# Calculating Solubility Product of AgCl from electrode potential

I have to use a measured electrode potential to find the solubility product of $$\ce{AgCl}$$. In our experiment, we first mixed one drop of $$\pu{1.0 M}$$ $$\ce{AgNO3}$$ with $$\pu{10 mL}$$ of $$\pu{1.0 M}$$ $$\ce{NaCl}$$ to precipitate $$\ce{AgCl}$$. For the purposes of our experiment, we are to assume that the concentration of $$\ce{Cl-}$$ remains $$\pu{1.0 M}$$, and use the Nernst equation to determine the value of $$\ce{[Ag+]}$$. The reaction that occurred in our cell was between $$\pu{1 mL}$$ of the solution of $$\ce{Ag+}$$ and $$\ce{Cl-}$$ left behind from the precipitation reaction described above, and $$\pu{1 mL}$$ of of $$\pu{0.10 M}$$ $$\ce{Zn^2+}$$. The reactions were all carried out at $$\pu{25 ^\circ C}$$, and the voltage obtained by our group was $$\pu{0.85 V}$$.

# My Method

The half-reactions going on in this cell were: \begin{align} &&\ce{2Ag+(aq) + 2e- &-> 2Ag(s)} & (E^\circ &= 0.80)\\ +&&\ce{Zn(s) &-> Zn^2+(aq) + 2e-} & (E^\circ &= 0.76)\\\hline &&\ce{2Ag+(aq) + Zn(s) &-> Zn^2+(aq) + 2Ag(s)} & (E^\circ_\mathrm{cell} & = 1.56) \end{align}

From here, I used the Nernst equation to solve for $$\ce{[Ag+]}$$, with $$n=2$$ moles of electrons transferred and $$Q = \ce{[Zn^2+]}/\ce{[Ag+]^2}$$:

$$0.85 = 1.56 - \frac{0.0592}{2}\log\frac{0.10}{[\ce{Ag+}]^2}$$

Solving this for $$\ce{[Ag+]}$$ yields approximately $$3.21 \times 10^{-13}$$.

I then applied the law of mass action to determine the solubility product of $$\ce{AgCl}$$: $$\ce{[Ag+][Cl^{-}]} = [3.21 \times 10^{-13}] \times [1.0] = 3.21 \times 10^{-13}$$

However, this number is quite different from the expected value of $$1.8 \times 10^{-10}$$. I am wondering if my math or approach is wrong or if my group's data is simply wrong?