Copper Chloride Experiment

Copper Chloride Experiment

Introduction

The velocity in a chemical reaction can vary, depending on the concentration of the substance, the temperature under a reaction, and the area on the substance. For an example when water turns into steam. If the temperature rise until the water starts to boil it will evaporate faster than if when it stays outside in a warm and sunny day. The water will transform into steam sooner or later, but it will not be as fast as when boiling.

Questions

Question - I recently carried out a percent yield lab in which iron (limiting reagent) was reacted with copper(II)chloride solution and then the students collected the copper precipitate. The reaction given for this lab in the textbook predicted Copper and iron(II)chloride as the products. The theoretical yield from 1.00g of iron (steel wool actually) was 1.14g of copper. The samples were allowed to dry for several days and then weighed. Much more copper was produced that expected. I performed the lab myself and recovered 1.72g. My question is ....is the product actually iron(III)chloride? if so then the amount produced would make sense. The filtrate had a greenish-blue color. ferrous ions give a pale green color while ferric ions a yellow-brown color, however is it possible that the excess cupric ions are masking ferric ions and that they are produced and not ferrous ions? If not then what other possible reason would there be for so much product. In some cases there was 0.60 to 1.00g more that expected. The samples appeared to be quite dry and powdery so I cannot see that it is due to remaining water. --------------------------------------------------------------------------

Unless you take precautions to exclude air, the Fe(II) will readily be oxidized by atmospheric oxygen to Fe(III) oxides/hydroxides i.e.rust. The copper could mask this and could account for the excess weight of reaction product. My response was based upon my recollection that Cu(+1) is not stable under 'ordinary' conditions -- see sites below. Jim Swenson's explanation could be correct, I just don't know off the top of my head. http://nautarch.tamu.edu/class/anth605/File12.htm http://www.uncp.edu/home/mcclurem/ptable/copper/cu.htm Vince Calder =====================================================

Yvonne, If we check a Standard Reductions Table: Fe(2+) + 2e- = Fe(s) -0.41V Fe(3+) + 3e- = Fe(s) -0.04V Cu(2+) + e- = Cu(+) 0.16V Cu(2+) + 2e- = Cu(s) 0.34V Cu(+) + e- = Cu(s) 0.52V Fe(3+) + e- = Fe(2+) 0.77V This means that if you started with Fe(s)and Cu(2+)ions the most "spontaneous" reaction (the one having the most negative Gibb's Free Energy) is the transformation of Cu2+ to Cu(s) and Fe(2+) to Fe(s) - giving a voltage difference of 0.75V. Thus, on purely thermodynamic reasons the only possible reaction is that of: Fe(s) + Cu(2+) = Fe(2+) + Cu(s). It is not possible to collect iron(III) chloride because that will remain in solution as Fe(3+) and Cl(-) ions --the combination of these ions do not form a precipitate. Thus, the reason for your unexpectedly high mass yield must come from some other source. My guess would be that over time, the copper you collected converted to copper(II) oxide. There is such a large mass difference between Cu (s) and CuO that even just a little of it on the surface of your collected Cu that it could easily account for your mass discrepancy. I am also guessing that the copper you collected came out as small pellets so that there will be a large surface to volume ratio and a lot of the copper will be exposed to air. Greg (Roberto Gregorius) =====================================================

Yvonne- are you aware of Cu(I) Chloride? Although formation of Fe(III) would balance the reaction and explain some of your excess product, energetically it just can't happen with copper. Fe(III) is a moderately strong oxidizer. Cu+2 can't oxidize Fe+2 to Fe+3. FeCl3 is a standard copper etchant for making printed circuit boards! So if there was a significant amount of Fe(III) present: a) the solution would have lost free energy to make it: 2 Fe+2 + Cu+2 <-- 2 Fe+3 + Cu b) it would immediately spend itself dissolving some of your copper powder: 2 Fe+3 + Cu --> 2 Fe+2 + Cu+2 (True, almost saying the same thing twice. (a) is theory from my CRC, and (b) is highly empirical.) I can only think of one cogent reason why you are getting more product than expected: Your Cu(II) Chloride was really Cu(I) Chloride. Not all copper compounds have a stable Cu(I) solution, but Chloride is one that does. An old bottle of "Copper Chloride" crystals with a poor label might not validate an assumption that it is in the (II) state. "Cuprous" means Cu(I), "Cupric" means Cu(II). If you have any practical jokers present, I suppose a perfect accurate Cu(II) solution could be partly converted to Cu(I) by addition of an 0.3 gram bit of steel wool. Perhaps there would be complete dissolution of iron and no tell-tale precipitation of Cu(0), and not too much color change. In the course of running a normal, accurate experiment, you might observe that precipitation did not start until about half the iron is dissolved, because most or all of the Cu(II) might need to be converted to Cu(I) before any becomes Cu(0). Of course, to see this, you'd need to add your steel wool in about 3 smaller pieces, instead of one large piece. Only the first piece would dissolve completely without causing copper powder precipitate, leaving no solids in a dark but somewhat clear liquid. On the other hand, Cu(I) Chloride solution, left for a long time in air, would try to convert itself to Cu(II) solution. Lacking a balancing amount of Cl-, it would end up being CuClOH. Cu(OH)2 is quite insoluble, so some precipitate would probably form. Given some HCl, the precipitate would not need to form. Then Cu(I) chloride would be cleanly converted to Cu(II) chloride by shaking with air or by standing a long time. This kind of thing might help explain some of the variability in your yield, without resorting to "jokers". How much exposure to air does your reaction meth...