Question

Question: A student decides to measures the solubility of lead sulphate in water and set up the electrochemical cell:

$\mathit{P}\mathit{b}\left|PbS{O}_4\right|\mathit{S}{{\mathbf{O}}^{\mathbf{2}\mathbf{\text{-}}}}_{\mathbf{4}}\left(aq,0.0500M\right)\mathbf{\left|}\mathbf{\right|}\mathit{C}{\mathit{l}}^{\mathbf{\text{-}}}\left(aq,1.00M\right)\left|AgCl\right|\mathit{A}\mathit{g}$

At 25⁰C the student finds the cell potential to be 0.546 V, and from the appendix E the student finds

$\mathit{A}\mathit{g}\mathit{C}\mathit{l}\left(s\right)\mathbf{+}\mathit{e}\mathbf{\rightleftarrows }\mathit{A}\mathit{g}\left(s\right)\mathbf{+}\mathit{C}{\mathit{l}}^{\mathbf{\text{-}}}\left(aq\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{222}\mathit{V}$

What does he find for the K_sp of PbSO₄?

Short answer

Answer:

$AgCl\left(s\right)+e\leftrightharpoons Ag\left(s\right)+C{l}^{\text{-}}\left(aq\right)$

Step-by-step solution

Step-1: The reaction taking place at cathode and anode.

The anode reaction involved is:

$Pb\left(s\right)+S{O_4}^{2\text{-}}\left(aq\right)\rightleftharpoons PbS{O}_4\left(s\right)+2e^{\text{-}}$

The cathode reaction is:

$Agcl\left(s\right)+e^{\text{-}}\rightleftharpoons Ag\left(s\right)+C{l}^{\text{-}}\left(aq\right).,{E^{0_{}}}_{cathode}=0.222V$

Step-2: Calculating the reduction potential of the anode.

The cell potential as 0.546V

Thus,

$$\begin{gathered} E^0={E^0}_{cathode}\text{-}{E^0}_{anode} \\ \Rightarrow 0.546=0.222V\text{-}{E^0}_{anode} \\ \Rightarrow {E^0}_{anode}=\text{-}0.324V \end{gathered}$$

Step-3: To calculate the solubility Ksp

We have to calculate the solubility product for . So we are going to use the cell potential for anode in our calculations.

In certain case, if we take K_eq= K_sp, then we can calculate K_sp.

For the reaction,

$Pb\left(s\right)+S{O_4}^{2\text{-}}\left(aq\right)\rightleftharpoons PbS{O}_4\left(s\right)+2e^{\text{-}}$

$$\begin{gathered} {\log }_{10}{K}_{eq}=\frac{n}{0.0592}{E^0}_{anode} \\ \Rightarrow {\log }_{10}{K}_{eq}=\frac{2}{0.0592}\times \left(-0.324\right) V \\ \Rightarrow {\log }_{10}{K}_{eq}=-10.94 \end{gathered}$$

Here, n =2 which is the number of electrons in the anode reaction.

$K_{eq}=1.13\times {10}^{\text{-}11}$

Hence, the solubility product of PbSO₄comes out to be

_{$K_{ep}=K_{eq}=1.13\times {10}^{\text{-}11}$}