2: Potentiometric Titrations (Experiment)

Pre-lab Questions

1. All electrolyte solutions are non-ideal. From the point of view of thermodynamics, this means that the Nernst equations used above should be written in terms of the effective concentration or activity of each ion, not its true or analytical concentration. Define activity and analytical concentration. Compare and contrast these two terms.
2. Assuming that you have weighed 0.8500g of \( \ce{Fe(NH4)2(SO4)2 * 6H2O} \), determine the volume required to reach the equivalence point of the titration. The concentration of the permanganate solution is 0.02M.

Introduction

In Experiment 1, we examined electrochemical cells in which half-reactions were combined to convert the stored Gibbs energy of the constituent half-reactions into electrical work. In this experiment, we will use the electrochemical cell as a method of chemical analysis called potentiometric titration. A potentiometric titration is similar to the acid-base titrations you have performed; however, the means of determining the concentration of a species is dependent upon an oxidation-reduction reaction rather than a neutralization reaction. The redox species of known concentration serves as the titrant, while the redox species of unknown concentration is present in the sample solution being titrated. When titrant is added to the sample solution, the redox reaction occurs. The electrodes are used to monitor this reaction.

In a potentiometric titration, the two electrodes are identified as indicator and reference electrodes rather than cathode and anode. In fact, reduction does occur at one electrode and oxidation at the other, but in this system, attention is focused on analysis, so the electrode identification is different.

Methodology

The titrant in this part of the experiment is the permanganate ion \( \ce{[MnO_4^{-}]} \), an oxidizing agent. The balanced half-reaction and the standard electrochemical potential under acidic conditions are:

\[ \ce{MnO_4^{-} + 8H^{+} + 5e^{-} -> Mn^{2+} + 4H2O \space\space\space E^{∘} = }1.51 \space V \]

The sample being titrated is a \( \ce{Fe^{2+}} \) solution represented by the following half-reaction. Notice that this reaction is written as an oxidation. (Does this need to be balanced?)

\[ \ce{Fe^{2+} -> Fe^{3+} + e^{−}} \]

A Nernst equation can be written for each of the half-reactions independently. The Nernst equation for the permanganate is given below. What is the Nernst equation for the iron half-reaction?

\[ \ce{E_{PM} = E_{PM}^{∘} - \frac{RT}{5F} \ln{\frac{[Mn^{2+}]}{[MnO_4^{-}][H^{+}]^8}}} \]

The potential difference developed in the cell is defined by the equation

\[ \ce{ E_{cell} = E_{ind} − E_{ref}} \]

where \( \ce{E_{cell}} \) is the electrochemical potential of the entire cell, \( \ce{E_{ind}} \) is the electrochemical potential of the indicator electrode (cathode), and \( \ce{E_{ref}} \) is the electrochemical potential of the reference electrode (anode). You will be collecting cell potential values, but it is important to be aware of the setup of this experiment and how this differs from Experiment 1.

The discussion below goes into further detail about each of these potentials and explains how to calculate each. Following this discussion, a derivation of the equilibrium cell potential for the redox reaction between iron and permanganate will be given. You are expected to read through this discussion; not all the details are given, so you may need to read a little more carefully.

The Reference Electrode

The reference electrode is a system whose potential is maintained constant and that is essentially isolated from the system in which analysis occurs. Electrical contact is maintained through a liquid junction that plays a role similar to the salt bridge employed in Experiment 1. The reference electrode is an \( \ce{Ag-AgCl} \) electrode. The half-reaction for this system is

\[ \ce{AgCl_{(s)} + e^{-} -> Ag_{(s)} + Cl_{(aq)}^{-}} \]

where \( \ce{AgCl(s)} \) is an insoluble precipitate on the surface of \( \ce{Ag} \) metal. The electrochemical shorthand for the electrode is

\[ \ce{Ag \space | \space AgCl \space | \space sat'd \space KCl \space |} \]

The electrolyte used in the liquid junction is saturated \( \ce{KCl} \) because it minimizes the liquid junction potential between the electrolyte in the reference electrode and the external test solution (see Figure \(\PageIndex{1}\)).

The Nernst equation for the \( \ce{Ag-AgCl} \) electrode is

\[ \ce{E_{ref} = E - \ln{[Cl^{-}]} = }0.197 \space V \]

where the standard potential, \( \ce{E_{ref}^{∘}} \), for this half-reaction is 0.222 V and the concentration of \( \ce{Cl^{-}} \) ion in saturated \( \ce{KCl} \) at 25 °C is approximately 4 M. For additional information about reference electrodes, see the following article: Journal of Chemical Education. 1999, 79, 97.

The Indicator Electrode

The indicator electrode in the experimental cell is a platinum wire. It acts as a source or a sink for electrons and does not participate in the electrode half-reaction chemically. However, the equilibrium potential corresponding to the half-reaction in the cell is established at this electrode. For example, if the redox equilibrium in the cell is

\[ \ce{Fe^{3+} + e^{-} -> Fe^{2+}} \]

then the potential of the indicator electrode is

\[ \ce{E_{ind} = E_{Fe^{3/2}} = E -\ln} \]

where \( \ce{E} \) is the standard potential for the \(\ce{Fe(III)/Fe(II)}\) half-reaction (0.68V in a concentrated \( \ce{H2SO4} \) solution). If the concentrations of \( \ce{Fe^{3+}} \) and \( \ce{Fe^{2+}} \) are equal, then the potential of the indicator electrode is equal to the standard potential of the \( \ce{Fe^{3+} / Fe^{2+}} \) redox couple.

Given the information about the indicator and reference electrodes for the example discussed above \( \ce{ \bf{ E_{cell}}} \) is equal to 0.46V (Prove this to yourself).

Recall that \( \ce{ E_{cell} = E_{ind} − E_{ref}} \).

 

Figure \(\PageIndex{1}\): Schematic Diagram for Potentiometric Titrations with \( \ce{Ag-AgCl} \) as Reference Electrode and a \( \ce{Pt} \) as Indicator Electrode

Derivation of the Equilibrium Cell Potential

The two half-reactions for this experiment, as stated before, are:

 \[ \ce{MnO_4^{-} + 8H^{+} + 5e^{-} -> Mn^{2+} + 4H2O} \]

\[ \ce{Fe^{2+} -> Fe^{3+} + e^{-}} \]

The full oxidation-reduction reaction can be described by the chemical equation (check to make sure it is balanced)

 \[ \ce{MnO_4^{-} + 5Fe^{2+} + 8H^{+} -> Mn^{2+} + 5Fe^{3+} + 4H2O} \]

We start with an unknown sample of \( \ce{Fe^{2+}} \). Initially, the concentration of \( \ce{Fe^{3+}} \) in the sample is vanishingly small, and therefore, the ratio \( \ce{[Fe^{2+}]/[Fe^{3+}]} \) is very large. Since the potential of the indicator electrode is represented by the equation as shown in the previous section

\[ \ce{E_{ind} = E_{Fe^{3/2}} = E -\ln} \]

this means the potential of the indicator electrode will be very negative. As \( \ce{MnO_4^{-}} \) is added from the titration burette, the concentration of \( \ce{[Fe^{3+}]} \) increases, and the potential of the indicator electrode moves in the positive direction. In this experiment, we will follow (and graph) \( \ce{E_{ind}} \), the indicator potential, as a function of the volume of \( \ce{MnO_4^{-}} \) (in mL) solution added.

The equivalence point is reached when 1 mole of \( \ce{MnO_4^{-}} \) has been added for every 5 moles of \( \ce{Fe^{2+}} \) originally present in the unknown sample (cf. Equation \(\PageIndex{12}\)). At this point

\[ \ce{[Fe^{2+}] = 5[MnO_4^{-}]} \]

Although the concentration of \( \ce{Fe^{3+}} \) and \( \ce{Mn^{2+}} \) are vanishingly small because of the equilibrium shown in Equation \(\PageIndex{12}\), the equality

\[ \ce{[Fe^{3+}] = 5[Mn^{2+}]} \]

also holds throughout the titration.

At equilibrium, the electrochemical potential of the oxidation and reduction half-reactions in the titrated solution should be approximately equal (PM is permanganate).

\[ \ce{E_{Fe^{3/2}} = E_{PM}} \]

As a result

\[ \ce{E - \ln = -\ln} \]

Equation \(\PageIndex{12}\) can be used to write an expression for the equilibrium constant of the reversible reaction

\[ \ce{K_{eq} = \frac{[Mn^{2+}][Fe^{3+}]^{5}}{[H^{+}][Fe^{2+}]^{5}[MnO_4^{-}]} = 10^{161}} \]

The extremely large equilibrium constant indicates that the reaction is strongly shifted toward the products side and essentially goes to completion. Rearranging Equation \(\PageIndex{17}\), the logarithm (note that we have switched from the "natural-log" in Equation \(\PageIndex{17}\) to "log" in Equation \(\PageIndex{19}\) of the equilibrium constant is

\[ \ce{log\frac{[Mn^{2+}][Fe^{3+}]^{5}}{[H^{+}]^{8}[Fe^{2+}]^{5}[MnO_4^{-}]} = } \frac{5\ce{F}}{2.303\ce{RT}} \ce{(E_{PM}^{∘} - E_{Fe^{3/2}}^{∘}) = 161} \]

To estimate the value of \( \ce{E_{ind}} \) at the equivalence point, we add 5 times Equation \(\PageIndex{3}\) to Equation \(\PageIndex{13}\). Keeping in mind that the equality in Equation \(\PageIndex{16}\) is valid at any point during the titration, we obtain

\[ \ce{E_{Fe^{3/2}} + 5E_{PM} = 6E_{ind} = E + 5 -\ln} \]

Applying the conditions for the equivalence point

\[ \ce{E_{equiv} = E_{ind} = \frac{E_{Fe^{3/2}}^{∘} + 5E_{PM}^{∘}}{6} - \frac{RT}{6F} \ln{\frac{1}{[H^{+}]^{8}}}} \]

When the concentration of \( \ce{H^{+}} \) is 1 M, the second term of Equation \(\PageIndex{21}\) goes to zero and

\[ \ce{E_{equiv} = } \frac{0.68 + (5 \times 1.51 \, V)}{6} = 1.37 \, V \]

We have just derived the equilibrium potential for the redox reaction of this experiment, assuming that the concentration of \( \ce{\bf{H^{+}}} \) is 1 M. When you carry out the titration, you will find that the potential of the indicator electrode changes very rapidly in the region of the equivalence point. It is important to anticipate this factor and titrate more slowly in the region of the equivalence point potential to obtain enough data to characterize this region of the plot.

Experimental

The \( \ce{KMnO4} \) solution provided by the storeroom has an approximate concentration of 0.02 M. You will be given the exact concentration so that you can complete the necessary calculations. Use a 50 mL beaker to get 40 mL potassium permanganate solution. Transfer the solution into the burette provided.

We will determine the concentration of \( \ce{Fe^{2+}} \) in a \( \ce{Fe(NH4)2(SO4)2} \) sample. This material is obtained as a hydrate, so its composition varies somewhat depending on the amount of water associated with each formula unit of \( \ce{Fe(NH4)2(SO4)2} \). Weigh out approximately 0.8-1.0 g of the iron salt and dissolve it in exactly 100 mL of 1 M \( \ce{H2SO4} \) within a 250 mL beaker. Accurately measure \( \ce{H2SO4} \) with a 100 mL graduated cylinder.

Connect the voltage probe to the LabQuest interface and the wires in accordance to Figure \(\PageIndex{1}\). Start Logger pro, select Experiment then click on Data Collection. Change the selection in Mode to Events with Entry, type in "mL" in the Column Name box then click Done.

Prepare a rinse beaker with DI water. Uncap the reference \(\ce{Ag/Cl}\) electrode and platinum wire and rinse both in DI water before adding them to your sulfuric acid solution. Add a magnetic stir bar then start stirring. Now you are ready to carry out the titration. Click Collect then click on Keep then enter 0 (mL) for the initial reading. Titrate 2 ml press Keep then enter 2 (ml). Continue titrating and recording (input only the total value added (mL)) every 2 mL increment up to the point that brings you closest to the equivalence point (~20 mL). Stir the solution until the \( \ce{MnO_4^{-}} \) color disappears and then measure \( \ce{E_{cell}} \). Use 0.5 mL increments until you come within a fraction of an mL from the equivalence point (~22 mL). The final part of the titration should be carried out drop-wise. Click Stop to end collection and save your data.

Based on the weight of the iron sample and the concentration of the \( \ce{KMnO4} \) titrant, estimate the number of milliliters of the \( \ce{KMnO4} \) solution that are required to reach the equivalence point.

Permanganate is a particularly good redox reagent because it provides its own color indicator. Note the volume added when the pink color remains for at least 1 minute. Does this correspond to the calculated equivalence point? Continue adding \( \ce{\bf{MnO_4^{-}}} \) beyond the equivalence point in 1 mL increments so that you can see the total shape of the titration curve.

Enter your data on a spreadsheet and use a numerical technique to determine when the change in \( \ce{E_{cell}} \) with the volume of titrant is at a maximum. The data analysis section will give you more direction about the conclusion and calculations to be performed.

Clean Up

All \( \ce{Fe} \) and \( \ce{KMnO4} \) solutions must be disposed of in the proper waste bottles.

Data Analysis

1. Plot a graph of \( \ce{E_{cell}} \) against the volume of titrant. Graph and calculate \( \ce{E_{equiv}} \) from the 1^st & 2^nd derivatives of \( \ce{E_{cell}} \) vs. volume. On the \( \ce{E_{cell}} \) vs. volume graph, identify \( \ce{E_{equiv}} \) and the point where the color change remained permanent. 
2. Based on the change in \( \ce{E_{cell}} \) with the volume of titrant at the equivalence point, estimate the precision of the titration. This is found from the maximum change in \( \ce{E_{cell}} \) for the addition of 1 drop (0.05 mL) of titrant (1^st & 2^nd derivatives). In other words, how much does the potential change at the equivalence point for a given volume?
3. Calculate the concentration of \( \ce{Fe^{2+}} \) in your sample based on your titration curve. To do this, consider how the equivalence point is defined. When you did acid/base titrations, why did you do calculations at the equivalence point instead of a "random" point of the curve?
4. Determine the concentration of \( \ce{Fe^{2+}} \) in your sample based on the mass of \( \ce{Fe(NH4)2(SO4)2} \) weighed, assuming the molecular weight on the bottle of iron salt is correct.
5. You just used two different methods (steps 3 and 4) to calculate the concentration of iron in your sample. Discuss any deviations between the answers obtained by these two calculations. If your answers were the same, why might there be a difference? In addition, you estimated the equivalence point of the titration during the experiment. Was this calculated volume to reach the equivalence point the same as the equivalence point based on your graph? Discuss any reasons why there is or why there might be a difference.
6. The analyst needs to find the exact concentration of the unknown, i.e. the \( \ce{Fe^{2+}} \) concentration. The cell is responding to the activity or effective concentration of \( \ce{Fe^{2+}} \). Discuss how this affects the analysis in the present experiment. In other words, how or why might using concentration instead of activity affect the outcome of the calculations?