Given the following standard reduction potentials, $\begin{array}{ll}\mathrm{Pb}^{2+}(a q) +2 e^{-} \rightarrow \mathrm{Pb}(s) & E^{\circ}=-0.126 \mathrm{~V} \\\mathrm{PbSO}_{4}(s) +2 e^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) +\mathrm{SO}_{4}^{2-}(a q) & E^{\circ}=-0.355 \mathrm{~V}\end{array}$
Determine K_sp for PbSO₄(s) at 25°C.(R = 8.31 J/K?mol,F = 96480 C/mol e?)

Question

Given the following standard reduction potentials,

\begin{array}{ll}\mathrm{Pb}^{2+}(a q) +2 e^{-} \rightarrow \mathrm{Pb}(s) & E^{\circ}=-0.126 \mathrm{~V} \\\mathrm{PbSO}_{4}(s) +2 e^{-} \rightarrow \mathrm{Pb}(\mathrm{s}) +\mathrm{SO}_{4}^{2-}(a q) & E^{\circ}=-0.355 \mathrm{~V}\end{array}

Determine K_sp for PbSO₄(s) at 25°C.(R = 8.31 J/K?mol,F = 96480 C/mol e?)

Quizplus · Accepted Answer

The solubility product Ksp can be calculated using the Nernst equation and the difference in standard reduction potentials. The difference in potentials (ΔE°) is -0.126 V - (-0.355 V) = 0.229 V. Using the relation ΔG° = -nFE° and ΔG° = -RTlnK, we find Ksp = exp((nFE°)/(RT)) = exp((2 * 96480 * 0.229)/(8.31 * 298)) ≈ 1.8 × 10^-8.

Given the Following Standard Reduction Potentials Determine Ksp for PbSO4(s)at 25°C

Given the Following Standard Reduction Potentials Determine K_sp for PbSO₄(s)at 25°C