Question

The potential of a silver electrode is measured relative to an Ag-AgCl electrode for the titration of 100.0 mL of 0.100 M Cl- with 0.100 M Ag^+. What is the potential after 75.00 mL of titrant is added. E^o = 0.799 V for Ag+, E = 0.197 V for the Ag-AgCl electrode and Ksp = 1.8 × 10⁻¹⁰.

A) 0.493 V

B) 1.070 V

C) 0.521 V

D) 1.267 V

E) 0.134 V

Answer is E, can someone show me the work leading to this answer.

Answer 1

The titration reaction will be:

Ag⁺ + Cl^- ---> AgCl(s)

[Cl-] = 0.1 M

volume = 100 mL

[Ag+] = 0.1 M

volume of Ag+ = 75 mL

At this point, 75% of chloride ions will get precipitated and 25% will remain in solution.

initial moles of Chloride = molarity X volume = 0.1 X 0.1 = 0.01 moles

moles reacted = 0.1 X 0.075 = 0.0075 moles

moles remain unreacted = 0.0025 moles

[Cl- ] left = moles / total volume = 0.0025 X 1000 / (100+75) = 0.0143 M

Ksp = 1.8 × 10⁻¹⁰ = [Ag+][Cl-]

[Ag+ ]= 1.8 × 10⁻¹⁰ / 0.0143 M = 1.26 X 10⁻⁸ M

The equation for silver electrode and cell is

E= 0.799 + 0.059 X log[Ag⁺] - (0.197) =0 .602 + 0.0592 log(1.26 X 10⁻⁸)= 0.602 -0.468 = 0.134 V