Question

In order to prepare the solution for the left half-cell, mix appropriate volumes of the 0.1 M silver nitrate and 0.2 M potassium chloride stock solutions.

Ag│AgCl(s), Cl− (0.05 M)║Ag+ (0.1 M)│Ag

You can assume that Ag+ and Cl− will react quantitatively to form AgCl, which precipitates out of the solution. Carefully consider what should be the limiting reagent in this reaction if you want to have “unreacted” [Cl− ] = 0.05 M remaining in solution after this reaction is complete and you are ready to begin the electrochemical reaction. Show your work and describe how you will achieve this.

Answer 1

Equation

AgNO₃ + KCl ------------> AgCl + KNO₃

Ionic equation

Ag⁺ + Cl^- + K⁺ + NO3^-   ---------------> AgCl + K⁺ + NO3^-

since concentration of [Cl-] = 0.05 M, the limiting reagent must be KCl

concentration of Ag⁺ ion before reaction is 0.1 M

concentration of NO₃^- ion before reaction is 0.1 M

Concentration of K⁺ ion before reaction is 0.2 M

Concentrartion of Cl^- ion before reaction is 0.2 M

after reaction remaining concentration of Cl^- ion will be 0.1 M

Therefore limiting reagent will be Potassium chloride (KCl)