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Topic: Acids in redox  (Read 3244 times)

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Offline Big-Daddy

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Acids in redox
« on: September 03, 2013, 07:55:07 PM »
Solutions of two weak polyprotic acids , HxA and HyB, are each placed in contact with standard hydrogen electrode at 298 K. When a cell is constructed by interconnecting them through a salt bridge find the emf of the cell.

Since E° will be 0 as both electrodes establish the SHE equilibrium, can we go on to say that

[tex]E = \frac{RT}{2F} \cdot ln(\frac{c_{H^+,A}^2}{c_{H^+,B}^2})[/tex]

Where I have set A as the right-hand electrode and B as the left hand electrode, and now we just need to work out the concentrations of H+A and H+B, i.e. the proton concentrations in each compartment, to find the value of the emf E?

Offline Borek

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Re: Acids in redox
« Reply #1 on: September 04, 2013, 02:51:22 AM »
E° doesn't mater, it always cancels out for concentration electrodes.
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Offline Big-Daddy

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Re: Acids in redox
« Reply #2 on: September 04, 2013, 07:37:59 AM »
Ok, and then all I need to do is calculate the concentrations of H+ in each compartment? Using the standard equilibrium calculations, if the system is known to be at equilibrium, and then I find the EMF?

Offline Big-Daddy

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Re: Acids in redox
« Reply #3 on: September 13, 2013, 03:35:08 PM »
Hi, I just had a thought - surely at equilibrium the EMF will be 0, so there actually is no way of estimating what the H+ concentrations will be (except for at equilibrium when they will have to end up summing to EMF = 0 anyway), so what does this question really mean?

Originally I had paraphrased it - it's actually "two weak acids HA and HB (initial concentrations and Ka values given) are each placed in contact with standard hydrogen electrode at 298 K. When a cell is constructed by interconnecting them through a salt bridge find the emf of the cell." I managed to attain a potential solution by using equilibrium calculations to find [H+]_A and [H+]_B, but neither does this work out to EMF = 0, as it should at equilibrium, nor does it apply if EMF does not = 0. So what could I do? This is a textbook problem, must be some way of solving it?

Offline Borek

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Re: Acids in redox
« Reply #4 on: September 13, 2013, 03:42:48 PM »
You have separate acid dissociation equilibria in each half cell, but total system is not at equilibrium.
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Offline Big-Daddy

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Re: Acids in redox
« Reply #5 on: September 13, 2013, 05:07:14 PM »
So the HA -> H+ + A- and HB -> H+ + B- reversible reactions are at equilibrium, hence my equilibrium calculations are satisfactory to find the concentrations of H+ required for the original equation in the first post - but the SHE reversible reaction (1/2 H+ + e- -> H2) established in each solution has not reached equilibrium? But how can this be, as this means [H+] must still be changing for the SHE reaction to be nearing equilibrium still, whereas according to the acid dissociation equilibria, the equilibrium point has already been reached so [H+] should now be constant?

Offline Borek

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Re: Acids in redox
« Reply #6 on: September 13, 2013, 06:07:59 PM »
[H+] would be changing only if you would allow current to flow, no current, no reaction, system doesn't go to equilibrium.
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Offline Big-Daddy

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Re: Acids in redox
« Reply #7 on: September 13, 2013, 06:46:52 PM »
Ok, got it, thanks.

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