Consider the titration of 100.0mLof $\mathbf{0}\mathbf{.}\mathbf{0100}{\mathbf{MCe}}^{\mathbf{4}\mathbf{+}}$ in $\mathbf{1}{\mathbf{MHClO}}_{\mathbf{4}}$by $\mathbf{0}\mathbf{.}\mathbf{0400}{\mathbf{MCu}}^{\mathbf{+}}$ to give ${\mathbf{Ce}}^{\mathbf{3}\mathbf{+}}$ and ${\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}$ , using Pt and saturated Ag | AgCl electrodes to find the end point.

(a) Write a balanced titration reaction.

(b) Write two different half-reactions for the indicator electrode.

(c) Write two different Nernst equations for the cell voltage.

(d) Calculate Eat the following volumes of${\mathbf{Cu}}^{\mathbf{+}}\mathbf{:}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{24}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{25}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{30}\mathbf{.}\mathbf{0}$ and 50.0 mL. Sketch the titration curve.

Titration is a method or procedure for estimating the concentration of a dissolved material by calculating the least quantity of known-concentration reagent necessary to produce a particular effect in interaction with a known volume of the test solution.

a)

Titration of${\mathrm{Ce}}^{4+}$ with ${\mathrm{Cu}}^{+}$.

The titration reaction is $C{e}^{4+}+C{u}^{+}\to C{e}^{3+}+C{u}^{2+}$.

b)

Now, in Appendix H, write two reduction half-reactions and determine their standard reduction potentials:

$$\begin{gathered} {\mathrm{Ce}}^{4+}+{\mathrm{e}}^{-}\to {\mathrm{Ce}}^{3+}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{E}}^{°}=1.70\mathrm{V} \\ {\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\mathrm{E}}^{°}=0.161\mathrm{V} \end{gathered}$$

c)

Using these two equations, the Nernst equation for reducing$C{e}^{4+}$is:

$$\begin{gathered} E=E_{+}-E_{-} \\ {E}_{+}=E^{°}-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]} \end{gathered}$$

$E^{-}$ is the potential of standard Ag - AgCl electrode (0.197 V).

By combining the first and second equations,

$$\begin{gathered} E=E^{°}-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-E_{-} \\ E=1.70V-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-0.197V.........\left(1\right) \end{gathered}$$

In the same way, the Nernst equation for the reduction of :

$$\begin{gathered} E=E_{+}-E_{-} \\ {E}_{+}=E_{+}^{°}-0.05916log\frac{\left[C{u}^{+}\right]}{\left[C{u}^{2+}\right]} \end{gathered}$$

By combining the first and second equations,

$$\begin{gathered} E=E^{°}-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-E_{-} \\ E=0.161V-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-0.197V.........\left(1\right) \end{gathered}$$

d)

Calculate the equivalency point using data from the job before proceeding with potential calculations for each extra volume:

$$\begin{gathered} V\left(C{u}^{+}\right)=\frac{c\left(C{e}^{4+}\right)·V\left(C{u}^{+}\right)}{c\left(C{u}^{+}\right)}V\left(C{u}^{+}\right) \\ =\frac{0.01M·100ML}{0.04M}V\left(C{u}^{+}\right) \\ =25ML \end{gathered}$$

Now determine three titration regions:

1. Before the equivalence point - 1.00,12.5 and 24.5 mL.

2. At the equivalence point - 25.0mL.

3. After the equivalence point - 25.5,30.0 and 50.0 mL.

There is an excess of $C{e}^{4+}$in the solution at 1.00 mL of added $C{u}^{+}$before equivalence point, so use equation ( 1 ) to compute the potential by simply replacing concentrations of Ce ions with their volume of excess$C{e}^{4+}$ is 24/25.

Now use equation ( 1 ) to calculate potential:

$$\begin{gathered} E=1.70V-0.05916\log \frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-0.197V \\ =1.70V-0.05916\log \frac{{\displaystyle \frac{1}{25}}}{{\displaystyle \frac{24}{25}}}-0.197 \\ =1.58V \end{gathered}$$

The volume of $C{u}^{+}$is one-half of the quantity required for the equivalence point at 12.5 mL, therefore the log term is 0 and the potential equation is:

$E=E^{°}-E_{-}$

For the task that means:

$$\begin{gathered} E=E^{°}\left(C{e}^{4+}\right|C{e}^{3+})-E_{-} \\ E=1.70V-0.197V \\ E=1.50V \end{gathered}$$

Apply the same calculation used at 24.5 mL, substituting the volume of produced and surplus iron:

$$\begin{gathered} E=1.70V-0.05916log\frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}-0.197V \\ E=1.70V-0.05916log\frac{{\displaystyle \frac{24.5}{25}}}{{\displaystyle \frac{0.5}{25}}}-0.197 \\ E=1.40V \end{gathered}$$

At 25 mL, the equivalence point,$C{e}^{3+}$ and$C{u}^{+}$ are at equilibrium and$C{e}^{4+}$ and$C{u}^{2+}$ also so write:

$$\begin{gathered} \left[C{e}^{3+}\right]=\left[C{u}^{2+}\right] \\ \left[C{e}^{4+}\right]=\left[C{u}^{+}\right] \end{gathered}$$

Now calculate the potential by adding the equations together:

$2E_{}=1.70-0.05916log\frac{\left[C{e}^{3+}\right]}{\left[C{e}^{4+}\right]}+0.161-0.05916\log \frac{\left[C{u}^{+}\right]}{\left[C{u}^{2+}\right]}$

By rearranging the equation,

$2E_{+}=1.861V-0.05916\log \frac{\left[C{e}^{3+}\right]\left[C{u}^{+}\right]}{\left[C{e}^{4+}\right]\left[C{u}^{3+}\right]}$

By inserting the concentrations of cerium,

role="math" localid="1664860739226" $2E_{+}=1.861V-0.05916\log \frac{\left[C{e}^{3+}\right]\left[C{u}^{4+}\right]}{\left[C{e}^{4+}\right]\left[C{u}^{3+}\right]}$

Now the log term is and calculate :

$$\begin{gathered} 2E_{+}=1.861V/\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}:2 \\ {E}_{+}=0.931V \end{gathered}$$

The total potential at 25 mL is:

$$\begin{gathered} E=E_{+}-E_{-} \\ E=(0.931-0.197)V \\ E=0.734V \end{gathered}$$

There is an excess of $C{u}^{+}$in the solution after equivalence point at 25.5 mL of additional $C{u}^{+}$, therefore use equation (2) to determine the potential by simply replacing concentrations of copper ions with their corresponding volume.

For example, if $25.5mLofC{u}^{+}$was added, the volume of formed$C{u}^{2+}$ stays 25 mL and the volume of excess$C{u}^{+}$ is 0.5/25.

Now use equation (2) to calculate potential:

$$\begin{gathered} E=0.161V-0.05916log\frac{\left[C{u}^{+}\right]}{\left[C{u}^{2+}\right]}-0.197V \\ E=0.161V-0.05916log\frac{{\displaystyle \frac{1}{25}}}{{\displaystyle \frac{25}{25}}}-0.197V \\ E=0.065V \end{gathered}$$

At 30 mL, use the same equation and substitute the volume of excess cerium:

$$\begin{gathered} E=0.161V-0.05916log\frac{\left[C{u}^{+}\right]}{\left[C{u}^{2+}\right]}-0.197V \\ E=0.161V-0.05916log\frac{{\displaystyle \frac{5}{25}}}{{\displaystyle \frac{25}{25}}}-0.197V \\ E=0.005V \end{gathered}$$

The increased volume is double the value of equivalency volume at 50 mL, the log term is 0 , and we may apply this equation:

$E=E^{°}-E_{-}$

For the task that means:

$$\begin{gathered} E=E^{°}\left(C{u}^{+}\right|C{u}^{2+})-E_{-} \\ E=0.161V-0.197V \\ E=-0.036V \end{gathered}$$

Now sketch the titration curve with added volume of $C{u}^{+}$on x-axis and potential on y-axis: