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Topic: nernst equation Q  (Read 5524 times)

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bowersst

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nernst equation Q
« on: November 30, 2004, 07:21:29 PM »
can someone show me how to use the nerst equation? I have about 30 of these questions, but I don't feel like posting all of them here. :P I will just post this one, and hopefully I can figure out the rest from it.

 What is the pE value in a solution in equilibrium with air (21% O2 by volume) at pH 6.00?

^The main part of my confusion involves the fact that there are different versions of the nernst equation, and I also don't know which parts are supposed to come from a book, and other need calculating.

Offline Donaldson Tan

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Re:nernst equation Q
« Reply #1 on: November 30, 2004, 07:39:21 PM »
use the nerst equation for electrochemistry. it relates electrode potential to concentration of reduced and oxidised species.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Demotivator

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Re:nernst equation Q
« Reply #2 on: December 01, 2004, 10:42:03 AM »
use the nerst equation for electrochemistry. it relates electrode potential to concentration of reduced and oxidised species.
The nernst equation can be  used for any redox process to calculate potential for reaction.

E = E(0) + (RT/nF)ln([oxidized]/[reduced])
For standard temp, RT/F is a constant, and converting ln to log yields:
E = E(0) + (.0591/n)log([oxidized]/[reduced])
The standard reduction half rxn (from tables) is:
4H+ + O2(g)  + 4e -> 2H2O  E(0) = 1.23 V
The oxidized part [oxidized] of the equation is:
([H+]^4 [pO2]
The [reduced] part is [H2O]^2 but it is not included because it is the solvent. n=4 (# of e)

E = 1.23 + (.0591/4)log([H+]^4 [pO2])
where [H+] = 10^-6 because ph=6
and pO2(g) = 0.21 atm  because that's the partial pressure of oxygen since it's 21% by volume.

then
E = 1.23 + (.0591/4)log([10^-6]^4 [0.21])
E= 1.23 - .36 = .87 V
PE = -log(E) = -log(.87) = .06

« Last Edit: December 01, 2004, 01:09:16 PM by Demotivator »

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