1. 1 Electrochemistry Chapter 17 Seneca Valley SHS

2. 2 17.1Voltaic (Galvanic) Cells: Oxidation-Reduction Reactions Oxidation-Reduction Reactions Zn added to HCl yields the spontaneous reaction Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g). The oxidation number of Zn has increased from 0 to 2+. The oxidation number of H has reduced from 1+ to 0. Thus, Zn is oxidized to Zn 2+ while H + is reduced to H 2. H + causes Zn to be oxidized and is the oxidizing agent. Zn causes H + to be reduced and is the reducing agent. Note: the reducing agent is oxidized and the oxidizing agent is reduced.

3. 3 17.1 Voltaic (Galvanic) Cells Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit. If a strip of Zn is placed in a solution of CuSO 4, Cu is deposited on the Zn and the Zn dissolves by forming Zn 2+. Zn is spontaneously oxidized to Zn 2+ by Cu 2+. The Cu 2+ is spontaneously reduced to Cu 0 by Zn. The entire process is spontaneous.

4. 4 Voltaic Cells Voltaic cells consist of –Anode: Zn(s)  Zn 2+ (aq) + 2e - –Cathode: Cu 2+ (aq) + 2e -  Cu(s) –Salt bridge (used to complete the electrical circuit): cations move from anode to cathode, anions move from cathode to anode. The two solid metals are the electrodes (cathode and anode). As oxidation occurs, Zn is converted to Zn 2+ and 2e -. The electrons flow towards the cathode where they are used in the reduction reaction.

5. 5 Voltaic Cells We expect the Zn electrode to lose mass and the Cu electrode to gain mass. “Rules” of voltaic cells: 1. At the anode electrons are products. (Oxidation) 2. At the cathode electrons are reagents. (Reduction) Electrons flow from the anode to the cathode. Therefore, the anode is negative and the cathode is positive. (anode and oxidation begin with vowels; cathode and reduction begin with consonants) Electrons cannot flow through the solution, they have to be transported through an external wire.

6. 6 Voltaic Cells 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 counterion of Cu is in excess). Anions move into the anodic compartment to neutralize the excess Zn 2+ ions formed by oxidation.

7. 7 Voltaic Cells

8. 8 17.1 Voltaic Cells: Potential Difference The flow of electrons from anode to cathode is spontaneous. Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. Potential difference: is the difference in electrical potential. Potential difference is measured in volts. One volt is the potential difference required to impart one joule of energy to a charge of one coulomb:

9. 9 Cell EMF Electromotive force (emf) is the force required to push electrons through the external circuit. Cell potential: E cell is the emf of a cell. For 1M solutions at 25  C (standard conditions), the standard emf (standard cell potential) is called E  cell. Standard Reduction Potentials Convenient tabulation of electrochemical data( found in the appendix--pg. A26). Standard reduction potentials, E  red are measured relative to the standard hydrogen electrode (SHE) which defined as 0 V.

10. 10 Cell EMF Standard Reduction Potentials The SHE is the cathode. It consists of a Pt electrode in a tube placed in 1 M H + solution. H 2 is bubbled through the tube. For the SHE, we assign 2H + (aq, 1M) + 2e -  H 2 (g, 1 atm) E  red of zero. The emf of a cell can be calculated from standard reduction potentials: E  cell = E  red (cathode) - E  red (anode)

11. 11 Cell EMF Standard Reduction Potentials Consider Zn(s)  Zn 2+ (aq) + 2e -. We measure E cell relative to the SHE (cathode): E  cell = E  red (cathode) - E  red (anode) -0.76 V = 0 V - E  red (anode). Therefore, E  red (anode) = -0.76 V. Standard reduction potentials must be written as reduction reactions: Zn 2+ (aq) + 2e -  Zn(s), E  red = -0.76 V. Since E  red = -0.76 V we conclude that the reduction of Zn 2+ in the presence of the SHE is not spontaneous.

12. 12 Cell EMF (Sample Ex. 17.1a) Consider a voltaic (galvanic) cell based on the reaction : Al 3+ (aq) + Mg(s) --> Al(s) + Mg 2+ (aq) Give the balanced half cell reactions and calculate E° for the cell.

13. 13 Cell EMF (Sample Ex. 17.1b) A galvanic cell is based on the reaction: MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) --> ClO 4 - (aq) + Mn 2+ (aq) + H 2 O(l) Give the balanced cell reaction and calculate E° for the cell.

14. 14 Spontaneity of Redox Reactions In a voltaic (galvanic) cell (one that runs spontaneously) E  red (cathode) is more positive than E  red (anode) since E  cell = E  red (cathode) - E  red (anode) More generally, for any electrochemical process E  = E  red (reduction process) - E  red (oxidation process). A positive E  indicates a spontaneous process A negative E  indicates a nonspontaneous process. This can be used to understand the activity series.

15. 15 Spontaneity of Redox Reactions Line Notation: –This is a convenient way to quickly describe an electrochemical cell. –In this notation the anode components are listed on the left and the cathode components are listed on the right – separated by double vertical lines (as shown below) Ex: Mg(s) | Mg 2+ (aq) || Al 3+ (aq) | Al(s) Practice: Draw the line notation for the redox reaction: Fe 3+ (aq) + Cu(s)  Cu 2+ (aq) + Fe 2+ (aq)

16. 16 17.3 EMF and Free-Energy Change We can show that  G = -nFE  G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. We define: 1 F= 96,500 C/mol = 96,500 J/V  mol Since n and F are positive, if  G > 0 then E < 0.

17. 17 Effect of Concentration on Cell EMF A voltaic cell is functional until E = 0 at which point equilibrium has been reached. The point at which E = 0 is determined by the concentrations of the species involved in the redox reaction. The Nernst Equation The Nernst equation relates emf to concentration using and noting that

18. 18 Effect of Concentration on Cell EMF (Sample Ex. 17.3) Using the standard cell potential data, calculate ∆G° for the reaction: Cu 2+ (aq) + Fe(s) --> Cu(s) + Fe 2+ (aq) Is this reaction spontaneous? Use:

19. 19 Effect of Concentration on Cell EMF The Nernst Equation This rearranges to give the Nernst equation: The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K: Remember that n is number of moles of electrons.

20. 20 Effect of Concentration on Cell EMF Concentration Cells We can use the Nernst equation to generate a cell that has an emf based solely on difference in concentration. One compartment will consist of a concentrated solution, while the other has a dilute solution. Example: 1.00 M Ni 2+ (aq) and 1.00  10 -3 M Ni 2+ (aq).

21. 21 Effect of Concentration on Cell EMF Cell EMF and Chemical Equilibrium A system is at equilibrium when  G = 0. From the Nernst equation, at equilibrium:

22. 22 Batteries Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal grid in sulfuric acid: PbO 2 (s) + SO 4 2- (aq) + 4H + (aq) + 2e -  PbSO 4 (s) + 2H 2 O(l) Anode: Pb: Pb(s) + SO 4 2- (aq)  PbSO 4 (s) + 2e -

23. 23 Batteries Lead-Acid Battery The overall electrochemical reaction is PbO 2 (s) + Pb(s) + 2SO 4 2- (aq) + 4H + (aq)  2PbSO 4 (s) + 2H 2 O(l) for which E  cell = E  red (cathode) - E  red (anode) = (+1.685 V) - (-0.356 V) = +2.041 V. Wood or glass-fiber spacers are used to prevent the electrodes from touching.

24. 24 Batteries Common Dry Cell Batteries Anode: Zn cap: Zn(s)  Zn 2+ (aq) + 2e - Cathode: MnO 2, NH 4 Cl and C paste: 2NH 4 + (aq) + 2MnO 2 (s) + 2e -  Mn 2 O 3 (s) + 2NH 3 (aq) + 2H 2 O(l) The graphite rod in the center is an inert cathode. For an alkaline battery, NH 4 Cl is replaced with KOH. Anode: Zn powder mixed in a gel: Zn(s)  Zn 2+ (aq) + 2e - Cathode: reduction of MnO 2.

25. 25 Batteries Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. On Apollo moon flights, the H 2 -O 2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen: 2H 2 O(l) + O 2 (g) + 4e -  4OH - (aq) Anode: 2H 2 (g) + 4OH - (aq)  4H 2 O(l) + 4e -

26. 26 Corrosion Corrosion of Iron Since E  red (Fe 2+ ) < E  red (O 2 ) iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l). Anode: Fe(s)  Fe 2+ (aq) + 2e -. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3.xH 2 O(s). Oxidation occurs at the site with the greatest concentration of O 2.

27. 27 Corrosion Corrosion of Iron

28. 28 Corrosion Preventing the Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc. Zinc protects the iron since Zn is the anode and Fe the cathode: Zn 2+ (aq) +2e -  Zn(s), E  red = -0.76 V Fe 2+ (aq) + 2e -  Fe(s), E  red = -0.44 V With the above standard reduction potentials, Zn is easier to oxidize than Fe.

29. 29 Corrosion Preventing the Corrosion of Iron

30. 30 Corrosion Preventing the Corrosion of Iron To protect underground pipelines, a sacrificial anode is added. The pipe is turned into the cathode and an active metal is used as the anode. Often, Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg(s), E  red = -2.37 V Fe 2+ (aq) + 2e -  Fe(s), E  red = -0.44 V

31. 31 Electrolysis Electrolysis of Aqueous Solutions Nonspontaneous reactions require an external current in order to force the reaction to proceed. Electrolysis reactions are nonspontaneous. In voltaic and electrolytic cells: –reduction occurs at the cathode, and –oxidation occurs at the anode. –However, in electrolytic cells, electrons are forced to flow from the anode to cathode. –In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.)

32. 32 Electrolysis Electrolysis of Aqueous Solutions

33. 33 Electrolysis Electrolysis of Aqueous Solutions Example, decomposition of molten NaCl. Cathode: 2Na + (l) + 2e -  2Na(l) Anode: 2Cl - (l)  Cl 2 (g) + 2e -. Industrially, electrolysis is used to produce metals like Al. Electrolysis with Active Electrodes Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating.

34. 34 Electrolysis Electrolysis with Active Electrodes

35. 35 Electrolysis Electrolysis with Active Electrodes Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO 4 : Anode: Ni(s)  Ni 2+ (aq) + 2e - Cathode: Ni 2+ (aq) + 2e -  Ni(s). Ni plates on the inert electrode. Electroplating is important in protecting objects from corrosion.

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