Question

The solubility product constant for ${\mathrm{PbI}}_2$is $7.1\times {10}^{-9}$at ${25}^{\circ }\mathrm{C}$. Calculate $E^0$for the process

${\mathrm{PbI}}_2\left(s\right)+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pb}\left(s\right)+2{\mathrm{I}}_2$

Short answer

The calculated value for$E^0$for the process${\mathrm{Pbl}}_2\left(s\right)+2e^{-}\rightleftharpoons \mathrm{Pb}\left(s\right)+2{\mathrm{l}}_2$is$-0.367\mathrm{V}$.

Step-by-step solution

Given information

The solubility product constant for ${\mathrm{PbI}}_2$is $7.1\times {10}^{-9}$.

The solubility product constant, $K_{sp}$, is the equilibrium constant for a stable substance disbanding in an aqueous explanation. It denotes the class at which a solute disbands in explanation. The additional soluble a substance is, the more elevated the $K_{sp}$value it has.

Calculation

The calculation is shown below:

${\mathrm{Pb}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pb}\left(\mathrm{s}\right),E^0\left({\mathrm{Pb}}^{2+}/\mathrm{Pb}\right)=-0.126\mathrm{V}$

$E^0\text{process}=E^0\left({\mathrm{Pb}}^{2+}/\mathrm{Pb}\right)-\frac{0.0592}{n}\log \frac{1}{Ksp}$

$E^0\text{process}=-0.126\mathrm{V}-\frac{0.0592}{2}\log \frac{1}{7.1\times {10}^{-9}}$

$E^0\text{process}=-0.126\mathrm{V}-0.241\mathrm{V}$

$E^0\text{process}=-0.367\mathrm{V}$