Ascorbic acid (0.0100M)was added to 10.0mL of $\mathbf{0}\mathbf{.}\mathbf{0200}{\mathbf{MFe}}^{\mathbf{3}\mathbf{+}}$at pH 0.30, and the potential was monitored with Pt and saturated Ag | AgClelectrodes.

Dehydroascorbic acidrole="math" localid="1664865837362" $\mathbf{+}\mathbf{2}{\mathit{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{\to }\mathit{a}\mathit{s}\mathit{c}\mathit{o}\mathit{r}\mathit{b}\mathit{i}\mathit{c}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}{\mathit{E}}^{\mathbf{°}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{390}\mathit{V}$

(a) Write a balanced equation for the titration reaction,

(b) Using ${\mathit{E}}^{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{767}\mathit{V}$ for the role="math" localid="1664865912877" $\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}}\mathbf{\mid }\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ couple, calculate the cell voltage when 5.0,10.0 and 15.0 mL of ascorbic acid have been added. (Hint: Refer to the calculations in Demonstration 16 - 1.)

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)

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

$$\begin{gathered} F{e}^{3+}+e^{-}\to F{e}^{2+}{E}^{°}=0.767V \\ dehydroascorbicacid+2H^{+}+2e^{-}\to ascorbicacid+H_{2}O\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{E}^{°}=0.390V \end{gathered}$$

To create the balanced equation, multiply the first equation by 2 and the second equation backwards:

$$\begin{gathered} 2F{e}^{3+}+2e^{-}\to 2F{e}^{2+} \\ ascorbicacid+H_{2}O\to dehydroascorbicacid+2H^{+}+2e^{-} \end{gathered}$$

The sum of these two reactions is:

$$\begin{gathered} 2F{e}^{3+}+2e^{-}+ascorbicacid+H_{2}O\to 2F{e}^{2+}+dehydroascorbicacid+2H^{+}+2e^{-} \\ 2F{e}^{3+}+ascorbicacid+H_{2}O\to 2F{e}^{2+}+dehydroascorbicacid+2H^{+} \end{gathered}$$

b)

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

$$\begin{gathered} n(ascorbicacid)=2n\left(F{e}^{2+}\right) \\ V(ascorbicacid)=2\cdot \frac{\left(c\right(F{e}^{2+})\cdot V(F{e}^{2+}2+)}{c(ascorbicacid)} \\ V(ascorbicacid)=2\cdot \frac{0.02MH\cdot 10mL}{0.01,M} \\ V(ascorbicacid)=10mL \end{gathered}$$

Now determine three titration regions:

1. Before the equivalence point 5.00 mL.

2. At the equivalence point - 10.00 mL.

3. After the equivalence point - 15.00 mL.

The Nernst equations for each reduction reaction are:

$$\begin{gathered} E=0.767V-0.05916log\frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}-0.197"V......"\left(1\right) \\ E=0.390V-\frac{0.05916}{2}\log \frac{\left[ascorbicacid\right]}{\left[dehydroascorbicacid\right][H^{+}{]}^2}-0.197V.......\left(2\right) \end{gathered}$$

Because there is an excess of $F{e}^{3+}$in the solution before the equivalence point, we can utilise equation (1) to determine the potential by simply inserting Fe ion concentrations with their associated volume. The volume of ascorbic acid is one-half of the quantity required for the equivalence point at 5.00 mL, therefore the log term is 0. In addition, E_ of Ag - AgCl may be found in appendix H. The potential equation is as follows:

$E=E^{°}-E_{-}$

For the task that means:

$$\begin{gathered} E=E^{°}(F{e}^{3+}\mid F{e}^{2+})-E_{-} \\ E=0.767V-0.197V \\ E=0.57V \end{gathered}$$

To compute potential at equivalent volume, multiply the ascorbic acid equation by 2 so that the logarithm components may be added on:

$$\begin{gathered} 2E_{+}=2\left(0.390V-\frac{0.05916}{2}\log \frac{[ascorbicacid]}{[dehydroascorbicacid][H^{+}{]}^2}\right) \\ 2E_{+}=0.78V-0.05916log\frac{[ascorbicacid]}{[dehydroascorbicacid][H^{+}{]}^2} \end{gathered}$$

For each reduction reaction, the Nernst equations are as follows:

$$\begin{gathered} E=0.767V-0.05916log\frac{\left[F{e}^{2+}\right]}{\left[F{e}^{3+}\right]}-0.197V..........\left(1\right) \\ E=0.390V-\frac{0.05916}{2}log\frac{[ascorbicacid]}{[dehydroascorbicacid][H^{+}{]}^2}-0.197V.......\left(2\right) \end{gathered}$$

At the equivalence point,$F{e}^{2+}$ and dehydroascorbic acid, as well as$F{e}^{3+}$ and ascorbic acid, are at equilibrium, therefore,

$$\begin{gathered} \left[F{e}^{2+}\right]=2[dehydroascorbicacid] \\ \left[F{e}^{3+}\right]=2[ascorbicacid] \end{gathered}$$

By inserting the concentrations in the sum of two reactions,

$3E_{+}=1.547V-0.05916log\frac{[ascorbicacid]\left[F{e}^{2+}\right]}{[dehydroascorbicacid]\left[H^{+}{]}^2\right[F{e}^{3+}]}$

By substituting iron concentrations,

$3E_{+}=1.547V-0.05916log\frac{2[ascorbicacid]dehydroascorbicacid]}{2\left[dehydroascorbicaeid\right][ascorbicacid[H^{+}{]}^2}$

The equation now looks like this:

$3E_{+}=1.547V-0.05916log\left(\frac{1}{[H^{+}{]}^2}\right)$

Concentration of$H^{+}$ can be calculated from pH :

$$\begin{gathered} \left[H^{+}\right]={10}^{-pH} \\ \left[H^{+}\right]={10}^{-0.30} \\ \left[H^{+}\right]=0.501M \end{gathered}$$

Now calculate $E_{+}$:

$$\begin{gathered} E_{+}=\frac{1.547V-0.05916\log \left({\displaystyle \frac{1}{0.{501}^2}}\right)}{E} \\ {E}_{+}=0.504V \end{gathered}$$

The total potential at 10 mL is:

$$\begin{gathered} E=E_{+}-E_{-} \\ E=(0.504-0.197)V \\ E=0.307V \end{gathered}$$

There is an excess of ascorbic acid in the solution after equivalence point at 15 mL of additional ascorbic acid, therefore use equation (2) to determine the potential by simply replacing concentrations of ions with their corresponding volume.

For example, if 15 mL of ascorbic acid was added, the volume of formed dehydroascorbic acid is 10/10 and the volume of excess ascorbic acid is 5/10.

Now use equation (2) to calculate potential:

$$\begin{gathered} E=0.390V-\frac{0.05916}{2}log\frac{[ascorbicacid]}{[dehydroascorbicacid][H^{+}{]}^2}-0.197V \\ E=0.390V-\frac{0.05916}{2}log\frac{{\displaystyle \frac{5}{10}}}{{\displaystyle \frac{10}{10}}.0.{501}^2}-0.197V \\ E=0.184V \end{gathered}$$

The cell voltage is 0.184 V.