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Topic: how do i solve pH given an electrode potential?  (Read 10976 times)

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Offline polly

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how do i solve pH given an electrode potential?
« on: April 13, 2007, 08:22:16 AM »
hi all,

I just need to know where to start if I had to calculate the pH of a solution given a potential of the solution relative to a standard hydrogen electrode?

any starting point would be good
Cheers

Offline Borek

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Re: how do i solve pH given an electrode potential?
« Reply #1 on: April 13, 2007, 09:53:14 AM »
First of all - what solution, what electrode, what redox system?

If pH electrode - http://www.ph-meter.info/pH-electrode-potential

But if that's a platinum electrode and MnO4-/Mn2+ system situation is completely different.
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Offline polly

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Re: how do i solve pH given an electrode potential?
« Reply #2 on: April 13, 2007, 04:55:01 PM »
the exact question is as follows

Calculate the pH of a solution whose potential at 298K, measured with a standard hydrogen electrode at atmospheric pressure of 1.012 atm.(corrected for the vapour pressure of water at 298K) is -0.324V (vs. NHE)

I have only the Nernst equation as notes but nothing on how to work out the pH as asked in this question. No idea what is meant by NHE either.

No idea where to begin.

Thanks
« Last Edit: April 14, 2007, 07:41:14 AM by polly »

Offline Borek

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Re: how do i solve pH given an electrode potential?
« Reply #3 on: April 13, 2007, 06:05:30 PM »
NHE = Normal Hydrogen Electrode.

Nernst equation can be easily modified to contain pH (hint: think about the pH definition).
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Offline polly

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Re: how do i solve pH given an electrode potential?
« Reply #4 on: April 14, 2007, 08:35:26 AM »
Hi Borek, ok I did what you asked..., can you please tell me if the following answer is correct?

the cell potential of the solution is measured in reference to the SHE, which has a value of 0 by definition.

using                                          E(cell) = -0.0592 log[H+]

and thus,                                    pH =E(cell) / -0.0592 = -0.324 / -0.0592 = 5.47

Is this correct?

Thanks a lot.





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