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Topic: Redox potentials  (Read 9107 times)

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Offline Shadow

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Redox potentials
« on: June 26, 2012, 04:48:59 AM »
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?

Offline UG

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Re: Redox potentials
« Reply #1 on: June 26, 2012, 05:16:26 AM »
I do believe this is a potentiometric titration so an indicator electrode and a reference electrode would have been used.

Offline Borek

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Re: Redox potentials
« Reply #2 on: June 26, 2012, 05:26:15 AM »
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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Offline Shadow

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Re: Redox potentials
« Reply #3 on: June 26, 2012, 06:30:22 AM »
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?

Offline Shadow

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Re: Redox potentials
« Reply #4 on: July 01, 2012, 11:12:05 AM »
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?

Offline Borek

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Re: Redox potentials
« Reply #5 on: July 01, 2012, 12:51:34 PM »
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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Offline Shadow

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Re: Redox potentials
« Reply #6 on: July 01, 2012, 01:05:55 PM »
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?

Offline Borek

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Re: Redox potentials
« Reply #7 on: July 01, 2012, 01:49:41 PM »
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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Offline Shadow

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Re: Redox potentials
« Reply #8 on: July 01, 2012, 02:21:14 PM »
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?

Offline Borek

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Re: Redox potentials
« Reply #9 on: July 01, 2012, 04:46:26 PM »
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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Offline Shadow

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Re: Redox potentials
« Reply #10 on: July 02, 2012, 04:21:25 AM »
Why, when calculating the electrode potential I have to multiply by 5 one half-reaction, and when calculating el.-mot. force not?

Offline Borek

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Re: Redox potentials
« Reply #11 on: July 02, 2012, 05:02:25 AM »
I don't see it being true.
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Offline Shadow

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Re: Redox potentials
« Reply #12 on: July 02, 2012, 05:21:26 AM »
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?

Offline Borek

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Re: Redox potentials
« Reply #13 on: July 02, 2012, 05:44:37 AM »
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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Offline Shadow

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Re: Redox potentials
« Reply #14 on: July 02, 2012, 06:06:44 AM »
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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