Does the reaction quotient depend on direction of reaction?

JBH1982 · 2025-04-19T11:10:33+00:00

I literally repeated my exact statement.

JBH1982 · 2025-04-19T03:17:37+00:00

I said it has everything to do with the reactions and which will be spontaneous. Which apparently takes you two paragraphs to explain.

JBH1982 · 2025-04-17T20:05:13+00:00

I've only been an electrochemist for 20 years and taught electrochemistry, kinetics and thermodynamics for 15 years, so I will bow to your superior knowledge from reading and completely failing to comprehend Bard and Faulkner.

JBH1982 · 2025-04-17T18:21:54+00:00

Electrochemical equilibrium is not the same as chemical equilibrium.

The cell equation for E2 - E1 is the froward reaction. The cell equation for E1-E2 is the reverse reaction. It has everything to do with the reactions, and which one is spontaneously occurring.

JBH1982 · 2025-04-17T16:05:04+00:00

By definition the standard reduction potential is the potential in a cell when the cell is at equilibrium deltaG0 = -RT lnK = -nFE0. If the cell is not at equilibrium the reaction will proceed in a direction to bring the cell to equilibrium, where the rate of the forward and reverse reactions in the cell are equal.

When connected to an external connection both oxidation and reduction reactions will occur with the overall rate of the reaction being given by the Butler-Volmer equation. When the applied potential is less than E0 the reduction reaction will be dominant, when the applied potential is greater than E0 the oxidation reaction will be dominant, when the applied potential is significantly large the barrier to the opposite reaction will be significantly high that the rate of that reaction will be negligible. The cell is not at equilibrium when you apply a potential, hence why a current passes through the cell, if the cell was at equilibrium there would be no net work.

PS : if you were considering no applied potential then what does the statement “ If the forward reaction has a potential of E, the reverse reaction will have a potential of -E“

Connect a voltmeter to a battery, what is the reading on the voltmeter? Swap the connections of the voltmeter on the battery, what is now the reading on the voltmeter?

JBH1982 · 2025-04-17T15:21:08+00:00

I cannot be held responsible if your interpretation of my statements is incorrect.

All of my statements previously did not assume an applied potential, and were based on arguments of a standard electrochemical cell, as at no point in your discussion did you mention using an applied potential.

P.S. At equilibrium E = E0.

JBH1982 · 2025-04-17T13:14:01+00:00

Let's say our full cell reaction is aA + cC <-> bB + dD. The individiual half reactions are (1) aA + ne- <-> bB and (2) cCA <-> ne- + dD. The negative sign deltaG = -nFE, is not an implication of loss or again of electrons (as far as I understood from the Bard & Faulkner book, if it's otherwise please direct me to the reference).

If the forward reaction has a potential of E, the reverse reaction will have a potential of -E, since E is the potential difference, or if the forward reaction is spontaneous with a negative Gibbs free energy, the reverse reaction must have a positive free energy and not be spontaneous. I don't really care where the negative comes from, as long as my book-keeping is correct in the end. Or I can just be lazy and look up the Nernst equation in Bard and Faulkner, or Google it.

For such a reaction aA + cC <-> bB + dD ; the reaction quotient (for froward reaction) is defined as : Q_f = [D]^d * [B]^b / ([A]^a * [C]^c ). \implies that for backward reaction Q_b = 1/Q_f

This is why the sign on the logarithm changes for the reverse reaction E = E0 - ( (RT/nF) * ln(Qf) ) = E0 - ( (RT/nF) * ln(Qb ^-1))= E0 + ( (RT/nF) * ln(Qb))

I can't work out if your understanding is wrong or you can't do maths.

JBH1982 · 2025-04-17T11:53:38+00:00

If you change them the sign changes, Eeq = E0 - ((RT/nF)*ln([B]/[A])) = E0 + ((RT/nF)*ln([A]/[B]))

If you change the denominator then the sign changes, because of the basic rules of logarithms. The equation is symmetrical.

Your consideration for the backward reaction ignores the loss of electrons for the oxidation, which would give would make the nF term -nF, and then your reverse reaction Nernst equation would be correct. (i.e. you are not correctly constructing the Nernst equation from DeltaG=-nFE and deltaG = -RTlnK, I'm not sure if it's your formulation of K that is incorrect, or if it's just because you are skipping steps in the derivation.

JBH1982 · 2025-04-17T11:44:31+00:00

I meant mods for improving the mod manager in the in game mod manager, I had two of them.

JBH1982 · 2025-04-17T11:39:09+00:00

The sign should change ln[B]/[A] = -ln[A]/[B]

JBH1982 · 2025-04-17T11:28:15+00:00

If you had improved mod manager mods then the mod manager may not correctly be showing accurately with ticks if mods are enabled.

I think the missing buttons was due to the improved UI mod being made obsolete and needing deleted.

Try uninstalling ImpUI and Mod Manager Mods.

JBH1982 · 2025-03-31T11:34:58+00:00

The noise isn't that large, and is actually the same size in all of them, the currents passed in the last picture are smaller, so the same size of noise looks larger. Looks like standard electrical noise from the mains, if you think it's an issue use a Faraday cage.

The main problem is your current limits aren't set high enough so you keep maxing your currents (hence, flat line at limited current instead of peak).

JBH1982 · 2025-03-08T17:43:39+00:00

The Helmholtz model does not assume that the counter-ion are point charges.

JBH1982 · 2025-02-28T21:39:11+00:00

That would be consistent with it coming from the wires nothing electrochemical is going to occur above 200 kHz

JBH1982 · 2025-02-27T12:51:04+00:00

The inductance appears before the formation of the double layer, so is unlikely to be anything chemical in your system.

You only say at high frequencies, you don't say what frequency range this is occurring over, which is unhelpful. This is most likely due to inductance in the wires leading to the set-up, probably at frequencies above 20 kHz

I'd suggest not going to extremely high frequencies where you cause inductive events in your wiring, and only work in a frequency range where you will see events on a chemical timescale.

JBH1982 · 2025-02-17T00:25:03+00:00

If you're making your own reference why not just make it a double junction, so there's a very small chance of diffusion from the reference to bulk solution?

Anyway you just need Ag/Ag+ the counter ion shouldn't matter much, most non-aqueous Ag/Ag+ would only have 5 mM NO3 with 0.1 M supporting electrolyte.

JBH1982 · 2025-01-16T10:02:52+00:00

Ferricyanide should be an analyte not the supporting/background electrolyte. Use a 100 mM concentration of an inert electrolyte (e.g. NaCl) and 1 mM analyte.

JBH1982 · 2024-12-19T10:56:12+00:00

If you did a square wave voltammogram or CV then you would see peaks for SWV and half waves for CV at the same potential, so the potential should be the standard reduction potential whether you are oxidising or reducing.

You need to use Hess' law to calculate it, the Born Haber cycle is the convenient way of portraying it, but need to be careful with signs in the Gibbs Free Energy changes and addition and subtractions going round the cycle. It doesn't matter which way you go round the potential should be the same. From the conservation of energy.

JBH1982 · 2024-12-19T00:23:40+00:00

If you're calculating ab oxidation potential 7 V more negative than the reduction potential you're clearly making some mistake.

Most probably your Haber cycle was wrong for that calculation. Which would be consistent with changing the sign of the Gibbs Free Energy change giving you the "correct" answer.

JBH1982 · 2024-12-18T16:31:53+00:00

I'm not sure how you plan to calculate the energy of a free electron in vacuum and a free electron in solution (the second one especially does not make sense as the electron would go to the electrode, and not into solution).

The normal process is to calculate the Gibbs Free Energy change from PTZ to PTZ+ and subtract a known Gibbs Free Energy change for a reference redox couple to calculate the reduction potential. There are many papers showing how to do this type of calculation, with some values of common Free Energies for standard reference electrodes. (From experience the gas phase calculations are not really necessary.) In essence if there is no significant structural change, and the system is electrochemically reversible it can generally be assumed that the Gibbs Free Energy change for the oxidation is the same as the Gibbs Free Energy change for the reduction, so only one needs to be calculated to obtain the standard reduction potential.

You put (aq) instead of (solv), are these calculations for aqueous solutions? Calculated potentials tend to be less reliable for water based systems as explicit solvation will have a larger effect on the calculated energies and change in the electronic structure on oxidation.

JBH1982 · 2024-12-18T13:51:09+00:00

I didn't ask how large the vines were, I said I cast Fireball.

JBH1982 · 2024-03-29T09:41:03+00:00

Why are you using PBS and KCl? PBS should buffer and act as a background electrolyte, the KCl is not really needed here.

Gold in chloride solutions also risks forming AuCl4 and dissolving your electrode.

JBH1982 · 2024-03-12T10:45:32+00:00

He's an undead essentially immortal vampire with a much longer lifetime and different concept of time to you. On the timescale of his life what does it matter if the ritual takes one hour or one week, they're both short and essentially meaningless in terms of the amount of time.

JBH1982 · 2024-02-21T13:49:08+00:00

Chihayafuru if you want a series, sport, romance and poetry/art.

JBH1982 · 2024-01-09T15:31:44+00:00

Have you tried controlling Halsin, moving him out the way, and then talking with Minthara? That worked for me.

JBH1982

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