Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Shadow on June 26, 2012, 04:48:59 AM
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Calculate the electrode potential of a solution at the equivalence point, where Fe2+ was titrated with KMnO4 in acid enviroment, under pH=1.5. E0Fe3+/Fe2+=0.68V, E0MnO4-/Mn2+=1.51V
Before going to the calculation, first, I don't understand the concept, what means the calculated electrode potential if there is no electrode in the solution. There are no anode nor cathode so there is no electromotive force, right?
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I do believe this is a potentiometric titration so an indicator electrode and a reference electrode would have been used.
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Whenever there is a redox system present, it has a redox potential. Electrode system with a voltmeter is just a measuring device.
Just like the air in my room have some temperature even if there is no thermometer around.
But I agree the wording is lousy. It would be better to ask "what is the redox potential of the system" or "what would be the electrode potential if the electrode were put into the solution".
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Ok, then I assume what would be the electrode potential if I had an electrode in the solution. In the answer it is written that I have to calculate the potentials for the half reactions, then the potential of the manganese redution to multiply by 5 and add the potential of the iron oxidation. Why is it calculated this way, and what is the difference with the electromotive force?
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If I put one electrode in the soluition then the potential is measured as a sum(paying attention on the half-reactions because some have to be multyplied), but if I put two electrodes then the electromotive force is created so I have to deduct the potentials. Is this true?
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If you put one electrode into solution, you can't measure anything. You always need two electrodes, one will be the reference, the other one is a measuring electrode.
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Then why is the electrode potential calculated the way I wrote?
Ok, then I assume what would be the electrode potential if I had an electrode in the solution. In the answer it is written that I have to calculate the potentials for the half reactions, then the potential of the manganese redution to multiply by 5 and add the potential of the iron oxidation. Why is it calculated this way, and what is the difference with the electromotive force?
The two half-reactions merge. Why is this different than the el.-mot. force?
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When you try to measure the potential in the solution you create another circuit - one in which solution is a half cell and reference electrode is another half cell - and you measure EMF in this circuit.
When you mix two redox systems in the solution, they will react till potential of each (as given by the Nernst equation) is identical to the potential of the other. As concentrations of ions (molecules) from both systems are combined by the stoichiometry and fact that number of electrons exchanged must be identical on both sides (technically that's just a charge conservation) you can relatively easily derive formula for the final potential of the mixture.
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Then Nernst equation is used to calculate the potential of half cells which would be Fe3+/Fe2+ and MnO4-/Mn2+, but to calculate the potential of the whole solution another procedure is used. What does that potential represent, what is it used for, why is it important?
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but to calculate the potential of the whole solution another procedure is used
It is not another procedure. The difference is you assume reaction reached the equilibrium (which is hardly surprising - if you mix reagents they will react till they reach the equilibrium).
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Why, when calculating the electrode potential I have to multiply by 5 one half-reaction, and when calculating el.-mot. force not?
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I don't see it being true.
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You mean that the EMF to be calculated, the number of electrons that take part in the half reactions should be the same, so again it must be multyplied.
Why is the electrode potential at the equilibrium a sum? It should be 0?
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At equilibrium potentials of both half cells are identical, but the potentials itself are not zero - and the electrode put into solution will have this potential.
In other words EMF is zero, but the potentials are not zero. Not that it makes sense to speak about EMF in this context.
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Ok, just one question: Why is the potential of the MnO4-/Mn2+ half-reaction multylpied by 5, shouldn't the other reaction be multyplied by 5 (Fe3+/Fe2+)?
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Why is the potential of the MnO4-/Mn2+ half-reaction multylpied by 5
It is not, you are misunderstanding something.
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Then it is a mistake in the source I read this. Now, in my book there are some very simple examples, to calculate the EMF for Mg/Zn, Cu/Pb... cells. In these problems all the metals are equivalent so I didn't have trouble to do it, but in the problems, like the one I posted, it is more difficult for me to understand how to calculate it and to differentiate it from the electrode potential. I thank you for the help you gave me , it would be probably too much to ask from you, but:
Do you maybe know a good site where those more complex problems are correctly and more deeply explained?
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For permanganate potential the only place 5 would be present is the number of electrons:
[tex]E = E_0 + \frac {RT} {5F} \ln \frac{[\textrm{MnO}_4^-][\textrm{H}^+]^8}{[\textrm{Mn}^{2+}]}[/tex]
But number of electrons is always in the same place, no matter of what redox system you are dealing with.
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The source I was reading from says:
E1=E0+0.059/5*log[MnO4-][H+]8/[Mn2+]
E2=E0+0.059*log[Fe3+]/[Fe2+]
The electrode potential is: E=5E1+E2. Using the given data it is 1.253V.
You say that this is wrong. How should it be calculated then? And how would be the EMF calculated? Just E1-E2=0?
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This is only partial information. Please elaborate - what is the problem that they are trying to solve, what are initial conditions. Without context it is impossible to tell what is going on.
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I wrote it in my first post:
Calculate the electrode potential of a solution at the equivalence point, where Fe2+ was titrated with KMnO4 in acid enviroment, under pH=1.5. E0Fe3+/Fe2+=0.68V, E0MnO4-/Mn2+=1.51V
Before going to the calculation, first, I don't understand the concept, what means the calculated electrode potential if there is no electrode in the solution. There are no anode nor cathode so there is no electromotive force, right?
This is how they solved:
The source I was reading from says:
E1=E0+0.059/5*log[MnO4-][H+]8/[Mn2+]
E2=E0+0.059*log[Fe3+]/[Fe2+]
The electrode potential is: E=5E1+E2. Using the given data it is 1.253V.
You say that this is wrong. How should it be calculated then? And how would be the EMF calculated? Just E1-E2=0?
Forgot to add this:
[MnO4-]/[Mn2+]=1
[Fe3+]/[Fe2+]=1
I don't understand this procedure.
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OK, we drifted a little bit and I forgot what the original problem was.
See if this page helps:
http://www.titrations.info/potentiometric-titration-equivalence-point-calculation
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Why, when calculating the electrode potential I have to multiply by 5 one half-reaction
You have the number 1/5 there to multiply because you have a 5 electron transfer for every Mn7+ ion you have, which is in one of your half reactions. It shouldn't be multiplied by just 5. Look at the oxidation states of Mn you have.
Mn7+ --> Mn2+
Your other half reaction that has iron is just multiplied by 1 since there's only 1 electron being transferred.
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Why, when calculating the electrode potential I have to multiply by 5 one half-reaction
You have the number 1/5 there to multiply because you have a 5 electron transfer for every Mn7+ ion you have, which is in one of your half reactions. It shouldn't be multiplied by just 5. Look at the oxidation states of Mn you have.
Mn7+ --> Mn2+
Your other half reaction that has iron is just multiplied by 1 since there's only 1 electron being transferred.
Shouldn't the second equation be multyplied by 5, so the number of electrons will be equal in both half reactions?
OK, we drifted a little bit and I forgot what the original problem was.
See if this page helps:
http://www.titrations.info/potentiometric-titration-equivalence-point-calculation
Thanks for the link, I will examine it through the day.
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One last question: I read that gasses are also included in Nernst equation, actually their partial pressure is used. In a problem I saw that it is expressed in bars, do I have to do the calculations with bars or Pa, because with Pa I don't get the right result and bar isn't a SI system unit?
EDIT: I think I understood what you wrote CopperSmurf, after the 1/5 is transformed to 1 it is much easier to calculate, so it isn't multyplied by 5 because of the electron change, only because of easier calculations.
And here was the mistake:
The source I was reading from says:
E=5E1+E2.
It is 6E=5E1+E2.