[Beetle]

Can anybody tell me what the relationship is between entropy and enthalpy and the EMF of a galvanic cell

ie if enthalpy and entropy are both +ve what happens to the Eo of the cell?

[Beetle]

I think i might have figured something out ... if enthalpy and entropy are both +ve then the reaction is only spontaneous at high temperatures

so therefore at low temperatures the cell will have a negativ Eo and it will increase as the temperature increases

is that right

sorry i didnt think of that 5 mins ago when i posted

[mike]

E⁰ is a standard potential so must be at standard temperature and pressure.

[mike]

The Nernst equation can be used to find emf of non-standard conditions.

[Borek]

E⁰ is a standard potential so must be at standard temperature and pressure.

More precisely for conditions to be standard activities of all species reacting must be equal to 1.

It means (approximately) that all concentrations must be 1M and all partial pressures (if gases are involved) must be 1 atm.

[mike]

More precisely for conditions to be standard activities of all species reacting must be equal to 1.

It means (approximately) that all concentrations must be 1M and all partial pressures (if gases are involved) must be 1 atm.

Yes, standard temperature, pressure AND 1M concentration (you can't forget about temperature!)

[Borek]

Yes, standard temperature, pressure AND 1M concentration (you can't forget about temperature!)

When I was composing my post I have asked myself - will it be pointed out as a mistake if I will not write about temperature? Oh well, it was already mentioned and I am just adding to precision, don't bother, no problem...

KABOOOM!  

Pressure of 1 atm is not enough - PARTIAL pressures must be 1 atm. It means that in case of the cell where more than one gas reacts total pressure may be higher.

[mike]

Sorry Borek I didn't mean to hurt your feelings

I think between the two of us maybe we came up with the right conditions, and probably confused the original poster (as we didn't really answer the question anyway )

Keep up the good work  

I think i might have figured something out ... if enthalpy and entropy are both +ve then the reaction is only spontaneous at high temperatures

Do you mean for deltaG = deltaH - TdeltaS ?

I think if deltaH and deltaS are both positive then it may also just depend on the magnitude of their value.

[sdekivit]

Sorry Borek I didn't mean to hurt your feelings

I think between the two of us maybe we came up with the right conditions, and probably confused the original poster (as we didn't really answer the question anyway )

Keep up the good work  :)Do you mean for deltaG = deltaH - TdeltaS ?

I think if deltaH and deltaS are both positive then it may also just depend on the magnitude of their value.

the relation between the free Gibbs energy and the emf of a galvanic cell is:

delta G = -nFE

thus the emf of a galvanic cell must be positive, so that thefree Gibbs energy is negative, crucial for a spontaneous reaction.

In accordance when the emf is negative, delta G is negative and delta S is positive:

Delta G = delta H - Tdelta S

[mike]

In accordance when the emf is negative, delta G is negative and delta S is positive:

isn't this the opposite of what you just said?

[mike]

In accordance when the emf is negative, delta G is negative and delta S is positive:

Delta G = delta H - Tdelta S

what do you mean?

[Beetle]

You have all managed to thoroughly confuzzle me now!

I was trying to answer an example question and it says

For a particular rxn in a voltaic cell both deltaHo and deltaSo are positive. Which of the following statements is true?

a) Eo cell will increase with an increase in temperature
b) Eo cell will decrease with an increase in temperature
c) Eo cell will not change when the temperature increases
d) deltaGo > 0 for all temperatures
e) none of the above statements are true

I've ruled out (d) cos deltaG will be -ve at high temperatures.

And then b'se Eo must be at standard temp and pressure then Eo cell will not change when temperature increases.

so it must be (c) i guess.

[sdekivit]

isn't this the opposite of what you just said?

yes sorry, i meant E must be positive.

And then, because delta G is negative delta S >0 and this is in accordance to the second law of thermodynamics, that every process wants to create the highest entropy.