Question

The following cell was found to have a potential of 0.124 V:

$\mathrm{Ag}\overline{)\mathrm{AgCl}}\left(\mathrm{sat}’\mathrm{d}\right){\mathrm{IICu}}^{2+}\left(3.25\times {10}^{-3}\mathrm{M}\right)\overline{)\mathrm{membrane}\mathrm{electrode}\mathrm{for}{\mathrm{Cu}}^{2+}}$

When the solution of known copper activity was replaced with an unknown solution, the potential was found to be 0.055 V. What was the $pCu$of this unknown solution? Neglect the junction potential.

Short answer

The value of $pCu$isrole="math" localid="1649245856932" $4.82$.

Step-by-step solution

Given information

The given cell is as follows:

$\mathrm{Ag}\overline{)\mathrm{AgCl}}\left(\mathrm{sat}’\mathrm{d}\right){\mathrm{IICu}}^{2+}\left(3.25\times {10}^{-3}\mathrm{M}\right)\overline{)\mathrm{membrane}\mathrm{electrode}\mathrm{for}{\mathrm{Cu}}^{2+}}$

The Ecell at the junction potential

The $E_{cell}$at the junction potential by the copper cations is as follows:

role="math" localid="1649246932363" $$\begin{gathered} E_{cell}=E_J-\frac{0.0592}{2}\left(-\log \left[{\mathrm{Cu}}^{2+}\right]\right) \\ =E_J+\frac{0.0592}{2}\log \left[{\mathrm{Cu}}^{2+}\right].........\left(1\right) \end{gathered}$$

The electrode potential for the cell is given as data-custom-editor="chemistry" $0.124\mathrm{V}$.

The concentration of copper ions is given as data-custom-editor="chemistry" $3.25\times {10}^{-3}\mathrm{M}$.

Calculate junction potential 

Substitute the $E_{cell}$and concentration of copper ions in equation (1) as follows:

$$\begin{gathered} E_{cell}=E_J+\frac{0.0592}{2}\log \left[{\mathrm{Cu}}^{2+}\right] \\ 0.124\mathrm{V}=E_J+0.0296\left[\log \left(3.25\times {10}^{-3}\right)\right] \\ 0.124\mathrm{V}=E_J+0.0296\left[\log 3.25-3\log 10\right] \\ 0.124\mathrm{V}=E_J+0.0296\left(0.512-3\right) \\ 0.124\mathrm{V}=E_J-0.0736 \\ {E}_J=0.124\mathrm{V}+0.0736 \\ =0.1976\mathrm{V} \end{gathered}$$

The value of pCu

The $pCu$is as follows:

role="math" localid="1649247541745" $$\begin{gathered} E_{cell}=E_J-\frac{0.0592}{2}pCu \\ pCu=\frac{2\left(E_J-E_{cell}\right)}{0.0592}........\left(2\right) \end{gathered}$$

pCu of this unknown solution

Substitute the values in the equation (2) as follows:

$$\begin{gathered} pCu=\frac{2\left(E_J-E_{cell}\right)}{0.0592} \\ =\frac{2\left(0.1976V-0.055V\right)}{0.0592} \\ =\frac{0.1426}{0.0296} \\ =4.82 \end{gathered}$$