Calculate the standard free-energy change for the reaction at 25 ∘C. Refer to the list of standard reduction potentials.
2Au3+(aq)+3Mg(s)↽−−⇀2Au(s)+3Mg2+(aq).
Why is the E°cell=E°cathode - E°anode (1.498)-(-2.38)?
Why do we not multiply each value by the coefficient of each reaction like (2*1.498)-(3*-2.38)?
free energy change
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 38
- Joined: Mon Jan 09, 2023 2:23 am
Re: free energy change
Hi! Because cell potential is an intensive property, its value does not depend on the physical quantity or amount of matter present, so we would never multiply by a factor of 2 since it remains the same no matter the coefficients.
Hope this helps!
Hope this helps!
Re: free energy change
E isn't a factor that depends on the mass/size of its reactants/factors. Therefore, no matter if you multiply a reaction by a certain factor to get it to cancel out - you do not multiply the E naught value. You would, however, change the sign if you flip the equation around.
-
- Posts: 35
- Joined: Mon Jan 09, 2023 8:58 am
Re: free energy change
E/Eaught is an intensive property so we don't ever manipulate its value unless we are flipping it (and thus making it negative) for an oxidized reaction :)
Re: free energy change
E naught is intensive, it is a set of values already determined under standard conditions therefore the coefficients have no effect on its value
-
- Posts: 35
- Joined: Mon Jan 09, 2023 9:05 am
Re: free energy change
Like everyone else is saying, we learned that E naught is an intensive property which means that the mass of the molecules does not come to play at all in this question and the values you are given are not to be multiplied since these values are not affected by any external measurements.
-
- Posts: 34
- Joined: Mon Jan 09, 2023 9:38 am
Re: free energy change
Hello, remember that E knot is an intensive property. This means that you do not multiply the value of E knot. The only time we manipulate it is when we are flipping it to make it negative and we would do this for an oxidized reaction. Hope this helps:)
Return to “Work, Gibbs Free Energy, Cell (Redox) Potentials”
Who is online
Users browsing this forum: No registered users and 5 guests