1. For a half-reaction, the more (+) the E o red value, the greater the tendency for that reaction to “go” in that direction (i.e., reduction). Strongest oxidizer is… Other strong oxidizers are… the other halogens and oxyanions in which the central atom has a large ___ charge. e.g., MnO 4 –, Cr 2 O 7 2–, ClO 3 –, etc. (Cl 2, Br 2, I 2 ) F 2 (g) + 2 e –  2 F – (aq) E o red = +2.87 V (+) Strong oxidizers LOVE their own e – s (and they want everybody else’s, too). F2F2

2. Poorest oxidizer is… Li. Li + (aq) + e –  Li(s) E o red = –3.05 V (–) sign indicates poor tendency to “go” in this direction, but large magnitude (i.e., 3.05 V) shows strong tendency to “go” in other direction (i.e., oxidation). Lithium batteries take advantage of lithium’s strong tendency to BE oxidized (i.e., to REDUCE other stuff.) Poor oxidizers HATE their own e – s (and have no interest in accepting anyone else’s e – s, either.)

3. In comparing the reduction potentials of two half- reactions, consider the scale shown. The “higher- up” reaction is the reduction half-cell; the “lower- down” reaction is the oxidation half-cell. 0 V (+) V (–) V A– C– A– B– C– Red. Ox. Red. Ox. Red. Ox.

4. Spontaneity of Redox Reactions E o cell = E o red,cath – E o red,an (same equation as before) -- The Activity Series is based on standard reduction potentials. If E o (or E, or emf) is +…spontaneous. –…nonspontaneous. standard conditions nonstandard conditions Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb H 2 Sb Bi Cu Hg Ag Pt Au “e – haters” “e – lovers” Activity Series for Metals

5. Relationship between E and  G…  G = –nFE In standard states…  G o = –nFE o n = # of mol of transferred e – F = Faraday’s constant = 96,500 J V. mol e – Michael FaradayJosiah Gibbs (1839–1903)(1791–1867) C mol e – ( )

6. For… 5 Fe 2+ + MnO 4 – + 8 H +  5 Fe 3+ + Mn 2+ + 4 H 2 O (a) What is n? (b) Find  G o. (a) n = 5 (b) Fe 3+ + e –  Fe 2+ E o red = 0.77 V MnO 4 – + 8 H + + 5 e –  Mn 2+ + 4 H 2 O E o red = 1.51 V E o cell = E o red,cath – E o red,an E o cell = 1.51 V – 0.77 V = 0.74 V  G o = –nFE o = –5 mol e – (0.74 V) 96,500 J V. mol e – () = –357 kJ (SPONTANEOUS)

7. Effect of Concentration on Cell EMF Cell emf drops gradually due to changing concentrations of reactants and products. When emf = 0 V, cell is “dead.” Nernst equation: At 25 o C (298 K): Walther Nernst (1864–1941)

8. If [Cd 2+ ] = 2.0 M and [Fe 2+ ] = 0.030 M… = –0.01 V Fe(s) + Cd 2+ (aq)  Cd(s) + Fe 2+ (aq) Find emf at 25 o C when [Cd 2+ ] = 0.030 M and [Fe 2+ ] = 2.0 M. Fe 2+ (aq) + 2 e –  Fe(s) E o red = –0.44 V Cd 2+ (aq) + 2 e –  Cd(s) E o red = –0.40 V E o = +0.04 V = 0.09 V nonspont. spont.