Question

0. (15 points) An analytical chemist conducted the titration of 50.00 mL of 0.100 M Ag using 0.150 M CT solution. Silver wire is the indicator electrode and the saturated Ag-AgCl electrode is the reference electrode. E = 0.197 V for the saturated Ag-AgCl electrode, EºAg+/Ag= 0.1993 and Ksp = 1.8 x 10-8 for AgCI a) Calculate E at the volume of Cl' is 25.0 mL. Show all your work. b) Calculate E at the volume of Cl' is 50.0 mL. Show all your work.

Answer 1

Formula: E_cell = E^o_cell - 0.059 V/n * Log(1/[Ag⁺]) = E^o_cell + 0.059 V/n * Log[Ag⁺]

Here, n = no. of electrons tranferred in the cell reaction = 1

Now, Ksp = [Ag⁺][Cl^-], then [Ag⁺] = Ksp/[Cl^-]

(a) [Cl^-] = 25 mL * 0.15 M/(25+50) mL = 0.05 M

i.e. E = 0.799 V + 0.059 V/1 * Log(1.8*10^-8/0.05) = 0.419 V

(b) [Cl^-] = 50 mL * 0.15 M/(50+50) mL = 0.075 M

i.e. E = 0.799 V + 0.059 V/1 * Log(1.8*10^-8/0.075) = 0.408 V