Ask A Scientist Chemistry Archive

 
name         Yvonne D.
status       educator
age          40s

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.
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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 method allow?
Running your stir-bar in a closed bottle would help keep out air.

About your filtrate color:
Most Copper(II) solutions are blue to green and intense.  (Think of copper 
sulfate.)
If iron is the limiting reagent, then by implication you have an excess of 
copper(II).
So there must be some left-over copper(II), which can explain the 
blue-green all by itself.
The Fe(II) may or may not be contributing a significant amount of 
greenishness.

Fe(III) is pretty strongly colored.  If there was much of it in the solution,
it's reddish color mixed with the blue-green would make something dark, 
ugly, muddy-looking.
So I think that's direct evidence you are not getting Fe(III).

Thanks for including so much detail.
That makes it much easier to guess relevantly on your behalf.

Jim Swenson
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