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Topic: Equilibrium Potential of Half Cell  (Read 4544 times)

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Offline DUDE778

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Equilibrium Potential of Half Cell
« on: March 23, 2008, 09:53:48 PM »
I cant seem to get this and i dont know why :'(
Ce4+ + e− > Ce3+    E° = 1.72 V   
Fe3+ + e− > Fe2+    E° = 0.771 V

If a solution is prepared by mixing 5.0 mL of 0.30 M Fe2+ with 3.0 mL of 0.12 M Ce4+.
What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)?

I have three chances to do this and i put..
0.949 and
2.491
I AM LOST  ???

Offline Borek

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Re: Equilibrium Potential of Half Cell
« Reply #1 on: March 24, 2008, 05:02:04 AM »
Show how you get these values. Or at least one of them.
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Offline DUDE778

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Re: Equilibrium Potential of Half Cell
« Reply #2 on: March 24, 2008, 07:12:31 PM »
i added the two potentials together the first time, then for my second try I subtracted the two potentials...I don't know what's wrong..

Offline Borek

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Re: Equilibrium Potential of Half Cell
« Reply #3 on: March 24, 2008, 07:26:53 PM »
You need to account for the concentrations, that's what the Nernst equation is for. You have to combine stoichiometry of the reaction with Nernst equations (for each half-cell).
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