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### Topic: Redox potentials  (Read 9106 times)

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#### Borek ##### Re: Redox potentials
« Reply #15 on: July 02, 2012, 06:22:19 AM »
Why is the potential of the MnO4-/Mn2+ half-reaction multylpied by 5

It is not, you are misunderstanding something.
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#### Shadow

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• Mole Snacks: +5/-11 ##### Re: Redox potentials
« Reply #16 on: July 02, 2012, 07:16:07 AM »
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?

#### Borek ##### Re: Redox potentials
« Reply #17 on: July 02, 2012, 07:45:41 AM »
For permanganate potential the only place 5 would be present is the number of electrons:

$$E = E_0 + \frac {RT} {5F} \ln \frac{[\textrm{MnO}_4^-][\textrm{H}^+]^8}{[\textrm{Mn}^{2+}]}$$

But number of electrons is always in the same place, no matter of what redox system you are dealing with.
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#### Shadow

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• Mole Snacks: +5/-11 ##### Re: Redox potentials
« Reply #18 on: July 02, 2012, 08:12:34 AM »
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?

#### Borek ##### Re: Redox potentials
« Reply #19 on: July 02, 2012, 08:24:11 AM »
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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#### Shadow

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• Mole Snacks: +5/-11 ##### Re: Redox potentials
« Reply #20 on: July 02, 2012, 08:54:37 AM »
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?
[MnO4-]/[Mn2+]=1
[Fe3+]/[Fe2+]=1
I don't understand this procedure.

#### Borek ##### Re: Redox potentials
« Reply #21 on: July 02, 2012, 10:45:24 AM »
OK, we drifted a little bit and I forgot what the original problem was.

http://www.titrations.info/potentiometric-titration-equivalence-point-calculation
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#### CopperSmurf

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• Gender:  ##### Re: Redox potentials
« Reply #22 on: July 02, 2012, 09:46:28 PM »
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.

#### Shadow

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• Mole Snacks: +5/-11 ##### Re: Redox potentials
« Reply #23 on: July 03, 2012, 05:10:53 AM »
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.

http://www.titrations.info/potentiometric-titration-equivalence-point-calculation
Thanks for the link, I will examine it through the day.

#### Shadow

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• Mole Snacks: +5/-11 ##### Re: Redox potentials
« Reply #24 on: July 03, 2012, 09:42:48 AM »
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.