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Topic: pe vs. pH  (Read 1791 times)

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Offline Shirro

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pe vs. pH
« on: May 16, 2012, 06:47:41 AM »
I have a following problem:

In unaerobic enviroment, there is Manganese dioxide and hydrogen sulfide.
[Mn2+]=10-3 M
[SO42-]=10-5 M
[H2S]=102-
SO42- + 10 H+ + 2 e-  <-> H2S + 4 H2O     EH0 = + 0,31 V
MnO2(s) + 4 H+ + 2 e- <-> Mn2+ + 2 H2O         EH0= + 1,29 V

I'm asked to do pe vs. pH diagram. Can anyone tell me  how to do this?

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