Cell Voltages

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. - DGro DEo = n F DEo is the standard cell voltage or potential of a standard cell - when all reactants and products are in their standard state. DGro = - n F DEo DEo > 0 ; spontaneous cell reaction under standard conditions

3. A standard Cr3+(aq)|Cr(s) and a standard Co2+(aq)|Co(s) half cell are connected to make a galvanic cell. The voltage of the cell equals 0.464 V at 25oC. Write an equation to represent the reaction taking place in the cell and calculate its DGor. Cr(s)  Cr3+(aq) + 3e- anode Co2+(aq) + 2e- Co(s) cathode Overall cell reaction 2Cr(s) + 3 Co2+(aq)  2Cr3+(aq) + 3 Co(s) DEo = 0.464 V DGor= - n F DEo = - (6 moles) (9.64853 x 104 coulomb/mole) (0.464 V) = - 2.69 x 105 J or - 269 kJ

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

5. Cr3+(aq) + 3e-  Cr(s) Eo(Cr3+|Cr) = - 0.744 V Co2+(aq) + 2e- Co(s) Eo(Co2+|Co) = - 0.28 V Eo(Co2+|Co) > Eo(Cr3+|Cr)

6. Define the following half reaction to be the reference 2 H+(aq, 1 M) + 2e- -> H2 (g, P = 1 atm) Eo = 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 < 0 => 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) | Zn2+ (aq) || Sn4+ (aq), Sn2+ (aq) | Pt (s) is + 0.91 V DEo = Eo (cathode) - Eo (anode) + 0.91 V = Eo (Sn4+ (aq), Sn2+ (aq) ) - Eo (Zn(s) | Zn2+ (aq) ) Eo (Sn4+ (aq), Sn2+ (aq) ) = + 0.91 V + Eo (Zn(s) | Zn2+ (aq) ) = 0.91V - 0.76 V = + 0.15 V

11. Using standard reduction potentials Strong oxidizing agents - have large positive standard reduction potentials Examples: F2, MnO4-, H2O2 O2 (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)  M2+(aq) + H2(g) spontaneous if Eo(M2+|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 Fe2+(aq) in its standard state unstable with respect to disproportionation at 25oC? Fe3+(aq) + e- Fe2+(aq) Eo = 0.771 V Fe2+(aq) + 2e- Fe(s) Eo = -0.477 V Overall disproportionation reaction 3 Fe2+(aq)  2 Fe3+(aq) + Fe(s) DEo = -0.477 - 0.771 = -1.218 V No disproportionation

14. Effect of Concentration on DE DGr = DGro + RT ln Q DGr = - n FDE DGro = - n FDEo - n FDE = - n FDEo + RT ln Q DE = DEo - (RT/ n F ) ln Q Nernst Equation relates cell voltage with concentrations of reactants and products (through Q)

15. At 25.00oC (298.15 K), R T / F = 0.025693 V DE = DEo - (0.025693 / n ) ln Q The reduction potential of a non-standard half cell is: E = Eo - (RT/ nhcF ) ln Qhc For Zn2+ (aq) + 2 e- -> Zn(s) E = Eo - (RT/ 2 F ) ln (1 / [Zn2+(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 H3O+(aq) concentration, and the other electrode serves as a reference. A calomel electrode has a reduction half reaction Hg2Cl2 (s) + 2 e- -> 2 Hg(l) + 2 Cl- (aq) Eo = +0.27 V When combined with the H+(aq)/H2(g) electrode, the overall cell reaction is: Hg2Cl2 (s) + H2 (g) -> 2 H+ (aq) + 2 Hg(l) + 2 Cl- (aq)

17. Q = [H+(aq)]2 [Cl- (aq)]2 / PH2 If PH2 is held at 1 atm then Q = [H+(aq)]2 [Cl- (aq)]2 DE = DEo - (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. DE depends only on [H+(aq)]. Other electrodes are selectively sensitive to ions such as Ca2+, NH4+, Na+, S2-.