Question

The following cell was found to have a potential of -1.007 V:

${\mathrm{SCEIIX}}^{-}\left(0.0200\mathrm{M}\right),{\mathrm{CdX}}_2\left(\mathrm{sat}’\mathrm{d}\right)\overline{)\mathrm{Cd}}$

Calculate the solubility product of ${\mathrm{CdX}}_2$, neglecting the junction potential.

Short answer

The solubility product for ${\mathrm{CdX}}_2$is $2.75\times {10}^{-16}$.

Step-by-step solution

Given information

The given cell is as follows:

${\mathrm{SCEIIX}}^{-}\left(0.0200\mathrm{M}\right),{\mathrm{CdX}}_2\left(\mathrm{sat}’\mathrm{d}\right)\overline{)\mathrm{Cd}}$

The cell potential is given as$-1.007\mathrm{V}$.

Half-cell reaction  

The half-cell reactions involved in the given cell are shown below:

$$\begin{gathered} \mathrm{Cathode}:{\mathrm{CdX}}_2\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cd}\left(\mathrm{s}\right)+2{\mathrm{X}}^{-} \\ \mathrm{Anode}:2\mathrm{Hg}\left(\mathrm{l}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{l}\right)\to {\mathrm{Hg}}_2\mathrm{Cl}2+2{\mathrm{e}}^{-} \end{gathered}$$

Therefore, the reduction potential of anodic and cathodic half-cell reactions are $0.244\mathrm{V}$and $-0.403\mathrm{V}$.

Step 3: Ecell expression 

The $E_{cell}$for the reaction is as follows:

${E^o}_{cell}=E_{cathode}^o-E_{anode}^o............\left(1\right)$

Substitute reduction potential for half-cell reaction in the above equation (1) as follows:

$$\begin{gathered} {E^o}_{cell}=E_{cathode}^o-E_{anode}^o \\ =-0.403\mathrm{V}-0.244\mathrm{V} \\ =-0.647\mathrm{V} \end{gathered}$$

Decomposition of CdX2 

The equation is as follows:

${\mathrm{CdX}}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{X}}^{-}\left(\mathrm{aq}\right)$

The solubility product is given as follows:

$$\begin{gathered} K_{sp}=\left[{\mathrm{Cd}}^{2+}\right]{\left[{\mathrm{X}}^{-}\right]}^2 \\ \left[{\mathrm{Cd}}^{2+}\right]=\frac{K_{sp}}{{\left[{\mathrm{X}}^{-}\right]}^2}.......\left(2\right) \end{gathered}$$

The equation for the cathodic reaction 

The cathodic reaction can be written as follows:

${\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Cd}\left(\mathrm{s}\right)$

The Nernst equation for the above equation is as follows:

$E_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{\left[\mathrm{Cd}\right]}{\left[{\mathrm{Cd}}^{2+}\right]}..........\left(3\right)$

Cd is a solid substance. Therefore, the activity of Cd is taken as 1.

Substitute the concentration of each substance in the above equation (3) as follows:

$E_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{1}{\left[{\mathrm{Cd}}^{2+}\right]}........\left(4\right)$

Substitute equation (2) in equation (4) as follows:

role="math" localid="1649656499276" $E_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{{\left[{\mathrm{X}}^{-}\right]}^2}{K_{sp}}.......\left(5\right)$

Calculate solubility product for CdX2

Substitute the values in the above equation (5) as follows:

$$\begin{gathered} \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}{E}_{cell}=E_{cell}^o-\frac{0.0592}{2}\log \frac{{\left[{\mathrm{X}}^{-}\right]}^2}{K_{sp}} \\ -1.007\mathrm{V}=-0.647\mathrm{V}-\frac{0.0592}{2}\log \frac{{\left(0.02\right)}^2}{K_{sp}} \\ -1.007\mathrm{V}+0.647\mathrm{V}=-0.0296\left[2\log \left(0.02\right)-\log K_{sp}\right] \\ -0.36=0.0296\left[3.4+\log K_{sp}\right] \\ \log K_{sp}=\frac{-0.36}{0.0296}-3.4 \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\log K_{\mathit{s}\mathit{p}}=-15.56 \\ {K}_{sp}=2.75\times {10}^{-16} \end{gathered}$$