A cell has two lead electrodes. One dips into a saturated solution of lead chloride, PbCl₂, and the…

Step 1. The solubility. $\log\dfrac{0.50}{s} = \dfrac{0.0440}{0.02955} = 1.489$, so $\dfrac{0.50}{s} = 30.8$ and $s = 0.0162\ \text{mol/L}$.

Step 2. The solubility product. $[\text{Pb}^{2+}] = 0.0162\ \text{mol/L}$ and $[\text{Cl}^-] = 2s$. So $K_{sp} = 4s^3 = 4 \times (0.0162)^3 = 1.7 \times 10^{-5}\ \text{mol}^3/\text{L}^3$.

Step 3. Extra chloride. With $[\text{Cl}^-] = 0.10\ \text{mol/L}$, $[\text{Pb}^{2+}] = \dfrac{1.7 \times 10^{-5}}{(0.10)^2} = 1.7 \times 10^{-3}\ \text{mol/L}$.

Step 4. The new emf. $E = 0.02955\ \text{V} \times \log\dfrac{0.50}{1.7 \times 10^{-3}} = 0.02955\ \text{V} \times 2.469 = 0.0729\ \text{V}$.