At 298 K, E^°(Ag^+/Ag)=+0.80 V and E^°(AgCl/Ag,Cl^-)=+0.22 V. Take (2.303RT)/(F)=0.059 V. The solubility of AgCl in 0.010 M aqueous NaCl (ignore…

The silver chloride electrode is the $\mathrm{Ag^+/Ag}$ couple with $[\mathrm{Ag^+}]$ fixed by $K_{sp}$, so $E^\circ(\mathrm{AgCl/Ag})=E^\circ(\mathrm{Ag^+/Ag})+0.059\log K_{sp}$. Hence $\log K_{sp}=(0.22-0.80)/0.059=-9.83$ and $K_{sp}=1.48\times10^{-10}$. In pure water the solubility would be $\sqrt{K_{sp}}=1.2\times10^{-5}\ \mathrm{M}$ (option A, the common-ion-ignoring trap). In $0.010\ \mathrm{M}$ chloride, $s(0.010+s)=K_{sp}$, and since $s\ll0.010$, $s=K_{sp}/0.010=1.48\times10^{-8}\ \mathrm{M}$. Option B is just $K_{sp}$ divided by $1$, and D results from using $0.0001$ for the chloride concentration. The answer is C. As a check, the standard cell $\mathrm{Ag|AgCl|Cl^-\|Ag^+|Ag}$ has $E^\circ=0.80-0.22=0.58\ \mathrm{V}$ and $\log K=E^\circ/0.059$ for the reverse dissolution reaction gives the same $K_{sp}$.