1. Electrochemistry - the Science of Oxidation-Reduction Reactions 1.Constructing electrochemical cells - sketching cells which carry out redox reaction - electrodes and salt bridges 2.Diagramming the cell - cell conventions, anode and cathode 3.Calculating the cell potential and  G° 4.Galvanic and electrolysis cells - Faraday’s Laws

2. Constructing Electrochemical Cells Cu(s) Cu 2+ (aq) + 2 e - Ag + (aq) + e - Ag(s) Oxidation half-reaction on the left Reduction half-reaction on the right Overall cell reaction: Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Ag(s) Cu(s) Cu 2+ (aq) voltmeter salt bridge Ag + (aq) Cu(s) | Cu 2+ (aq) || Ag + (aq) | Ag(s)

3. The Standard Cell Potential Cu 2+ (aq) + 2 e - Cu(s) Ag + (aq) + e - Ag(s) Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E° = + 0.3442 V E° = + 0.7996 V E° = - (+0.3442) + (+.7996) = + 0.4554 V Ag(s) (cathode, positive) Cu 2+ (aq) salt bridge Ag + (aq) voltmeter Cu(s) (anode, negative)

4. Standard Reduction Potentials (Appendix E)

5. Relationship of Cell potential and  G Cu 2+ (aq) + 2 e - Cu(s) Ag + (aq) + e - Ag(s) Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E° = + 0.3442 V E° = + 0.7996 V  E° = - (+0.3442) + (+.7996) = 0.4554 V The cell potential directly measures  G for the reaction:  G = - n F  E n = moles of electrons transferred in the cell reaction F = 96,485 C-mol -1 (the Faraday, charge of 1 mol of protons)  G° = - (2) (96 485) (0.4554 V) = - 87880 kJ (We have assumed standard conditions, i.e., unit activities.)

6. Diagramming the Cell Ag(s) Cu(s) Cu 2+ (aq, 0.05 M) salt bridge Ag + (aq, 0.15 M) voltmeter Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Cu(s) | Cu 2+ (aq, 0.05 M) || Ag + (0.15M) | Ag(s) metal electrode (anode) phase boundary salt bridge phase boundary metal electrode (cathode)

7. Draw a Cell which Carries Out the Following Reaction Overall cell reaction: 4 Fe 3+ (aq) + 6 H 2 O(l) O 2 (g) + 4 H 3 O + (aq) + 4 Fe 2+ (aq) O 2 (g) + 4 H 3 O + (aq) + 4 e - 6 H 2 O (l) Fe 3+ (aq) + e - Fe 2+ (aq) Pt(s) voltmeter Fe 3+ (aq) Fe 2+ (aq) Pt(s) | O 2 (g, P O2 ) | H 3 O + (x M) || Fe 2+ (y M), Fe 3+ (z M) | Pt(s) Pt(s) H 3 O + (aq) salt bridge O 2 (g) Answer: If the cell is not a standard cell, you must specify partial pressures and concentrations.

8. Further Description of the Cell 2. Diagram the cell: 1. Calculate the standard cell potential: O 2 (g) + 4 H 3 O + (aq) + 4 e - 6 H 2 OE° = +1.229 V Fe 3+ (aq) + e - Fe 2+ (aq)E° = +0.770 V 3.Calculate  G° for the cell reaction: Is the reaction spontaneous as written? no Which electrode is the anode? the cathode?  G° = +177 kJ 4. Which electrode has a positive potential in the galvanic cell? In a galvanic cell, the anode has a negative potential, the cathode a positive potential In an electrolytic cell, the anode is positive, the cathode negative.. 4 Fe 2+ (aq) + 6 H 2 O(l) O 2 (g) + 4 H 3 O + (aq) + 4 Fe 3+ (aq) E° = - 0.459 V Pt(s) | Fe 2+ (aq), Fe 3+ (aq) || H 3 O + (aq) | O 2 (g, 1 atm) | Pt(s) M(-1) M(+1)

9. Analyzing Cells: a Step-by-Step Method 1.Write down the cell reaction. Note that the reaction may be spontaneous or non- spontaneous as written. 2.Calculate  E° cell. Write both half-reactions as reduction half-reactions (even if the half- reaction occurs as an oxidation in the cell reaction). Find the standard reduction potentials for both reduction half-reactions. Add the half-reactions to give the overall cell reaction. Notice how this is done in the example. 3.If  E° cell is positive, the reaction is spontaneous as written under standard conditions. If not, the reverse reaction is spontaneous under standard conditions. Cu 2+ (aq) + 2 e - Cu(s) Ag + (aq) + e - Ag(s) Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E° = + 0.3442 V E° = + 0.7996 V  E° = - (+0.3442) + (+.7996) = + 0.4554 V M(-1) M(+2) The multiplicative factor is negative, so E° is multiplied by (-1) The muliplicative of (+2) is NOT applied to E°. Cell potentials are an intrinsic, not extrinsic, property.

10. Galvanic Cells and Electrolysis Cells Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Ag(s) Cu(s) Cu 2+ (aq) salt bridge Ag + (aq) Galvanic cell: cell reaction runs in the spontaneous direction Electroytic cell: cell reaction runs in the non-spontaneous direction, driven by an applied electrical potential (a voltage) that is greater than the spontaneous potential galvanic: E°= 0.+4554 V electrolytic: E (applied) > E° positive negative cathode if galvanic anode if electrolytic anode if galvanic cathode if electrolytic

11. Cathode and Anode in Galvanic Cells and Electrolysis Cells Cu(s) (negative) Cu 2+ (aq) salt bridge Ag + (aq) By definition, the cathode is where reduction occurs and the anode is where oxidation occurs. In the galvanic cell, the copper electrode is the cathode, silver is the anode. In the electroytic cell, the silver electrode is the cathode, copper is the anode. Note that the sign of the potential doesn't change - copper is always negative, silver is always positive. But the designations "cathode" and "anode" change between galvanic and electroytic cells. Ag(s) (positive)

12. Analyzing Cells: a Step-by-Step Method 4.Draw a sketch of the cell. The anode (where the oxidation occurs) is the left hand compartment, the cathode is the right hand compartment. Note that if  E° cell is positive, the reaction is spontaneous as written under standard conditions. If not, the reverse reaction is spontaneous under standard conditions. Cu(s) Cu 2+ (aq) + 2 e - Ag + (aq) + e - Ag(s) Oxidation half-reaction on the left Reduction half-reaction on the right Ag(s) Cu(s) Cu 2+ (aq) voltmeter salt bridge Ag + (aq) sketch of the cell

13. Analyzing Cells: a Step-by-Step Method 5.Diagram the cell. Note that diagramming the cell is a formal procedure following well-defined rules. It is not the same as “sketching the cell”. Cu(s) | Cu 2+ (aq, 0.05 M) || Ag + (0.15M) | Ag(s) metal electrode (anode) phase boundary salt bridge phase boundary metal electrode (cathode) Ag(s) Cu(s) Cu 2+ (aq) voltmeter salt bridge Ag + (aq) sketch cell diagram

14. Analyzing Cells: a Step-by-Step Method 6.Calculate  G° 298 for the cell reaction using the general relationship:  G° = - n F  E° cell. If the cell potential is positive,  G° is negative and the reaction is spontaneous as written. If the cell potential is negative, the reaction as written is non-spontaneous. Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s)  E° = + 0.4554 V  G = - nF  E positive  E corresponds to a negative  G  G° 298 = - n F  E° cell = - 2 * 96485 *  E° 298 = - 87.9 kJ

15. Analyzing Cells: a Step-by-Step Method 7.Electroytic and Galvanic Cells Which electrode is the cathode (anode)? which is positive (negative)? By definition, the reduction occurs at the cathode, oxidation occurs at the anode. This is true in both galvanic and electrolytic cells. For a galvanic cell: The cell reaction occurs in the spontaneous direction. By convention, the sketch of the cell places the reduction half-reaction in the right-hand half-cell, which is therefore the cathode. The left-hand half- cell is the anode. For an electrolysis cell: The reaction runs in the non-spontaneous direction driven by an applied external voltage. The signs of the potentials of the two electrodes are unchanged, but the anode of the galvanic cell becomes the cathode of the electrolytic cell, and vice versa.

16. Analyzing Cells: a Step-by-Step Method 8.Standard and non-standard cells. In a standard cell, all activities are unity. When activities are not all unity, the cell is not a standard cell. In either case,  G is directly related to  E:  G° = - n F  E° cell (standard) or  G = - n F  E (non-standard) 9.The Nernst Equation describes the dependence of the cell potential on activities. since  G =  G° + RT ln e Q and  G = - n F E, it follows that  E =  E° - (RT/nF) ln e Q At T=298 K, this can be written in the form  E =  E° - (0.0592/n) log 10 Q(the Nernst Equation) Note that the Nernst Equation contains a base 10 logarithm, rather than base e. Thus  G and the cell potential contain precisely the same physical information. One can be calculated from the other. Also, since  G° = - RT ln e K, it follows that  E° cell = + (RT/nF) ln e K

17. Faraday's Laws of Electrolysis 3 Cu(s) + 2 Au 3+ (aq) 3 Cu 2+ (aq) + 2 Au(s) An electrical current of 1 ampere equals one coulomb per second - Q (coulombs) = I (amperes) * t (time) 1 mole of electrons has a charge of -96,485 C (the Faraday) moles of e- = (coulombs passed through cell) / 96485 = (I*t ) / 96,485 How much gold is deposited if 100 A is passed through this electrolysis cell for an hour? (Ans. 1.24 moles or 245 g) 1.Mass is proportional to electric charge passed through the cell. 2.Equivalent masses of different substances require equal amounts of electric charge passed through the cell.

18. Constructing Electrochemical Cells (1) Construct a cell which burns hydrogen and oxygen to water: (-2)* [2 H 3 O + (aq) + 2 e - H 2 (g) + 2 H 2 O(l) ] E° H+/H2 = 0.000 v (+1)* [O 2 (g) + 4 H 3 O + (aq) + 4 e - 6 H 2 O(l) ]E° O2/H2O = +1.229 v 2 H 2 (g) + O 2 (g) 2 H 2 O(l)  E° cell = +1.229 - (+0.000) = +1.229 volts Diagram the cell:Pt | H 2 (g) | H + (aq) || H + (aq) | O 2 (g) | Pt Draw the cell that carries out this reaction. Calculate Calculate K Is the cell reaction spontaneous as written? In the galvanic cell, what chemical reaction takes place at the anode? at the cathode? Which electrode is positive? How does this change in an electrolytic cell? For home practice:

19. Constructing Electrochemical Cells (2) Construct a cell to carry out the following redox reaction: (+1)* [PbO 2 (s) + SO 4 2- + 4 H 3 O + + 2 e - PbSO 4 (s) + 6 H 2 O ] E° = +1.685 v (-1)* [ I 2 (s) + 2 e - 2 I - (aq) ] E° = +0.535 v 2 I - (aq) + PbO 2 (s) + SO 4 2- + 4 H 3 O + I 2 (s) + PbSO 4 (s) + 6 H 2 O  E° cell = +1.685 - ( +0.535 ) = +1.150 volts Diagram the cell:Pt | I 2 (s) | I - (aq) || SO 4 2- (aq), H 3 O + | PbO 2 (s) | PbSO 4 (s) | Pt Draw the cell that carries out this reaction. Calculate In a galvanic cell, what chemical reaction takes place at the anode? at the cathode? Which electrode is positive? How does this change in an electrolytic cell? Is the cell reaction spontaneous as written? For home practice: