Levels Tought:
Elementary,Middle School,High School,College,University,PHD
Teaching Since: | Apr 2017 |
Last Sign in: | 234 Weeks Ago |
Questions Answered: | 12843 |
Tutorials Posted: | 12834 |
MBA, Ph.D in Management
Harvard university
Feb-1997 - Aug-2003
Professor
Strayer University
Jan-2007 - Present
1. Consider the following half-reaction, for which E° = +1.630 V: HOCl(aq) + H+(aq) + e− ⇌ ½Cl2(g) + H2O(l)
Â
a. (1 pt) Write the acid dissociation reaction for HOCl, including physical states. Although this is not a redox reaction, we can still calculate its potential. [See pages 43-44 of the Equation Guide.] Look up the pKa in Appendix G of Harris, then show that E° = −0.445 V at 25 °C. Repeat for the autoprotolysis of water reaction and calculate its E°.
Â
b. (1 pt) The half-reaction above is written for acidic conditions, with H+ and the protonated form of the weak acid. Convert it to basic conditions (writing it with OH− and the deprotonated form of the weak acid) by combining the half reaction above with the reactions you wrote in part a, then calculate its E°.
Â
c. (2 pts) Calculate E°′ (pH 7) for the original half-reaction.
Â
2. (1 pt) Consider the following half-reaction, for which E° = +0.222 V: AgCl(s) + e− ⇌ Ag(s) + Cl−(aq)Â
Â
With a saturated KCl fill solution, the potential of this half-reaction is +0.197 V. Use the Nernst equation to calculate the activity of the chloride ion in the fill solution. At 25 °C, saturated KCl is ~4.0 M. Why does the calculated activity differ from this concentration?
-----------