Lecture 26 11/04/05

Reactions with solution of: F 2 and Ag(s), F 2 and Ag +, F - and Ag(s),F - and Ag +

Nernst Equation for a Reaction 1. Write reduction half-reactions and E o 1.Multiply half-reactions, but not E o to balance e’s 2. Nernst equation for right half-cell (reduction) 3. Nernst equation for left half-cell (oxidation) 4. Find net cell voltage (E° cell = E° red - E° ox. ) 1.E° cell > 0 reaction goes right 2.E° cell < 0 reaction goes left 5. Balance net cell reaction

Latimer Diagram To find E o for unknown reduction half-reaction Add  G o ’s, not E o ’s  G o unknown =  G o 1 +  G o  G o # -n unkown FE o unknown = -n 1 FE o 1 + -n 2 FE o –nFE o # E o unknown = n 1 E o 1 + n 2 E o nE o # n unkown

MnO 4 2-  Mn 3+

Calomel Electrode Another type of reference electrode Hg 2 Cl 2 (s) + 2e -  2Hg(l) +2Cl -

Saturated Calomel Electrode (SCE) Most common of the calomel electrodes –0.1 M and 1 M are the other common ones Saturated with KCl –[Cl - ] doesn’t change if some evaporation E o = V at 25 o C Problems –KCl solubility is temperature dependent –Hg 2 Cl 2 breaks down at higher temps

Redox titrations Titrating 100 mL of 0.05 M Fe 2+ with 0.1 M Ce 4+ in 1 M HClO 4 Reference electrode: SCE Ce 4+ + Fe 2+  Ce 3+ + Fe 3+ Ce 4+ + e -  Ce 3+ E° = 1.70 V Fe 3+ + e -  Fe 2+ E° = V X-Axis: volume Ce 4+ added Y-Axis: potential (voltage vs. SCE)

Two things happening: First: Titration reaction (Ce 4+ + Fe 2+ ⇄ Ce 3+ + Fe 3+ ) –Goes to completion –Before equiv. point: excess Fe 2+ –After equiv. point: excess Ce 4+ Second: Reduction of Ce 4+ or Fe 3+ from cell potential –Negligible effect on concentration –Produces voltage for y-axis –2Fe 3+ +2Hg(l) +2Cl - ⇄ 2Fe 2+ +Hg 2 Cl 2 (s) –2Ce 4+ +2Hg(l) +2Cl - ⇄ 2Ce 3+ +Hg 2 Cl 2 (s)

At the equivalence point X-axis: CV=CV Y-axis: [Ce 3+ ]=[Fe 3+ ] and [Ce 4+ ]=[Fe 2+ ] Ce 3+ + Fe 3+ ⇄ Ce 4+ + Fe 2+ E cathode = ½(E° Ce + E° Fe ) E cell = ½(E° Ce + E° Fe ) E cell = ½( ) E cell = 0.99 V