[stewartgilligan]

Hi everyone...

I have a question regarding a potentiometric lab I have done.  I have been trying to figure out how to calculate the formal potential of each half reaction.  The lab was a titration of Fe(II) with Ce(IV), where a 25 mL aliquot of an unknown iron solution was titrated with standard cesium.  I understand that the nernst equation should be used here, and that the measured potential (from the potentiometer) is put into the equation, however, what about the concentrations?  I do not understand what the concentrations needed for completion of the equation are.  As well, 10 mL of mercury(II) chloride was added to the iron just before titrating with the cesium, what role does that play?  I have been at this for quite a bit, and it has grown incredibly frustrating..

Any help is greatly appreciated,

thanks

[Borek]

Cesium or cerium?

You don't need to know concentrations. Ratio of concentrations is what counts here - and this can be easily calculated. From the titration curve you should know the end-point. At 50% titration concentrations of both Fe²⁺ and Fe³⁺ are identical.

HgCl₂ is used to remove excess SnCl₂ which is in turn used to reduce Fe³⁺ to Fe²⁺. If you have not added SnCl₂ solution by yourself, my bet is that the Fe solution you were given already contained some tin (II) chloride to prevent iron oxidation to Fe³⁺ with atmospheric oxygen.

[stewartgilligan]

Hi again,

thanks for the reply.  I understand that the concentrations are identical, however, wouldn't this make the formal potential equal to the measured potential as the log of 1 is zero?  (From the nernst Equation)

E = E? - 0.0592/1 * log [Fe2+]/[Fe3+]. Where E? is the formal potential

[Borek]

wouldn't this make the formal potential equal to the measured potential as the log of 1 is zero?

You have just determined formal potential of Fe²⁺/Fe³⁺ reaction in your solution.

[stewartgilligan]

Hi,

thank you very much.  So if the iron concentrations are equal at 1/2 the titration point, then the concentrations of Ce3+ and Ce4+ would be equal at 2x the titration point?  Does that mean that the average of these two values would give the equivalence point potential?

[Borek]

So if the iron concentrations are equal at 1/2 the titration point, then the concentrations of Ce3+ and Ce4+ would be equal at 2x the titration point?

Sounds logical

Does that mean that the average of these two values would give the equivalence point potential?

Not always. In general it will be weighted average (weighted by numbers of electrons from both half-reactions). In this particular case both reactions use 1 electron, so normal average will be OK.

[stewartgilligan]

Hello,

thanks very much for all of your assistance.  I feel as if I have a handle on the concepts of this lab....

[CGriffins]

Hi,

I do not know if anyone will check an older thread, however, can someone please explain why the equivalence potential is equal to the average of the two formal potentials?  I am trying to understand that one, but it does not seem to be working

Thanks

[Borek]

I do not know if anyone will check an older thread, however, can someone please explain why the equivalence potential is equal to the average of the two formal potentials?  I am trying to understand that one, but it does not seem to be working

Try to derive it by yourself. You have two halfreactions and two Nernst equations. Both potentials must be identical. It is endpoint so you have added equivalent amount of titrant to titrated substance (don't forget to multiply them by numbers of electrons!). This gives you two additional equations for concentrations of Red and Ox forms of both titrant and titrated substance. You have a set of equations - solve it.

[sistermonica]

How would I calculate 10% titrated with the same setup?

[Borek]

How would I calculate 10% titrated with the same setup?

Assume there is only one redox system in the solution (titrated one) and calculate concentrations of red and ox forms from stoichiometry (assuming it reacted 100%). Then use Nernst equation.

[sistermonica]

so I would use .68-.0592log((100-10)/10)?

[sistermonica]

I'm actually pretty sure that what I put isn't right. I'm not sure where the 10% comes in the stochiometry. I have 3.921g of Fe(NH4)2(SO4)2 6H20 in 25 mL and 0.1 M Ce(NH4)2(NO3)6. After I find the molarity of the Fe solution, what would I do?

[lemonoman]

Assume there is only one redox system in the solution (titrated one) and calculate concentrations of red and ox forms from stoichiometry (assuming it reacted 100%). Then use Nernst equation.

This means that you have an initial concentration of Fe, [Fe²⁺]₀, but what you have at the 10% titration point is a concentration of [Fe²⁺] = (1 - 0.1)[Fe²⁺]₀ = 0.9[Fe²⁺]₀.  Hope that helps.

And by the way, I notice you revived an old, inactive thread.  That means you probably searched the chemicalforums database for similar questions...and not enough people do that.  Three cheers for you!   

[sistermonica]

Thank you so much! I started out with that yesterday but other people in my lab said otherwise.

I always try to research as much as I can so people don't have to keep repeating themselves 

Thank you again!