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Experiment 5

Experiment 5. Determining the Ksp of AgCl using Potentiometric Titrations. Adding together redox reactions. Ag + + e- ↔ Ag (s) E 0 1 AgCl (s) + e- ↔ Ag (s) + Cl - E 0 2. Ag + + e- ↔ Ag (s) E 0 + Ag (s) + Cl - ↔ AgCl (s) + e- E 0 - __________________________________

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Experiment 5

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  1. Experiment 5 Determining the Ksp of AgCl using Potentiometric Titrations

  2. Adding together redox reactions Ag+ + e- ↔ Ag(s) E01 AgCl(s) + e- ↔ Ag(s) + Cl- E02

  3. Ag+ + e- ↔ Ag(s) E0+ Ag(s) + Cl- ↔ AgCl(s) + e- E0-__________________________________ Ag+ + Cl- ↔ AgCl(s) E0cell = E0+ - E0-

  4. Ag+ + Cl- ↔ AgCl(s) E0cell = E0+ - E0- DG0 = -nFE0cell = -RT ln(Ksp) Ksp = e-DG0/RT = e(nF/RT)E0cell = eE0cell/59.16

  5. Adding together redox reactions Ag+ + e- ↔ Ag(s) E01 AgCl(s) + e- ↔ Ag(s) + Cl- E02

  6. Experimental Setup 0.1 M KCl in buret Double junction Ag/AgCl Reference Electrode Ag bullet Electrode Voltmeter [Ag+] = 0.10

  7. Potentiometric titration • Use very low current electrochemical cell to monitor the course of a redox tiratration • The reactions at the cathode do not change the bulk concentrations of reagents

  8. Experiment: Part 1 • Add 0.0500 M AgNO3 to 100 mL of 0.500 M KNO3. • Titration reaction; none. This is not a titration • Cathode reaction: Ag+ + e- ↔ Ag(s) • Cell Voltage increases because the Ag+ ion concentration increases • Ecell = E0(Ag+/Ag) – 59.16log(1/[Ag+]) - Eref

  9. Part 2 • Add a diluted solution of KCl • Titration reaction; Ag+ + Cl-→ AgCl(s) Equivalence point reached when just enough Cl- reacts with all the Ag+ • Cathode reaction before Equiv. point: Ag+ + e- ↔ Ag(s) • Cell Voltage decreases because the Ag+ ion concentration decreases • Ecell = E0(Ag+/Ag) – 59.16log(1/[Ag+]) - Eref

  10. At the Equivalence point • A sharp drop in cell voltage indicates when the equivalence point is reached • The equivalence point is used to determine the concentration of the diluted KCl solution

  11. After the Equivalence Point • In bulk: Adding excess Cl- • Reaction at cathode; AgCl(s) + e- ↔ Ag(s) + Cl- • Cell Voltage decreases because the Cl- ion concentration increases • Ecell = E0(AgCl/Ag) – 59.16log([Cl-]) - Eref

  12. Recap • Before Equivalence point: Ecell = E0(Ag+/Ag) – 59.16log(1/[Ag+]) – Eref Plot Ecell vs. log(1/[Ag+]) and intercept = E0(Ag+/Ag) - Eref

  13. Recap • After Equivalence point: Ecell = E0(Ag/AgCl) – 59.16log[Cl-] – Eref Plot Ecell vs. log [Cl-] and intercept = E0(Ag/AgCl) - Eref

  14. intercept 1 = E0(Ag+/Ag) - Eref intercept 2 = E0(AgCl/Ag) – Eref Int 1 – Int 2 = E0(Ag+/Ag) - E0(AgCl/Ag) Ksp = e(Int1 –Int 2)/59.16

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