n when calculating Gibbs Free Energy

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martha-1I
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Joined: Fri Sep 26, 2014 2:02 pm

n when calculating Gibbs Free Energy

Postby martha-1I » Tue Feb 03, 2015 12:23 pm

I understand that the n in G =-nFE is the moles of the reaction, but I don't understand how it is distinguished. For example in 13.9, we are asked to calculate the Gibbs Free Energy for 2Ce4+(aq)+3I-(aq) ----->2Ce3+(aq) +I3-(aq) with an Ecell of +1.08v. I don't see how the moles of this reaction is 2 when there are 3 moles of I-.

I remember from lecture that it was said that the moles of the oxidation reaction and reduction reaction should be the same, but I'm having a hard time making sense of this. Any clarification would be appreciated.

Regina Chi 2K
Posts: 51
Joined: Fri Sep 26, 2014 2:02 pm

Re: n when calculating Gibbs Free Energy

Postby Regina Chi 2K » Tue Feb 03, 2015 1:06 pm

If I remember correctly, we are not using the mols of an atom but only the stoichiometric coefficient that accompanies the electrons that get cancelled out in the balanced equation. Hope this helps.

Kayla Denton 1A
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Re: n when calculating Gibbs Free Energy

Postby Kayla Denton 1A » Tue Feb 03, 2015 1:07 pm

The key is to not just look at a given reactant or product, but the DIFFERENCE in charge between reactants and products.

For iodine, on the reactants side there are 3 moles of I- so a charge of -3. On the products side there is 1 mole of I3- so a charge of -1. The difference in charge is -2, so we need to add two moles of electrons to the products side in order for the charges to cancel out for this half-reaction.

Edit: there's actually more to it than this. We have to check the other half-reaction, 2Ce4+ --> 2Ce3+.
On the reactants side there is a charge of 2 moles x 4+ = +8. On the products side there is a charge of 2 moles x 3+ = +6. Once again, a difference of two moles of electrons, so we can definitely use n = 2 moles.

Chem_Mod
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Re: n when calculating Gibbs Free Energy

Postby Chem_Mod » Tue Feb 03, 2015 4:50 pm

The easiest way to find out n is to look how many electrons are transferred in your balanced HALF-reactions. For example, in the Copper/Zinc cell, n=2
In the Silver/Zinc cell, silver transfers 1 electron and zinc transfers 2. Since these don't match, the silver reaction must be multiplied by 2 before we could add up the half-reactions. In this case, n=2.
In an Aluminum/Zinc cell, aluminum transfers 3 electrons and zinc transfers 2. To make these match, aluminum reaction is multiplied by 2 and zinc by 3. Then, n=6.

General rule: Find the number of electrons in each balanced HALF-reaction. If they match, that is n (First example). If they don't match, take the lowest common multiple, and that is n (Second/third examples).

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: n when calculating Gibbs Free Energy

Postby AKatukota » Tue Feb 25, 2020 8:26 pm

Ohhh so in order to get n, you have to balance the half reactions and then subtract the electrons?

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: n when calculating Gibbs Free Energy

Postby AKatukota » Tue Feb 25, 2020 8:35 pm

Also, for part b of this question, how would you right out the reduction half reaction? I am confused on how to figure out what it is and how to write it out.

DLee_1L
Posts: 103
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Re: n when calculating Gibbs Free Energy

Postby DLee_1L » Tue Feb 25, 2020 11:22 pm

AKatukota wrote:Ohhh so in order to get n, you have to balance the half reactions and then subtract the electrons?

You don't have to subtract the electrons, all you have to do is make sure the number of electrons given off and received by both half reactions are the same and then that is your n moles of electrons. There are some cases where you have to be careful because the equation might have a divisible factor and that would decrease the amount of electrons. (Eg. 6N.7 with H2->2H+ + 2e- on the anode and cathode)


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