Electrochemical cells: Is gibbs free energy dependent upon concentration?

[Topher925]

Is the gibbs free energy of some reaction in an electrochemical cell dependent upon reactant concentration? The Nernst equation clearly states that open circuit potential is dependent upon concentration, and basic thermodynamics states that open circuit potential is directly proportional to gibbs free energy, so my intuition says yes. However, some sources (not the most credible) I have found say no. My electrochem text doesn't seem to give a definitive answer.

Approaching the question from another angle, I would think that entropy would be higher with a system with a lower concentration of reactants so given a constant temperature the gibbs free energy should be less via \DeltaG = \DeltaH - T\DeltaS.

 

[glassengineer]

Is the gibbs free energy of some reaction in an electrochemical cell dependent upon reactant concentration? The Nernst equation clearly states that open circuit potential is dependent upon concentration, and basic thermodynamics states that open circuit potential is directly proportional to gibbs free energy, so my intuition says yes. However, some sources (not the most credible) I have found say no. My electrochem text doesn't seem to give a definitive answer.

Approaching the question from another angle, I would think that entropy would be higher with a system with a lower concentration of reactants so given a constant temperature the gibbs free energy should be less via \DeltaG = \DeltaH - T\DeltaS.

You need to go back to the fundamentals. Entropy tends to increase unless energy is expended to reduce it within a closed system. Therefore, the more concentrated a solution is (be it solid liquid or gas) the less entropy it has. COncentrating and purifying takes enormous energy (heat, financial, & human). Think of it this way - a single crystal silicon, sapphire or diamond has less entropy than a multicrystalline form of the same material. It takes more effort to process these from impure natural materials states into single crystalline form than to make them into multicrystalline form. As they weather, age, or wear in service their entropy is increasing as the 2nd law of thermo holds true.

Please reorient yourself to see the forest through the trees. Take time to review and reflect on the fundamentals. Ask for help. If you are indeed truly overwhelmed, then perhaps you should orient your studies elsewhere. Only you know the answer to that...

 

[glassengineer]

looking at this another way. If as you mention delta T = 0 then delta H = 0. Therefore, delta G = (-1)* T * delta S, subsequently if concentration increases delta S is negative. Negative times negative is positive therefore delta G is positive! I hope that simplifies the concept and brings clarity for you. :)

 

[Topher925]

Please reorient yourself to see the forest through the trees. Take time to review and reflect on the fundamentals. Ask for help. If you are indeed truly overwhelmed, then perhaps you should orient your studies elsewhere. Only you know the answer to that...

How's the view from up there on your high horse? You've only just reiterated what I already said in my second paragraph.

Therefore, the more concentrated a solution is (be it solid liquid or gas) the less entropy it has.

If as you mention delta T = 0 then delta H = 0.

I didn't mention this.

 

[epenguin]

Is the gibbs free energy of some reaction in an electrochemical cell dependent upon reactant concentration? The Nernst equation clearly states that open circuit potential is dependent upon concentration, and basic thermodynamics states that open circuit potential is directly proportional to gibbs free energy, so my intuition says yes. However, some sources (not the most credible) I have found say no. My electrochem text doesn't seem to give a definitive answer.

Approaching the question from another angle, I would think that entropy would be higher with a system with a lower concentration of reactants so given a constant temperature the gibbs free energy should be less via \DeltaG = \DeltaH - T\DeltaS.

Maybe the 'no' comes from a confusion with 'standard potentials' and 'standard free energy', something like that?

Your intuition is right. E.g. eliminating complicating factors of different materials, imagine a 'concentration cell', so same electrode material both sides, same ions but 2 different concentrations. If the concentrations on both sides are equal there will be no current nor potential, so that answers you question, it must depend on concentration. For an electric current to flow is a way towards equalising the concentrations. So that will happen unless stopped by an opposing potential. Magnitude RT ln (c₁/c₂). (Ignoring junction potentials.) You worry about the sign! (However in any given setup this will be clear, so this is just about conventions.)

Same damned RT ln c that you get in every damned chemical thermodynamic thing.

I would not completely agree that anyone whose grasp of thermodynamics is shaky, unconfident, unstable should get out of the field because that would be a lot of people, also in any other field at all nearby they will still need the thermodynamics.

 

[Topher925]

Maybe the 'no' comes from a confusion...

THANK YOU! I knew that I wasn't crazy, even though some fellow colleges and some people of "academic authority" said otherwise. I really needed a sanity check.

I thought it would be a well understood concept given the popular equation RTln(K) = \DeltaG, but when everyone tells you "no" you start to question your intuition. That darn equation shows up in all of my electrochem, combustion, and stat thermo books.