Consider the titration of 25.0 mLof$\mathbf{0}\mathbf{.}\mathbf{0500}\mathit{M}\mathit{S}{\mathit{n}}^{\mathbf{2}\mathbf{+}}$with$\mathbf{0}\mathbf{.}\mathbf{100}\mathit{M}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}}$in 1MHCI to give$\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ and$\mathit{S}{\mathit{n}}^{\mathbf{4}\mathbf{+}}$, using Pt and calomel electrodes.

(a) Write a balanced titration reaction.

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

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

(d) Calculate Eat the following volumes of$\mathit{F}{\mathit{e}}^{\mathbf{3}}\mathbf{:}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{26}\mathbf{.}\mathbf{0}$and 30.0 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$S{n}^{2+}$with$F{e}^{3+}$.

The titration reaction is $S{n}^{2+}+2F{e}^{3+}\to S{n}^{4+}+2F{e}^{2+}$.

b)

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

$$\begin{gathered} S{n}^{4+}+2e^{-}\to S{n}^{2+}{E}^{°}=0.139V \\ F{e}^{3+}+e^{-}\to F{e}^{2+}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=0.732V \end{gathered}$$

c)

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

$$\begin{gathered} E=E_{+}-E_{-} \\ {E}_{+}=E^{°}-\frac{0.05916}{2}log\frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]} \end{gathered}$$

$E^{-}$ is the potential of standard calomel electrode (0.241 V).

By combining the first and second equations,

$$\begin{gathered} E=E^{°}+\frac{0.05916}{2}log\frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}-E_{-} \\ E=0.139V-\frac{0.05916}{2}log\frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}-0.241V.......\left(1\right) \end{gathered}$$

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

$$\begin{gathered} E=E_{+}-E_{-} \\ {E}_{+}=°-0.05916\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^3\right]} \end{gathered}$$

By combining the first and second equations,

$$\begin{gathered} E=E°-0.05916\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}-E_{-} \\ E=0.732V-0.05916\log \frac{\left[F{e}^{2+}\right]}{\left[F{e}^{2+}\right]}-0.241V..........\left(2\right) \end{gathered}$$

d)

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

$$\begin{gathered} V\left(T{I}^{3+}\right)=\frac{c\left(S{n}^{2+}\right).V\left(S{n}^{2+}\right)}{c\left(T{I}^{3+}\right)} \\ =\frac{0.01M.25mL}{0.05M} \\ =5mL \end{gathered}$$

Now determine three titration regions:

1. Before the equivalence point -1.00,12.50 and 24.0mL .

2. At the equivalence point -25.0mL .

3. After the equivalence point -26.0 and 30.0mL .

There is an excess of $S{n}^{2+}$in the solution at 1.00 mL of added $F{e}^{3+}$before equivalence point, so use equation to compute the potential by simply replacing concentrations of ions with their volume of excess $S{n}^{2+}$is 24/25 .

Now use equation to calculate potential:

$$\begin{gathered} E=0.139V-\frac{0.05916}{2}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}-0.241V \\ E=0.139V-\frac{0.05916}{2}\log \frac{{\displaystyle \frac{1}{25}}}{{\displaystyle \frac{24}{25}}}-0.241 \\ E=-0.061V \end{gathered}$$

The volume of $F{e}^{3+}$is one-half of the quantity required for the equivalence point at 12.5 mL , therefore the log term is and the potential equation is:

$E=E°-E_{-}$

For the task that means:

role="math" localid="1664871633983" $$\begin{gathered} E=E°\left(S{n}^{4+}|S{n}^{2+}\right)-E_{-} \\ E=0.139V-0.241V \\ E=-0.102V \end{gathered}$$

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

$$\begin{gathered} E=0.139V-\frac{0.05916}{2}log\frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}-0.241V \\ E=0.139V-\frac{0.05916}{2}\log \frac{{\displaystyle \frac{1}{25}}}{{\displaystyle \frac{24}{25}}}-0.241 \\ E=-0.061V \end{gathered}$$

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

$$\begin{gathered} 2\left[S{n}^{2+}\right]=\left[F{e}^{3+}\right] \\ 2\left[S{n}^{4+}\right]=\left[F{e}^{2+}\right] \end{gathered}$$

Now calculate the potential by adding the equations together:

role="math" localid="1664872109070" $$\begin{gathered} 2E_{+}=2\left(0.139V-\frac{0.05916}{2}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]}\right) \\ 2E_{+}=1.464V-0.05916\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{4+}\right]} \end{gathered}$$

By rearranging the equation,

$3E_{+}=1.01V-0.05916log\frac{\left[S{n}^{2+}\right]\left[F{e}^{4+}\right]}{\left[S{n}^{4+}\right]\left[F{e}^{2+}\right]}$

By inserting the concentrations of cerium,

$3E_{+}=1.01V-0.05916log\frac{\left[S{n}^{2+}\right]\left[F{e}^{2+}\right]}{\left[S{n}^{4+}\right]\left[F{e}^{3+}\right]}$

Now calculate $E_{+}$:

$$\begin{gathered} E_{+}=\frac{1.01V}{3} \\ {E}_{+}=0.337V \end{gathered}$$

The total potential at 25 mL is:

$$\begin{gathered} E=E_{+}-E_{-} \\ E=(0.337-0.241)V \\ E=0.096V \end{gathered}$$

There is an excess of $F{e}^{3+}$in the solution after equivalence point at 25 mL of additional $F{e}^{3+}$, therefore use equation (2) to determine the potential by simply replacing concentrations of Tl ions with their corresponding volume.

For example, if 26mL of $F{e}^{3+}$was added, the volume of formed $F{e}^{2+}$is 25/25 and the volume of excess$F{e}^{3+}$ is 1/25.

Now use equation (2) to calculate potential:

$$\begin{gathered} E=0.732V-0.05916log\frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}-0.241V \\ E=0.732V-0.05916\log \frac{{\displaystyle \frac{25}{25}}}{{\displaystyle \frac{1}{25}}}-0.241V \\ E=0.408V \end{gathered}$$

Now substitute the volume of extra$F{e}^{3+}\left(5mL\right)$ for 30mL of additional$F{e}^{3+}$:

$$\begin{gathered} E=0.732V-0.05916log\frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}-0.241V \\ E=0.732V-0.05916\log \frac{{\displaystyle \frac{25}{25}}}{{\displaystyle \frac{5}{25}}}-0.241V \\ E=-0.450V \end{gathered}$$

Now sketch the titration curve with added volume of $F{e}^{3+}$on x-axis and potential on y-axis: