons cannot flow through the solution; they have to be transported though an external wire. Anions and cations move through a porous barrier or salt bridge Cations move into the cathodic compartment to neutralize the excess negatively charged ions (cathode: Cu 2+ + 2e -  Cu, so the counter ion of Cu is in excess)

12. A build up of excess charge is avoided by movement of cations and anions through the salt bridge

13. The anions move into the anodic compartment to neutralize the excess Zn 2+ ions formed by the oxidation. Molecular View - “Rules” of voltaic cells * at the anode electrons are products oxidization occurs * at the cathode electrons are reactants reduction occurs

14. Cell EMF “the driving force” Reactions are spontaneous because the cathode has a lower electrical potential energy than the anode Potential difference: difference in electrical potential measured in volts One volt is the potential difference required to impart one joule (J) of energy to a charge of one coulomb (C) 1V = 1 J/C

15. Cell EMF “the driving force” Electromotive force (emf) is the force required to push electrons though the external circuit Cell potential: E cell is the emf of a cell E cell > 0 for a spontaneous reaction For 1 molar, 1 atm for gases at 25 0 C(standard conditions), the standard emf (standard cell potential) = E 0 cell

16. Standard Reduction Potentials & Cal. Cell Potentials Standard Reduction Potentials E 0 red are measured relative to a standard The emf of a cell is E 0 cell = E 0 red (cathode) – E 0 red (anode) The standard hydrogen electrode is used as the standard (standard hydrogen electrode) (SHE) 2H + (aq,1M)+2e -  H 2 (g,1atm) E 0 cell = 0 V

17. The standard hydrogen electrode The (SHE) is assigned a potential of zero Consider the following half reaction Zn (s)  Zn 2+ (aq) + 2e - We can measure E 0 cell relative to the SHE In the cell the SHE is the cathode It cons of a Pt electrode in a tube – 1M H + sol H2 is bubbled through the tube E 0 cell = E 0 red (cathode)-E 0 red (anode) 0.76V = 0V-E 0 red (anode) Therefore E 0 red (anode) = -0.76V

18. The standard hydrogen electrode Standard electrode potentials are written as reduction reactions Zn 2+ (aq) (aq,1 M)+ 2e -  Zn (s) E 0 = -0.76V Since the reduction potential is negative in the presence of the SHE the reduction of Zn 2+ is non-spontaneous However the oxidization of Zn 2+ is spontaneous with the SHE

19. Standard reduction potential The standard reduction potential is an intensive property Therefore, changing the stoichiometric coefficient does not affect E 0 red 2Zn 2+ (aq) + 4e -  2Zn (s) E 0 red = -0.76 V

20. E 0 red > 0 are spontaneous relative to the SHE E 0 red < 0 are non- spontaneous relative to the SHE The larger the difference between E 0 red values the larger the E 0 cell The more positive the E 0 cell value the greater the driving force for reduction

21. Oxidizing and Reducing Agents Consider the table of standard reduction potentials We use the table to determine the relative strength of reducing and oxidizing agents The more positive the E 0 red the stronger the oxidizing agent (written as the reactant) The more negative the E 0 red the stronger the reducing agent (written as the product)

22. Oxidizing and Reducing Agents We can use tables to predict if one reactant can spontaneously oxidize or reduce another Example F 2 can oxidize H 2 or Li Ni 2+ can oxidize Al (s) Li can reduce F 2

23. Spontaneity of Redox Reactions E 0 cell = E 0 red (red process) – E 0 red (oxid process) Consider the reaction Ni (s) + 2Ag + (aq)  Ni 2+ (aq) + 2Ag (s) The standard cell potential is E 0 cell = E 0 red (Ag + /Ag) – E 0 red(Ni 2+ /Ni) E 0 cell = (0.80 V) - (-0.28) E 0 cell = 1.08 V the value indicates the reaction is spontaneous

24. EMF and free energy change Delta G = -nFE where delta G is the change in free energy n = the number of moles of electrons transferred F = Faraday’s constant E = emf of the cell

25. EMF and free energy change F = 96,500 C/mole - = 96,500 J/(V)(mole - ) Since n and F are positive, if Delta G 0 and the reaction will be spontaneous. Effect of concentration on cell EMF the cell is function until E=0 at which point equilibrium has been reached and the cell is “dead” The point at which E=0 is determined by the concentrations of the species involved in the redox reaction

26. Walter Nernst (Nobel Prize 1920) Nernst Equation G = G 0 + RTlnQ -nFE = -nFE 0 + RTlnQ Solve the equation for E give the Nernst Eq E = E 0 - RT/nF lnQ Or for base 10 log E = E 0 - 2.3RT/nF logQ

27. The nernst eq at 298K = E = E 0 – 0.0592/n log Q Consider if you may Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) If [Cu 2+ ] = 5.0M and [Zn 2+ ] = 0.05 M E cell = 1.10 V – 0.0592/2 log 0.05/5 =1.16 V Cell emf and chemical equilibrium Log K = nE 0 /0.0592 thus if we know cell emf, we can calc the equilibrium constant