Electrochemistry-2

Electrochemistry Review Oxidation-Reduction (redox) reactions, assigning oxidation numbers. **LEO goes GER 0 +1-1 +2-1 0 Fe(s) + 2HCl(g) FeCl2(s) + H2(g) e- transferred from Fe to H+ Oxidizing agent (oxidant) = e- taker (gets reduced) Reducing agent (reductant) = e- giver (gets oxidized) Half Reactions 0 Fe Fe2+ + 2e- Oxidation 0 2H+ + 2e- H2 Reduction Balancing equations by the half reaction method. Often, all of the ions present at the beginning of a redox will not be represented, due to the fact that they are spectators and are not important to the redox process. Also, redox rxns may be accompanied by the terms "in acid solution or acidic environment" or "in base or basic environment." If acid, use H+ and H2O to balance the equation. If base, use OH- and H2O to balance the equation. Acidic env. MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g) Step 1: Write basic half reactions. MnO4- Mn2+ C2O42- CO2 Step 2: Each half rxn is balanced with respect to the a. elements oxidized and reduced and b. other atoms c. charge Reduction: 8H+ + MnO4- Mn2+ + 4H2O (atoms now balance net chg +7 net chg +2 using H+ and H2O) The e- change is 5. So: 5e- + 8H+ + MnO4- Mn2+ + 4H2O Oxidation:C2O42- 2CO2 (atoms balance) net chg -2 net chg 0 The e- change is 2. So: C2O42- 2CO2 + 2e- Step 3: Balance e- in the 2 half rxns by multiplying by correct values to give a common factor. e- lost must = e- gained. 2 x [5e- + 8H+ + MnO4- Mn2+ + 4H2O] 5 x [C2O42- 2CO2 + 2 e-] 10e- + 16H+ + 2MnO4- 2Mn2+ + 8H2O 71120017907000 5C2O42- 10CO2 + 10e- 16H+ + 2MnO4- + 5C2O42- 2Mn2+ + 10CO2+ 8H2O (e- cancel) If the reaction occurs in base, the balancing process is a little more involved. Here is a helpful method for balancing reactions in basic solution. Balance the half reactions as though they were in acid (using H+ and H2O). Add enough OH- ions to neutralize any H+ ions to both sides of either half-rxn. This keeps the half-rxns balanced and changes any H+ ions to water. You will automatically now have the OH- and H2O on the correct sides of the half-rxns. The oxidation number method (where each element in the basic equation is assigned its own oxidation number) can also be used effectively. +7 -2 +2 +3 -2 +4 -2 Red: MnO4- Mn2+ (5 e- gained) Ox: C2O42- 2CO2 (2 e- lost) The half-reaction method is usually preferred, because spectator ions are often left out of the reaction to be balanced, and the conditions may simply say “in acid” or “in base.” Electrochemical (aka voltaic or galvanic cells) Electron transfer takes place through an external path rather than directly. Find in Chpt. 17 of your text a classic diagram of a voltaic cell. Please review this diagram and the parts of the cell. There are also diagrams included in your notes (farther along in this unit.) 190502286000 Anode: oxidation occurs here (has - chg) Cathode: reduction occurs here (has + chg) 235521537719000"An ox and a red cat!" Think of Paul Bunyan when you work with redox cells. (Anode/oxidation, reduction/cathode) Ex. Anode: Zn(s) Zn2+(aq) + 2e- Cathode: Cu2+(aq) + 2e- Cu(s) Cell EMF: electromotive force 39643055207000This is the driving force or electrical "pressure" of a voltaic cell. Measured as cell potential in volts (V). 1V = 1J/C 1 coulomb = C = 1 amp second http://www.fotosearch.com/illustration/paul-bunyan.html Measured voltage is often less than calculated due to resistance in the cell and concentration changes near the electrodes. Assuming ideal behavior, emf is for 1M salt solutions. E symbolizes the emf generated by a cell. When conditions are thermodynamically standard, use E°: standard emf or standard cell potential. For Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) E° = +1.10V Emf depends on the reaction, the [ ]'s and the temperature. As before, E° suggests 1M [ ] and 298K. Standard Electrode Potentials: Ecell = Eox + Ered Each half rxn E is measured against a platinum/H2 electrode half-cell as a standard. For this reason: 2H+(1M) + 2e- H2(g)(1atm) w/ a Pt catalyst has E°red = 0V Standard reduction potential of 0 volts. If Zn(s) Zn2+(aq) + 2e- is measured against the platinum/H2 electrode, the half-cell potential of the Zn can be found. (Any voltage produced or required is assumed due to the non-H electrode.) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) E°cell = +0.76V E°cell = E°ox + E°red 0.76V = 0.76V + 0V *A half cell's reduction potential is equal to the opposite of the oxidation potential, so Zn2+(aq) + 2e- Zn(s) E°red = -0.76V (note sign change) There is a table of standard electrode potentials on P. 1018 (Appendix E) of the textbook. A table of electrode potentials is also supplied in the free response portion of the AP Exam. **Please note that tables such as these typically only show reduction potentials. To show oxidations, simply reverse the half rxn and change the sign on the voltage. The half-cell potentials are known for a great many redox half rxns. The emf for almost any redox can be calculated easily. Ex. Cr2O72- + 14H+ + 6I- 2Cr+3 + 3I2 + 7H2O What is the voltage generated by this cell? Red: Cr2O72- + 14H+ + 6e- 2Cr+3 + 7H2O E°red = +1.33V 406400017907000 Ox: 6I- 3I2 + 6 e- E°ox = -0.54V E°cell= +0.79V Oxidizing and Reducing Agents The more (+) the E° values are for a half rxn, the more likely the rxn is to occur. A (-) potential means the substance is more difficult to reduce than H+, and the less likely the reaction is to occur. F2(g) + 2e- 2F-(aq) E°red = +2.87V Very likely to occur. Elemental fluorine is the most aggressive oxidizing agent known. It takes electrons with great vigor. Li+(aq) + 1e- Li(s) E°red = -3.05V Not likely to occur at all. Poor oxidizing agent. However, if the equation is reversed: Li(s) Li+(aq) + 1e- E°ox = +3.05V Very likely, excellent reducing agent. A lithium/fluorine cell would produce the highest single cell voltage possible at +5.92V. Spontaneity and Extent of Rxn A +E°cell value indicates a spontaneous cell rxn (i.e. a rxn that goes more than 50% to completion. Electron flow is not forced but occurs by itself.) A -E°cell value is nonspontaneous. Emf and Free E Change Cell emf shows spontaneity and so must be related to ?G. ?G = -nFE where E = emf in V F = Faraday's constant = 96,500C/mol e- = 96,500 J/Vmol e- n = number of moles of e- transferred in the reaction F is the charge on 1 mol of e-, the Faraday. For reactants and products in standard states, ?G° = -nFE° Ex. Calculate ?G° for 2Br-(aq) + F2(g) Br2(l) + 2F-(aq) 2Br-(aq) Br2(l) + 2e- E°ox = -1.06V 218440017653000 F2(g) + 2e- 2F-(aq) E°red = +2.87V E°cell = +1.81V With a +E°cell value, ?G° should be (-), spontaneous. 2 mol of e- are transferred ([ ] = 1M). ?G° = -2 mol e- (96,500 J/V mol e-)(1.81V) = -349,000J = -349kJ Emf and Equilibrium Constant ?G° = -RTlnK and ?G° = -nFE°, so -nFE° = -RTlnK = -2.303RTlogK E° = RTlnK/nF or E° = 2.303RTlogK/nF If conditions are standard, the equation can be simplified: At 298K, E° = 2.303(8.314J/K mol)(298K)logK n(96,500J/V mol) = (0.0591V/n)logK *Please note that this constant is for log10. Ex. O2(g) + 4H+(aq) + 4Fe2+(aq) 4Fe+3(aq) + 2H2O(l) O2(g) + 4H+(aq) + 4e- 2H2O(l) E°red = 1.23V 4Fe2+(aq) 4Fe+3(aq) + 4e- E°ox =-0.77V 3098800-381000 E°cell = 0.46V Here n = 4 (mol e-) logK = nE°/0.0591V Log K = 4(0.46V)/0.0591V = 31.1 K = 1 x 1031.1 Rxn goes essentially to completion. Emf is affected by solution concentration Because ?G = ?G° + RTlnQ (or ?G = ?G° + 2.303RTlogQ) and ?G = -nFE and ?G° = -nFE°, -nFE = -nFE° + RTlnQ = -nFE° + 2.303RTlogQ Nernst Equation: E= E° - (2.303RT/nF)logQ = E° - (RT/nF)lnQ at 298K, E = E° - (0.0591/n)logQ Ex. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) E° = +1.10V Based on [ ] = 1M solution n = 2 moles of electrons E = 1.10V - (0.0591V/2)log([Zn2+]/[Cu2+]) *Note: if [Zn2+] = [Cu2+], log = 0 and E = E°. **Q applies to aq. slns and/or gases, not to solids or liquids. If the concentrations are changed to [Zn2+] = 0.050M and [Cu2+] = 5.00M E = 1.10V - (0.0591V/2)log(0.050/5.00) = 1.10V - (0.0591V/2)(-2.00) = 1.16V** Voltage increase. This makes sense with respect to Le Chatelier's Principle, too. As reactant conc. increases and product concentration decreases, the forward rxn is favored (equil. shifts right), the rxn is more spontaneous, and more energy is released. Commercial Voltaic Cells Line electricity typically costs $0.10-0.20/kwatt hr. Flashlight (D-cell) battery electricity costs $80-$120/kwatt hr. We really pay for portability. Lead storage battery: Pb and PbO2 electrodes in high conc. H2SO4 sln. Pb(s) + SO42-(aq) PbSO4(s) + 2e- E° = +0.356V 407670016573500PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- PbSO4(s) + 2H2O(l) E° = +1.685V E° = +2.041V So, if this battery only produces about 2 volts, and even a Li-F battery would produce about 6 volts, how are 9 and 12 volt batteries made? A car battery is a 12-V lead storage battery. Ni-Cad alkaline batteries: Anode: Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2e- E° = +0.403V 408940017335500Cathode: NiO2(s) + 2H2O(l) + 2e- Ni(OH)2(s) + 2OH-(aq) E° = +0.28V E° = +0.683V Rechargeable: no gases are produced, so the battery can be sealed, and recharged with line voltage and a DC transformer. Dry Cells: invented 1866 and use a Zn can anode and a central graphite cathode with a moist paste of NH4Cl and MnO2. Rxns: Zn Zn2+ + 2e- and 2NH4+ + 2MnO2 Mn2O3 + 2NH3 + H2O Electrolysis: Aq. slns: Na(s) can't be prepared by electrolyzing NaCl(aq). The reason is that H+ is easier to reduce than Na+. Cathode: 2H2O(l) + 2e- H2(g) + 2OH-(aq) E°red = -0.83V Na+(aq) + e- Na(s) E°red = -2.71V Anode: 2Cl-(aq) Cl2(g) + 2e- E°ox =-1.36V** 2H2O(l) 4 H+(aq) + O2(g) + 4e- E°ox =-1.23V **Unexpectedly, at least according to the numbers above, chlorine gas is produced at the anode rather than oxygen, even though the voltage for the oxygen half rxn is less negative and therefore supposedly more likely to take place. In systems where more than one oxidation or reduction are possible (as the sea water above), the set of half-rxns requiring the least voltage will be the rxns that occur. The discrepancy is attributed to a phenomenon called "overvoltage," the voltage that is actually required to make the rxn proceed. The overvoltage required to make oxygen is so high, that chlorine forms at the anode instead. Overvoltage may be due to very slow rxn rates (kinetics.) Below are diagrams of simple electrochemical and electrolytic cells. Please note the differences between them. -10985548895A N O D E - A N O D E + C A T H O D E + C A T H O D E - 00A N O D E - A N O D E + C A T H O D E + C A T H O D E - Electrochemical cell Electrolytic cell 254008128000 = direction of electron flow. Note that the electrons always flow from anode to cathode, no matter which type of cell it is. Important differences: Electrochemical cell has a porous barrier of some sort. The electrolytic cell does not. Why? In the electrochemical cell, the anode is the (-) pole and the cathode is (+). The electrolytic cell is just the opposite. The electrochemical cell has a power using/measuring device while the electrolytic cell must have a power source, or an “electron pump.” You may encounter the following symbols in simple electronic schematic diagrams. 230632057150020955002349500221424577470001968500615950078740000 = generator = battery (These 2 are power sources or e- pumps.) 1270007048500 = switch 8890014922500 = resistor (may be a light bulb as above) 3089275117475003429009080500 V = voltmeter (measures emf) A = ammeter (measures current) These last 3 are electrical power users. Electroplating: For electroplating, the cathode is the metal object to be plated (coated) with another metal. The anode is composed of the metal to be plated onto the cathode object. The solution is a salt sln of the plating metal. Ex. A nickel spoon is used as the cathode and placed into a bath of AgNO3(aq) with a solid silver anode. As Ag+(aq) ions are reduced at the surface of the nickel spoon cathode, they become solid and stick to the nickel, plating the spoon. At the anode, Ag atoms are oxidized to Ag+(aq) ions, replacing those that have been plated onto the spoon. Electrical work: ?G = wmax = -nFE wmax is calculated for spontaneous rxns. If E = +V, w = (-) work comes out of the system. For voltaic cells: wmax = -n x F x E J = (mol)(C/mol)(J/C) More e- = more work and more V = more work Electrolysis cell: wmin = -nFE calculated the same way as wmax. wmin is calculated for nonspontaneous rxns (-V). This is the minimum work required to make the rxn go. E = calculated voltage. Reality: It always takes more work (voltage) than was calculated to make an electrolytic cell run. E' = overvoltage. These rxns always need more energy than calculated. Electrical work is usually in units of power: watts (1 watt = 1J/s) x time thus, kwatt hrs. Corrosion: Corrosion is an important redox rxn that costs the world economy billions of dollars each year due to the corruption of structural metals. Corrosion can be avoided or delayed with the use of "sacrificial metals", i.e., metals that are more easily oxidized than the structural metal in question. Ex. Water heaters are composed of a steel tank coated with baked porcelain. Over time, cracks develop in the coating due to the expansion and contraction from temperature changes taking place in the tank. The iron in the steel begins to rust (oxidize). It is now only a matter of time before the tank rusts through somewhere and the basement is flooded. However, water heaters are equipped with a rod made of Mg. Being a more active metal than Fe, the Mg oxidized more easily than the Fe of the steel tank. So the Fe will not begin to oxidize to any great extent until the Mg rod is gone. The Mg is a sacrificial metal.