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
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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? 


