Question

The following cell was found to have a potential of 0.2897 V:

${\mathrm{SCEIIMg}}^{2+}\left(\mathrm{a}=3.2\times {10}^{-3}\mathrm{M}\right)\overline{)\mathrm{membrane}\mathrm{electrode}\mathrm{for}{\mathrm{Mg}}^{2+}}$

(a) When the solution of known magnesium activity was replaced with an unknown solution, the potential

was found to be 0.2041 V. What was the pMg of this unknown solution?

(b) Assuming an uncertainty of 60.002 V in the junction potential, what is the range of ${\mathrm{Mg}}^{2+}$activities within which the true value might be expected?

(c) What is the relative error in $\left[{\mathrm{Mg}}^{2+}\right]$associated with the uncertainty in $E_j$?

Short answer

(a) $\mathrm{pMg}$of unknown solution is$5.37$.

(b) ${\mathrm{Mg}}^{2+}$activities range from $3.64\times {10}^{-6}\mathrm{M}$to $4.97\times {10}^{-6}\mathrm{M}$.

(c) For role="math" localid="1649245032243" $K=0.3651$, Relative errorrole="math" localid="1649244996928" $=14.55\%$

For $K=0.3611$, Relative error $=16.67\%$

Step-by-step solution

Part (a) Step 1: Given information

The following cell was found to have a potential of $0.2897V$:

${\mathrm{SCEIIMg}}^{2+}\left(\mathrm{a}=3.2\times {10}^{-3}\mathrm{M}\right)\overline{)\mathrm{membrane}\mathrm{electrode}\mathrm{for}{\mathrm{Mg}}^{2+}}$

When the solution of known magnesium activity was replaced with an unknown solution, the potential was found to be $0.2041\mathrm{V}$.

Part (a) Step 2: Constant and ion activity relation

The relation is as follows:

$pX=-\log a_X=\frac{-n\left(E_{cell}-K\right)}{0.0592}........\left(1\right)$

$$\begin{gathered} n–\mathrm{number}\mathrm{of}\mathrm{electrons} \\ K–\mathrm{constant} \end{gathered}$$

Part (a) Step 3: pMg of the standard Mg2+ solution

Substitute the value in the above equation (1) as follows:

$$\begin{gathered} pMg=-\log a_{M{g}^{2+}}=\frac{-n\left(E_{cell}-K\right)}{0.0592} \\ =-\log \left(3.32\times {10}^{-3}\right)=\frac{-2\left(0.2897-K\right)}{0.0592} \\ K=\frac{2.48\times 0.0592}{2}+0.2897 \\ =0.3631 \end{gathered}$$

Part (a) Step 4: pMg of the unknown solution 

Substitute the value in the above equation (1) as follows:

$$\begin{gathered} pMg=-\log a_{M{g}^{2+}}=\frac{-n\left(E_{cell}-K\right)}{0.0592} \\ \frac{-2\left(0.2041-0.3631\right)}{0.0592}=-\log a_{M{g}^{2+}} \\ 5.37=-\log a_{M{g}^{2+}} \\ {a}_{M{g}^{2+}}=4.26\times {10}^{-6}\mathrm{M} \end{gathered}$$

Part (b) Step 1: Mg2+ activity when K=0.3631+0.002=0.3651

Substitute the value in the above equation (1) as follows:

$$\begin{gathered} pMg=-\log a_{M{g}^{2+}}=\frac{-n\left(E_{cell}-K\right)}{0.0592} \\ \frac{-2\left(0.2041-0.3651\right)}{0.0592}=-\log a_{M{g}^{2+}} \\ 5.44=-\log a_{M{g}^{2+}} \\ {a}_{M{g}^{2+}}=3.64\times {10}^{-6}\mathrm{M} \end{gathered}$$

Part (b) Step 2: Mg2+ activity when K=0.3631-0.002=0.3651 V

Substitute the value in the above equation (1) as follows:

$$\begin{gathered} pMg=-\log a_{M{g}^{2+}}=\frac{-n\left(E_{cell}-K\right)}{0.0592} \\ \frac{-2\left(0.2041-0.3611\right)}{0.0592}=-\log a_{M{g}^{2+}} \\ 5.304=-\log a_{M{g}^{2+}} \\ {a}_{M{g}^{2+}}=4.97\times {10}^{-6}\mathrm{M} \end{gathered}$$

Therefore, ${\mathrm{Mg}}^{2+}$activities range from $3.64\times {10}^{-6}\mathrm{M}$to $4.97\times {10}^{-6}\mathrm{M}$

Part (c) Step 1: Relative error for K=0.3651

$$\begin{gathered} \mathrm{Relative}\mathrm{error}=\frac{\left(3.64\times {10}^{-6}-4.26\times {10}^{-6}\right)}{4.26\times {10}^{-6}}\times 100\% \\ =14.55\% \end{gathered}$$

Part (c) Step 2: Relative error for K=0.3611

$$\begin{gathered} \mathrm{Relative}\mathrm{error}=\frac{\left(4.97\times {10}^{-6}-4.26\times {10}^{-6}\right)}{4.26\times {10}^{-6}}\times 100\% \\ =16.67\% \end{gathered}$$