Question

From the standard potentials

${\mathrm{Tl}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Tl}\left(s\right)E^0=-0.336\mathrm{V}$

$\mathrm{TlCl}\left(s\right)+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Tl}\left(s\right)+{\mathrm{Cl}}^{-}{E}^0=-0.557\mathrm{V}$

Calculate the solubility product constant for localid="1646120718290" $\mathrm{TlCl}$.

Short answer

The solubility product constant for$\mathrm{TICI}$is$1.84\times {10}^{-4}$

Step-by-step solution

Given information

The variables are stated below,

${\mathrm{Tl}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Tl}\left(\mathrm{s}\right)E^0=-0.336\mathrm{V}$

$\mathrm{TlCl}\left(s\right)+e^{-}\rightleftharpoons \mathrm{Tl}\left(s\right)+{\mathrm{Cl}}^{-}{E}^0=-0.557\mathrm{V}$

The overall reaction

$$\begin{gathered} {\mathrm{Tl}}^{+}+{\mathrm{Cl}}^{-}\rightleftharpoons \mathrm{TlCl}\left(\mathrm{s}\right) \\ {E}^0\mathrm{cell}=-0.336\mathrm{V}-(-0.557\mathrm{V}) \end{gathered}$$

$=0.221\mathrm{V}$

Finding the value

We know that

In $K=\frac{nF{E}^{0}cell}{RT}$

At ${25}^0\mathrm{C},\frac{F}{RT}=25.693\times {10}^{-3}\mathrm{V}$

So, $K=\frac{n{E}^0\mathrm{cell}}{25.693\times {10}^{-3}\mathrm{V}}$

If the $K$is a constant equilibrium

localid="1646120966096" $K=\frac{1\times 0.221\mathrm{V}}{25.693\times {10}^{-3}\mathrm{V}}$

$K=8.60$

$\Rightarrow K=e^{8.60}$

$K=5431.66$

The final calculation

Solubility product constant will be,

$Ksp=\frac{1}{K}$

$=\frac{1}{5431.66}$

$=1.84\times {10}^{-4}$