sankalpmittal
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Question based electrochemistry  electromotive force....
If excess metallic iron is added to an NCuSO_{4} solution, calculate the approximate concentration of Cu^{2+} when equilibrium is
established.
N is normality. It means that normality of solution:CuSO_{4} is 1 N.
I do not know where to begin. I tried applying nerst equation but to no avail..
Please help !!
Thanks in advanced..


Artemus Gordon
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I think this is a trick question. Iron is more reactive than copper, so if you dropped a sphere of iron into a beaker of 1N CuSO4, the Fe will be
oxidized and go into solution and the Cu++ ions will reduce to metallic copper and stick to the surface of the sphere. BUT once the sphere is fully
copper plated, the reaction will stop. OTOH, if the same amount of iron is in powder form, I think the reaction will proceed until the Cu++ is fully
consumed. So, unless you know the available surface area of the iron, this question can't really be answered.


Artemus Gordon
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BTW, how do you type superscripts and subscripts?


Artemus Gordon
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Please ignore my last post. I didn't know 'post reply' gave me more tools than 'quick reply'
CuSO_{4}
Cu^{2+}


Vargouille
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It's kinda a trick question, but not in the way that you think, Artemus. When metallic iron is mixed with copper (II) solutions, there may be an
equilibrium granted that the iron is not in excess. There's some math you can do with all of the possible reactions to find the answer for all of the
species in question, but that's not the case. When the iron is in excess, it will reduce all cupric ions to metallic copper. If it is specifically
mentioned that the surface area is significantly smaller than the volume of iron added, then you have to be given some way to calculate the surface
area to calculate the concentrations at equilibrium with some other math.


sankalpmittal
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Quote: Originally posted by Vargouille  It's kinda a trick question, but not in the way that you think, Artemus. When metallic iron is mixed with copper (II) solutions, there may be an
equilibrium granted that the iron is not in excess. There's some math you can do with all of the possible reactions to find the answer for all of the
species in question, but that's not the case. When the iron is in excess, it will reduce all cupric ions to metallic copper. If it is specifically
mentioned that the surface area is significantly smaller than the volume of iron added, then you have to be given some way to calculate the surface
area to calculate the concentrations at equilibrium with some other math. 
Ok, so I have the equation:
Fe+CuSO_{4}> FeSO_{4} + Cu
Which can be written as
Fe + Cu^{2+} > Fe^{2+} + Cu
Iron is being oxidized and copper is reduced. As iron is in excess, it will nearly reduce all cupric ions to metallic copper.
Nerst equation is
E_{cell} = E^{o}_{cell}  0.0591log{[Fe^{2+}]/[Cu^{2+}]}/n
From table,
E^{o}_{cell} = 0.78 for the above pairs.
At equilibrium, E_{cell} =0
Also we know that initially [Cu^{2+}] = 0.5 M
Because Molarity=Normality/valency factor = 1/2
What shall I do now ? Hints ?


Vargouille
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I may have been incorrect, actually. I think I know how they want you to get the answer, and it involves some algebra. Also, the unit of concentration
doesn't matter as long as you use the same one consistently. As long as we're assuming T=298K, let's go ahead.
Since normals are equivalents per liter, we can treat (mathematically) "[Fe^{+2}]" and "[Cu^{+2}]" as a number of equivalents without
changing the values themselves. If we say that [Fe^{+2}] = x, then [Cu^{+2}] = 1x, because one unit of cupric will react with one
unit of iron, and we start out with a concentration of 1N for cupric. So, you plug in E^{o}, figure out what n is (you know it, right?), solve
for x, and then for "1x", which is your answer in normals.


Artemus Gordon
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Quote: 
It's kinda a trick question, but not in the way that you think, Artemus. When metallic iron is mixed with copper (II) solutions, there may be an
equilibrium granted that the iron is not in excess. There's some math you can do with all of the possible reactions to find the answer for all of the
species in question, but that's not the case. When the iron is in excess, it will reduce all cupric ions to metallic copper. If it is specifically
mentioned that the surface area is significantly smaller than the volume of iron added, then you have to be given some way to calculate the surface
area to calculate the concentrations at equilibrium with some other math.

IANA expert, but I don't really understand you here. First, the posted question explicitly said the Fe is in excess. Second, is it common to assume in
chemistry class that surface area of a metal is not a limiting factor? Because in real life it usually is. Third, as I understand it, the Nermst eq.
is for calculating gradients of 2 halfcells separated by a membrane, but that is not the case here. Are you sure that an equilibrium will be
established? All of the Cu^{2+} ions are physically able to freely contact the Fe and become reduced, aren't they? (assuming no SA
limitation.) Do you think there is a reverse reaction that causes some of the metallic copper to be oxidized back to Cu^{2+}?
[Edited on 1982013 by Artemus Gordon]
[Edited on 1982013 by Artemus Gordon]


Vargouille
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To respond to your first point, I know. My point about the multiple species isn't important, since it doesn't apply to the matter at hand. For the
situation at hand, however, having done the math I suggested the OP do, the answer is so incredibly close to one, such that 1x is so close to 0, that
there should be no atoms of cupric left unreduced. To your second: yes, if it is not specifically mentioned in the question, and no information is
given to allow calculating with that in mind. To your third: it would depend on the form you use. The one I am familiar with is useful, if I remember
correctly and the textbook I just looked at wasn't misleading, for all aqueous redox reactions. In any case, the nature of the Nernst equation implies
that if the concentrations are disparate enough, it's possible for the reaction to go in reverse, even if just for a few atoms. That's the trick I
remembered from a couple of years back that made me pause. From my calculation, however, the conditions given in the OP weren't disparate, or exotic,
enough to keep it from going quantitatively to completion.


sankalpmittal
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Quote: Originally posted by Vargouille  I may have been incorrect, actually. I think I know how they want you to get the answer, and it involves some algebra. Also, the unit of concentration
doesn't matter as long as you use the same one consistently. As long as we're assuming T=298K, let's go ahead.
Since normals are equivalents per liter, we can treat (mathematically) "[Fe^{+2}]" and "[Cu^{+2}]" as a number of equivalents without
changing the values themselves. If we say that [Fe^{+2}] = x, then [Cu^{+2}] = 1x, because one unit of cupric will react with one
unit of iron, and we start out with a concentration of 1N for cupric. So, you plug in E^{o}, figure out what n is (you know it, right?), solve
for x, and then for "1x", which is your answer in normals. 
On solving as you said,
x/(1x) = 2.4885*10^{26}
I used n=2 in nerst equation due to transference of 2 (moles perhaps ?) electrons.
Now it is given that Iron was in excess, so it will reduce all the cupric ions to copper. We can approximate that x>1 or x=1
Which yields 1x=4*10^{27} equivalents.
[1x]= 2*10^{27} moles or Molar concentration.
(Equivalents/valency factor = Moles)
Here valency factor = 2..
But answer given is 3*10^{27} M concentration !!!
Where did I go wrong ??


Vargouille
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That's the right answer. Indeed n=2 moles of electrons. The thing is that the Nernst equation uses the Faraday constant, which is 96,485 C/mol
e^{}, and in the form it is used, the "moles of electrons" from n and F cancel out. As for your answer, the difference is just due to
rounding errors. Some textbook have different rounding rules, but I use the general rule of keeping four digits after the decimal when I'm not working
with a graphing calculator, which has the "ANS" key to remember the last answer and treat it like a constant. I didn't approximate anything, either, I
just did the algebra and solved for x. The way you did was treat it as y/1x, where y=1 but x=/=1 so that you could get a sensible answer. Your answer
was still close to the one I got, but not quite the same.


sankalpmittal
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Quote: Originally posted by Vargouille  That's the right answer. Indeed n=2 moles of electrons. The thing is that the Nernst equation uses the Faraday constant, which is 96,485 C/mol
e^{}, and in the form it is used, the "moles of electrons" from n and F cancel out. As for your answer, the difference is just due to
rounding errors. Some textbook have different rounding rules, but I use the general rule of keeping four digits after the decimal when I'm not working
with a graphing calculator, which has the "ANS" key to remember the last answer and treat it like a constant. I didn't approximate anything, either, I
just did the algebra and solved for x. The way you did was treat it as y/1x, where y=1 but x=/=1 so that you could get a sensible answer. Your answer
was still close to the one I got, but not quite the same. 
So what answer are you getting ? Why does the book give the answer different from mine ? The answer of textbook = 1.5 times of my answer which is
quite big a deal. Values given to us are quite small and that is why this much of difference is significant....


Vargouille
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1.5 times your answer isn't that big of a deal when you're dealing with rounding errors and your answer is a single SF answer of 2 to the whatever
power. If your value was 6 to the whatever power instead of 2, then you'd have a problem. My answer was 2.01x10^27 M. Granted, I used the equation
you gave in one of your posts rather than the one that uses natural log, which may have something to do with it. Not knowing the specific steps the
book used to get the answer, or how they rounded, I would get someone to check my math or find a solution guide.
EDIT: Having redone it using the natural log version, I got an answer of 2.07x10^27. I was playing around with Wolfram Alpha to see what might've
happened, and the error might have been that instead of 2.4485x10^26 for x/1x, they used 2x10^26, which gives 2.5x10^27 for the final molarity,
which they might've rounded up to 3x10^27. Good enough for government work, ey?
[Edited on 2082013 by Vargouille]


sankalpmittal
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Quote: Originally posted by Vargouille  1.5 times your answer isn't that big of a deal when you're dealing with rounding errors and your answer is a single SF answer of 2 to the whatever
power. If your value was 6 to the whatever power instead of 2, then you'd have a problem. My answer was 2.01x10^27 M. Granted, I used the equation
you gave in one of your posts rather than the one that uses natural log, which may have something to do with it. Not knowing the specific steps the
book used to get the answer, or how they rounded, I would get someone to check my math or find a solution guide.
EDIT: Having redone it using the natural log version, I got an answer of 2.07x10^27. I was playing around with Wolfram Alpha to see what might've
happened, and the error might have been that instead of 2.4485x10^26 for x/1x, they used 2x10^26, which gives 2.5x10^27 for the final molarity,
which they might've rounded up to 3x10^27. Good enough for government work, ey?
[Edited on 2082013 by Vargouille] 
I read the "preface" regarding the book and I discover that they use "log tables" to get the answer instead using calculator or other tools. I think
that made all the difference ? Right ?


Vargouille
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Quote: Originally posted by sankalpmittal  I read the "preface" regarding the book and I discover that they use "log tables" to get the answer instead using calculator or other tools. I think
that made all the difference ? Right ? 
Yeah, that'll do it, sure as anything.

