[maakii]

for an electrode with

H₂ (1atm) -> 2H⁺ (0.1M) + 2e-

and another with

H₂ (1atm) -> 2H⁺ (1M) + 2e-

is the overall reaction 2H⁺ (1M) -> 2H⁺ (0.1M), or can it be simplified into H⁺ (1M) -> H⁺ (0.1M) safely?

thanks lots!

[DevaDevil]

ok first of all as I read this you have an electrochemical cell with on one side a (platinum?) electrode with hydrogen (1 atm) in equilibrium with protons (0.1 M); and on the other side a (platinum?) electrode with hydrogen (1 atm) in equilibrium with protons (1M).

So: H₂/Pt |H⁺ (0.1M) || H⁺ (1M)| H₂/Pt
Am I correct in this assumption?


If yes, what will NOT happen is a reversal of the molarities
Yes, the 1M solution will receive electrons from the oxidation of hydrogen on the other side and the concentration of protons in solution will decrease; but on the other side hydrogen will be turned into protons, increasing the concentration of H⁺ there.
Since the cell will only work as long as there is a activity gradient (i.e. the concentration of protons and/or partial pressures of the gas on one side is bigger than on the other side), this cell will operate until the concentrations of H⁺ and pressure of hydrogen are equal on both sides of the cell.

in this case it would be 0.55 M H⁺ (Equal volumes on both sides assumed). Every proton you remove on the initial 1M side you create on the initial 0.1M side.
At about that concentration the reaction will have reached equilibrium and not continue until you decrease the concentration of protons on either side.

If the volumes differ significantly, the final concentration will be different. But fact is the final concentration will be equal on both sides, and greater than 0.1M (only at infinite volume on the 0.1M side the final concentration will be 0.1 M)

The final "reaction" is basically nothing but a concentration difference. You wouldn't write 2H⁺ --> 2H⁺ , but more something like [H⁺]_initial = 1M; [H⁺]_end = 0.XX, depending on whatever the final concentration will be. The same for the other side of the cell. Overall no real reaction takes place, both cancel each other out.

[maakii]

Hmm i understand what you are trying to say, that the reaction does not necessarily have to proceed to the extent of both reversing their concentrations, but i have checked many websites, and they all use this notation, so i think that H⁺ (1M) -> H⁺ (0.1M) is just a way of expressing the concentration gradient..

i was trying to find out if 2H⁺ (1M) -->2 H⁺ (0.1M) was correct, or should i use H⁺ (1M) -> H⁺ (0.1M) if i was plugging these values into the nerst equation, because if i were to use the 1st one, i would have to square the 0.1M and 1M, leading to a different answer as the 2nd one, where i wouldn't have to square the concentrations..

sorry for the misunderstanding!

[DevaDevil]

Nernst equation: E = E^o - RT/n F * ln (a_red/a_ox)

Since the halfcell equation here is: H₂ (g) --> 2H⁺ + 2e^-, the nernst eq. for the halfcell becomes for the (0.1M) side:

E = 0 (definition of hydrogen potential) - RT/2 F * ln (a_H2(g)/[H⁺]²)

Hence you use the squared value of the proton (or hydronium) concentration, but also n = 2

[maakii]

Oh i see..then you can find take the difference between both sides to get the cell potential..

thanks for the explaination!