1. Cell Voltages To compare cells compare voltages of cells in their standard state. Standard States For solids and liquids: the state of the pure solid or liquid at 1 atm and at a specified temperature For gases: the gaseous phase at 1 atm and at a specified temperature For solutions : Concentrations of 1 mol/L under a pressure of 1 atm and a specified temperature. (temperature typically specified at 298.15 K)

2.  E o is the standard cell voltage or potential of a standard cell - when all reactants and products are in their standard state.  G r o = - n F  E o  E o = -  G r o n F  E o > 0 ; spontaneous cell reaction under standard conditions

3. A standard Cr 3+ (aq)|Cr(s) and a standard Co 2+ (aq)|Co(s) half cell are connected to make a galvanic cell. The voltage of the cell equals 0.464 V at 25 o C. Write an equation to represent the reaction taking place in the cell and calculate its  G o r. Cr(s)  Cr 3+ (aq) + 3e - anode Co 2+ (aq) + 2e -  Co(s)cathode Overall cell reaction 2Cr(s) + 3 Co 2+ (aq)  2Cr 3+ (aq) + 3 Co(s)  E o = 0.464 V  G o r = - n F  E o = - (6 moles) (9.64853 x 10 4 coulomb/mole) (0.464 V) = - 2.69 x 10 5 J or - 269 kJ

4. Standard Reduction Potentials Under standard conditions:  E o = E o (right half cell) - E o (left half cell) For a galvanic cell:  E o = E o (cathode) - E o (anode) where the E o are the standard reduction potentials of the electrodes. For  E o > 0; spontaneous cell reaction To determine which of two half cells will be the anode and which the cathode compare standard reduction potentials

5. Cr 3+ (aq) + 3e -  Cr(s) E o (Cr 3+ |Cr) = - 0.744 V Co 2+ (aq) + 2e -  Co(s) E o (Co 2+ |Co) = - 0.28 V E o (Co 2+ |Co) > E o (Cr 3+ |Cr)

6. Define the following half reaction to be the reference 2 H + (aq, 1 M) + 2e - -> H 2 (g, P = 1 atm) E o = 0 V All standard reduction potentials are determined relative to this reference. If the standard reduction potential of a half reaction is > 0 => greater tendency to be reduced relative to H + (aq, 1 M) If standard reduction potential of a half reaction lower tendency to be reduced relative to H + (aq, 1 M)

7. In general, the more positive the standard reduction potential, the greater the electron-pulling power of the reduction half reaction, and therefore the more oxidizing the species The more negative the standard reduction potential, the greater the electron-donating power of the oxidation half reaction, and therefore the more reducing the species

9. Variation of standard reduction potentials. The most negative values occur in the s block and the most positive values occur close to fluorine.

10. The standard potential of an electrode can be determined by setting up a standard cell in which one electrode has a known standard potential and measuring the resulting cell voltage. For example, the standard potential of a zinc electrode is -0.76 V, and the standard emf of the cell Zn(s) | Zn 2+ (aq) || Sn 4+ (aq), Sn 2+ (aq) | Pt (s) is + 0.91 V  E o = E o (cathode) - E o (anode) + 0.91 V = E o (Sn 4+ (aq), Sn 2+ (aq) ) - E o (Zn(s) | Zn 2+ (aq) ) E o (Sn 4+ (aq), Sn 2+ (aq) ) = + 0.91 V + E o (Zn(s) | Zn 2+ (aq) ) = 0.91V - 0.76 V = + 0.15 V

11. Using standard reduction potentials Strong oxidizing agents - have large positive standard reduction potentials Examples: F 2, MnO 4 -, H 2 O 2 O 2 (in acidic medium) is a fairly strong oxidizing agent. Strong reducing agents - have large negative standard reduction potentials Examples: Na, Li

12. Electrochemical series: list of relative strengths of oxidizing and reducing agents. The strongest oxidizing agents are at the top of the table; the strongest reducing agents are at the bottom M(s) + 2H + (aq)  M 2+ (aq) + H 2 (g) spontaneous if E o (M 2+ |M) < 0

13. Disproportionation: a single species is both reduced and oxidized. Must be able to both give up and accept electrons Half reaction in which the species is reduced must have a larger reduction potential than the half reaction in which it is oxidized. Is Fe 2+ (aq) in its standard state unstable with respect to disproportionation at 25 o C? Fe 3+ (aq) + e -  Fe 2+ (aq)E o = 0.771 V Fe 2+ (aq) + 2e -  Fe(s)E o = -0.477 V Overall disproportionation reaction 3 Fe 2+ (aq)  2 Fe 3+ (aq) + Fe(s)  E o = -0.477 - 0.771 = -1.218 V No disproportionation

14. Effect of Concentration on  E  G r =  G r o + RT ln Q  G r = - n F  E  G r o = - n F  E o  - n F  E = - n F  E o + RT ln Q  E =  E o - (RT/ n F ) ln QNernst Equation relates cell voltage with concentrations of reactants and products (through Q)

15. At 25.00 o C (298.15 K), R T / F = 0.025693 V  E =  E o - (0.025693 / n ) ln Q The reduction potential of a non-standard half cell is: E = E o - (RT/ n hc F ) ln Q hc For Zn 2+ (aq) + 2 e - -> Zn(s) E = E o - (RT/ 2 F ) ln (1 / [Zn 2+ (aq)]

16. Ion-Selective Electrodes pH or concentration of ions can be measured by using an electrode that responds selectively to only one species of ion. In a pH meter, one electrode is sensitive to the H 3 O + (aq) concentration, and the other electrode serves as a reference. A calomel electrode has a reduction half reaction Hg 2 Cl 2 (s) + 2 e - -> 2 Hg(l) + 2 Cl - (aq)E o = +0.27 V When combined with the H + (aq)/H 2 (g) electrode, the overall cell reaction is: Hg 2 Cl 2 (s) + H 2 (g) -> 2 H + (aq) + 2 Hg(l) + 2 Cl - (aq)

17. Q = [H + (aq)] 2 [Cl - (aq)] 2 / P H 2 If P H 2 is held at 1 atm then Q = [H + (aq)] 2 [Cl - (aq)] 2  E =  E o - (RT/ n F ) ln [H + (aq)] 2 [Cl - (aq)] 2 The [Cl - (aq)] is held constant since the calomel electrode consists of a saturated solution of KCl.  E depends only on [H + (aq)]. Other electrodes are selectively sensitive to ions such as Ca 2+, NH 4 +, Na +, S 2-.