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Fe2+ or Fe3+?

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by

Vivian Ngo

on 22 May 2014

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Transcript of Fe2+ or Fe3+?

Hypothesis:

Based on the balanced chemical equations:

Fe (s) + CuSO4 (aq) ‡ Cu (s) + FeSO4 (aq)

2 Fe (s) + 3 CuSO4 (aq) ‡ 3 Cu (s) + Fe2(SO4)3 (aq)

We believe that since Fe3+ has no electron repulsion in its d subshell, more Fe3+ will be produced as time progresses while there will be little Fe2+ produced.
Procedure:


1. Mass an iron nail and record.

2. Mass about 5 g CuSO4 and dilute it to the appropriate concentration using proportions.

3. Place the nail in a large test tube and add 0.1 M CuSO4 until it is about 2-3 cm from the

top. Place a stopper in the tube and stop after 10 minutes.

4. Mass filter paper and record. Set up filtration apparatus and orient the filter paper

properly in the funnel.

5. Filter the solution into the paper and wash any residue from the test tube into the filter

paper.

6. Carefully remove the paper from the funnel and place it in a watch glass and place it in

the oven at 100 °C overnight.

7. Remove the watch glass and allow it to cool down room temperature.

8. Repeat steps 1-7, but instead of using 10 minutes in part 3, use 20 minutes.
Materials:


2 iron nails, 10 g Cu(SO4), filter paper, ring stand, clamp, 250-mL beaker, 100-mL graduated cylinder, 1 mL-pipette, oven, funnel, electronic weigh scale, test tube, rubber stopper, watch glass, sand paper.
Calculations 10 Minute Trial

10 Min Trial 20 Min Trial
Initial Mass of nail(g) 25.4396 g 25.8953 g

Mass of CuSO4 (g) 5.0000 g 5.0000 g

Mass of filter paper (g) 0.9225 g 0.9446 g

Mass of filter paper + Cu (g) 0.9277 g 0.9560 g

Final Mass of nail (g) 25.4351 g 25.8908 g

Volume of water (mL) 313.4 mL 313.4 mL
Important Chemical Equations
Conclusion
Analysis
Fe2+ or Fe3+?
Background:

There are numerous elements that form more than one oxidation state when forming ions.

Iron is one such element that can form Fe2+ or Fe3+.

Electron configuration is the distribution of electrons of an atom or molecule in atomic

or molecular orbitals. Electron configuration describes the arrangement of electrons in 4 distinct

orbitals around the nucleus (s,p,d,and f).

Oxidation states are determined by the amount of electrons that the atom loses or gains

when forming compounds or the ability for an atom to be oxidized or reduce other atoms in

a species. The oxidation states are usually defined by observing unpaired or paired electrons

present in the s,p,d,and f subshells.

Transition metals contain electrons in the d orbital. Since the d orbital has 5 suborbitals

in the subshells, it is further away from the nucleus and therefore have a lower attraction to the

nucleus. Electrons in the d subshell have a higher tendency to leave the atom because of its

distance from the nucleus making the electrons in the subshell to be unstable and highly eager

to bond with other species. This phenomena leads to a variety of oxidation states based on the

amount of electrons removed from the atom.


Purpose:


To determine if the ratio of Fe2+ : Fe3+ formed in the redox reaction between iron nails
and CuSO4 is dependent on time.



Data Table

Fe (s) + CuSO4 (aq) Cu (s) + FeSO4 (aq)

2 Fe (s) + 3 CuSO4 (aq) 3 Cu (s) + Fe2(SO4)3 (aq)



Calculations
20 Minute Trial
In this lab we were to determine the ratio of Fe2+ and Fe3+ produced as an iron nail was being oxidize by CuSO4.
Based on our calculations, Fe2+ was much more prominent in the solution while Fe3+ was not. As time progressed, however, more Fe3+ was produced and the ratio of Fe2+ : Fe3+ became less extreme (from 31:1 to 12:1). We can view this by looking at the electrons that occupy the d orbital of iron. Since Fe3+ is more stable (has 5 electrons evenly distributed in its d orbital shubshells), it is less reactive. Fe2+, however, is unstable due to a 6th electron that equates to an unequal distribution to electrons in the d orbital. Therefore Fe2+ will want to give away its electron and be more reactive.
Errors in this lab may include the un-thorough scraping of the iron nail to acquire all the copper (rust) produced. This will lead to a lower calculated moles of copper and lead to a decrease in percent yields for Fe2+.
ratio of percentages :
Fe2+ : Fe3+
96.89 : 3.11
31 : 1
ratio of percentages:
Fe2+ : Fe3+
92.27 : 7.73
12 : 1
The purpose of this lab was achieved through the means of determining the percent yield of Fe2+ and Fe3+ in a solution by the use of stochiometric calculations after gathering experimental data. From our calculations, we have determined that our hypothesis was only partially true. The concentration of Fe3+ will increase overtime but the concentration of Fe2+ in the solution is more prominent.
Arman Ahmed, Vivian Ngo, Justin Fabia
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