Potentiometric Titration Using a Platinum Indicator Electrode
Iron in iron salt has two main oxidation states: Fe3+ and Fe2+. During this titration, Fe2+ can react with the titrant, Ce4+, via the following reaction:
Fe2+(aq) + Ce4+ (aq) Fe3+(aq)+ Ce3+ (aq) (1)
To accomplish the titration, all iron in the iron sample has to be converted into Fe2+. The iron in the 3+ oxidation state must be reduced to Fe(II) by careful addition of tin(II) chloride. The tin(II) chloride will be in excess, but only very slightly. This excess tin(II) must be removed and this is accomplished by adding mercury(II) chloride which oxidizes it to tin(IV). This series of reactions can be represented by the following equations:
2 Fe3+ + Sn2+ (in excess) 2 Fe2+ + Sn4+ (2)
The excess Sn2+ is reacted with Hg2+:
Sn2++ 2 Cl– + 2 Hg2+ Sn4+ + Hg2Cl2 (white ppt) (3)
If too much Sn2+ is added, the excess will further reduce the Hg2Cl2 (Hg(I) to metallic mercury (Hg(0)):
Sn2+ + Hg2Cl2 Sn4+ + 2 Cl– + 2 Hg (black ppt) (4)
The mercury metal interferes with the titration.
The method of potentiometry consists of measuring the potential difference between an indicator electrode and a reference electrode at various intervals during the progress of a titration. The short hand notation of the poetniometric titration is:
Hg(l)/Hg2Cl2(s)/saturated KCl (aq)//Fe3+(aq), Fe2+(aq)/Pt.
Since the potential of the reference electrode (Hg(l)/Hg2Cl2(s)/saturated KCl (aq)) is fixed, the potential difference between the reference and Pt electrodes can be expressed by:
Where, E is the observed potential in mV or volts, k is a constant, R is the gas constant, T is absolute temperature, n = the numbers of electrons transferred in the electrochemical reaction, F is Faraday constant, aFe2+ and aFe3+ are the activities of the reduced and oxidized forms responsible for the equilibrium potential. In dilute solutions, the above equation is used with concentration replacing activity.
Due to reaction 1 in the titration, the concentration of Fe3+ will increase and the concentration of Fe2+ will decrease. Thus, according to equation (5), there is a change in potential during the titration. When the titration reaches the equivalence point, there will be a jump in potential. Once the equivalence point is obtained, the unknown sample concentration can be calculated on the base of equation 1.
0.5 M SnCl2·2H2O dissolved in 3.5 M HCl (with excess mossy tin added).
Saturated mercuric chloride solution.
Take approximately 7.5 grams of cerium (IV) ammonium sulfate dehydrate, add 12.5 mL of concentrated sulfuric acid and then carefully add 12.5 mL of distilled water. Stir the mixture for 30 minutes. Add 25 mL of distilled water and stir again for 10 minutes. Continue the addition of 25 mL portions of distilled water with intervening stirring until it is dissolved. Allow the solution to cool to room temperature, then dilute to 125 mL and store in a glass bottle. The resulting solution will be approximately 0.1 M in Ce (IV).
Weigh out the appropriate amount of primary standard ferrous ammonium sulfate. You should calculate the amount by assuming that you want your equivalence point to take about 25 mL of your Ce (IV) solution.
B. Standardization of Ce(IV) solution
Dissolve your standard ferrous ammonium salt in about 15 – 20 mL of 6 M HCl. Add drop by drop 0.5 M SnCl2 to the sample. You also need to swirl the solution at regular intervals. When the brown (or yellow) color turn colorless or a pale green, stop the addition of SnCl2 and let the solution sits on your desk for about 2 minutes. If the solution does not turn back to yellow, you are done with the reduction of the sample. If the solution does turn yellow, repeat the addition of SnCl2 until the solution is colorless or pale green. If the yellow color does not return, ADD AT MOST 2 EXTRA DROPS OF STANNOUS CHLORIDE TO THE SOLUTION and this sample can be titrated.
Add distilled water until the solution is approximately 75 mL, then quickly add 10 mL of saturated mercuric chloride solution which has been readied for this purpose. White precipitate should form, no matter how trace the amount. The amount of precipitate indicates how much Sn(II) is in excess. A small excess is best. However, if no precipitate is formed (waiting more than 2 minutes might help to have them formed), or that a heavy gray or black precipitate is formed instead (this is disastrous, nothing can be done about this), the sample has to be discarded.
Immediately immerse your platinum inlay electrode and reference electrode in the solution; record the initial mV and volume of your titrant; and start titrating your iron sample with the cerium (IV) solution. Add titrant in 0.2ml increment, record the mV and total volume of titrant after each addition. You should pass through the end point (big jump in mV) and titrate 2ml beyond it.
C. Sample Analysis
Precisely weigh out around 1.0 g of unknown sample.
Repeat the steps used in the standardization of the cerium (IV) solution for your unknown.
Do the sample anlaysis two times
Please explain why the potential of the solution keeps increasing when more titrant is added into the solution
If a student runs out of time and stops the titration as soon as the “big jump in mV” appears what happens and what might the consequences of this be if any? Is this considered an endpoint, equivalence point, or both? Explain.
In the redox reaction of Fe3+ and Sn2+, the lab manual directs students to add drops of SnCl2 solution until the brown-yellow solution turns colorless, then to add 2 extra drops even though the reaction seems complete. Why is this important? What quantitatively might happen if the two extra drops are not added? Explain.
There are two electrodes: the reaction occurs on the left electrode is : Ag(s) + Cl-( AgCl(s)+ e aq); and the reaction occurs on the right electrode is : Fe3+(aq) + e Fe2+(aq). If the concentration of Cl- is 0.0100M, Fe3+ is 0.0050M and Fe2+ is 0.0010M, please calculate the potential between the right and left electrodes (right over left)
Please list at least 5 sources which can directly affect the accuracy of the titration, please briefly explain why they can affect.