Consider the titration of 25.0mL of 0.0500MSn²⁺with 0.100MFe³⁺in 1MHCI to give Fe²⁺ and Sn⁴⁺, 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) Calculateat E the following volumes of Fe³⁺: 1.0,12.5,24.0,25.0,26.0 and 30.0mL. 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 Sn²⁺ with Fe³⁺.

The titration reaction is ${\mathrm{Sn}}^{2+}+2{\mathrm{Fe}}^{3+}\to {\mathrm{Sn}}^{4+}+2{\mathrm{Fe}}^{2+}$.

b)

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

$$\begin{gathered} {\mathrm{Sn}}^{4+}+2{\mathrm{e}}^{-}\to {\mathrm{Sn}}^{2+}\mathrm{E}°=0.139\mathrm{V} \\ {\mathrm{Fe}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\mathrm{E}°=0.732\mathrm{V} \end{gathered}$$

c)

Using these two equations, the Nernst equation for reducing Sn⁴⁺ 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 (0.241V) .

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${\mathrm{Fe}}^{3+}$:

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

By combining the first and second equations,

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

d)

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

$$\begin{gathered} V\left({\mathrm{Tl}}^{3+}\right)=\frac{c\left({\mathrm{Sn}}^{2+}\right)\cdot V\left({\mathrm{Sn}}^{2+}\right)}{c\left({\mathrm{Tl}}^{3+}\right)} \\ =\frac{0.01\mathrm{M}\cdot 25\mathrm{mL}}{0.05\mathrm{M}} \\ =5\mathrm{mL} \end{gathered}$$

Now determine three titration regions:

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

2. At the equivalence point$-25.0\mathrm{mL}$.

3. After the equivalence point$-26.0$and$30.0\mathrm{mL}$.

There is an excess of Sn²⁺ in the solution at 1.00mL of added Fe³⁺ before equivalence point, so use equation to compute the potential by simply replacing concentrations of Sn ions with their volume of excess Sn²⁺ is 24/25.

Now use equation to calculate potential:

$$\begin{gathered} E=0.139\mathrm{V}-\frac{0.05916}{2}\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left.{\mathrm{Sn}}^{4+}\right]}-0.241\mathrm{V} \\ \mathrm{E}=0.139\mathrm{V}-\frac{0.05916}{2}\log \frac{\frac{24}{25}}{\frac{1}{25}}-0.241 \\ \mathrm{E}=-0.143\mathrm{V} \end{gathered}$$

The volume of Fe³⁺ is one-half of the quantity required for the equivalence point at 12.5mL, therefore the log term is and the potential equation is:

$E=E°-E_{-}$

For the task that means:

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

Apply the same calculation used at 24.0mL, substituting the volume of produced and surplus iron:

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

At 25mL, the equivalence point, Sn²⁺ and Fe²⁺ are at equilibrium and Sn⁴⁺ and Fe³⁺ also so write:

role="math" localid="1667812750365" $$\begin{gathered} 2\left[S{n}^{2+}\right]=\left[F{e}^{3+}\right] \\ 2\left[S{n}^{4+}\right]=\left[F{e}^{2+}\right] \end{gathered}$$

Now calculate the potential by adding the equations together:

$$\begin{gathered} 2E_{+}=2\left(0.139\mathrm{V}-\frac{0.05916}{2}\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left[{\mathrm{Sn}}^{4+}\right]}\right) \\ 2E_{+}=1.464\mathrm{V}-0.05916\log \frac{\left[{\mathrm{Sn}}^{2+}\right]}{\left[{\mathrm{Sn}}^{4+}\right]} \end{gathered}$$

By rearranging the equation,

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

By inserting the concentrations of cerium,

$3E_{+}=1.01\mathrm{V}-0.05916\log \frac{\left[{\mathrm{Sn}}^{2+}\right]\left[{\mathrm{Sn}}^{4+}\right]}{\left[{\mathrm{Sn}}^{4+}\right]\left[{\mathrm{Sn}}^{2+}\right]}$

Now calculate $E_{+}$:

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

The total potential atis:

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

There is an excess of Fe³⁺ in the solution after equivalence point at 25mL of additional Fe³⁺, therefore use equation (2) to determine the potential by simply replacing concentrations of ions with their corresponding volume.

For example, if 26mL of Fe³⁺ was added, the volume of formed Fe²⁺ is 25/25 and the volume of excess Fe³⁺ is 1/25.

Now use equation (2) to calculate potential:

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

Now substitute the volume of extra Fe³⁺(5mL)for of additional Fe³⁺:

role="math" localid="1667813650443" $$\begin{gathered} E=0.732\mathrm{V}-0.05916\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}-0.241\mathrm{V} \\ \mathrm{E}=0.732\mathrm{V}-0.05916\log \frac{\frac{25}{25}}{\frac{5}{25}}-0.241\mathrm{V} \\ \mathrm{E}=0.450\mathrm{V} \end{gathered}$$

Now sketch the titration curve with added volume of Fe³⁺ on -axis and potential on -axis: