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Topic: Pourbaix diagram's line - how to decide the sides?  (Read 2734 times)

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Offline Silppuri

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Pourbaix diagram's line - how to decide the sides?
« on: February 23, 2018, 01:39:05 AM »
I'm wrestling with the corrosion of metals, more precisely the Pourbaix diagram. For those who understand it, I'd like to ask for some help. I haven't found the reason why the other species (e.g. Fe) goes always on the other side of -0,617 V and not the other, see the attached image.
If you add 1,00 litres of ethanol to 1,00 litres of water, you get 1,92 litres of solution.

Offline Borek

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Re: Pourbaix diagram's line - how to decide the sides?
« Reply #1 on: February 23, 2018, 03:07:08 AM »
First of all - it doesn't, not at -0.617.
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Offline Silppuri

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Re: Pourbaix diagram's line - how to decide the sides?
« Reply #2 on: February 23, 2018, 04:56:44 AM »
Is that calculated in a way that the concentration of Fe2+ is 10-6, as in my example?
If you add 1,00 litres of ethanol to 1,00 litres of water, you get 1,92 litres of solution.

Offline Borek

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Re: Pourbaix diagram's line - how to decide the sides?
« Reply #3 on: February 23, 2018, 06:23:58 AM »
Not sure what you are asking. Diagram shows what dominates the solution - it doesn't mean the other ions don't exist at a given pH/E point.
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Offline Silppuri

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Re: Pourbaix diagram's line - how to decide the sides?
« Reply #4 on: February 23, 2018, 07:04:21 AM »
Constructing a Pourbaix diagram involves creating imaginary lines where the other form is more dominant than the other on the other side, and these lines are created (in simplest forms) by using the Nernst equation, i.e. calculating. In my example we have an active species Fe2+ on the other side and the immune species Fe below and my question is, how do we know which species to put on which side?

On a side note: Your claim was that the line is not at -0,617 V. You provided a Pourbaix diagram which doesn't provide information about the concentration, so I wanted to know that do you have a proof for the claim, i.e. give the concentration and a calculatory example. Because in my example I think it really is located at -0,617 V.
If you add 1,00 litres of ethanol to 1,00 litres of water, you get 1,92 litres of solution.

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