Question

Let's say that the cell compartment for Ag+/Ag contains 100.0 mL of 1.0 M solution. Add 100.0 mL of 1.0 M NaCl to that compartment. Ksp of AgCl is 1.6 x 10^-10 so the Qsp > Ksp and solid AgCl will precipitate out of solution. This silver chloride is in equilibrium with its ions according to the Ksp equilibrium. Calculate ​[Ag+] (HINT: it will be very small) and then use your result to calculate the cell potential to the hundredths (2 decimal places) with this concentration of ​[Ag+].

E (Ag+) = 0.80V

E (Cu2+) = 0.34V

Answer 1

Ksp of AgCl = [Ag⁺][Cl^-]

Here, [Cl^-] = [NaCl] = 100 mL * 1 M/(100+100) mL = 0.5 M

i.e. 1.6*10^-10 = [Ag⁺] * 0.5 M

Therefore, [Ag⁺] = 3.2*10^-10 M

Now, the cell potential (E^o_cell) = E^o_Ag⁺/Ag  - E^o_Cu²⁺/Cu = 0.80 - 0.34 = 0.46 V