Calculating Gibbs free energy of 1/2 rxn


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205192823
Posts: 100
Joined: Wed Sep 18, 2019 12:19 am

Calculating Gibbs free energy of 1/2 rxn

Postby 205192823 » Thu Mar 12, 2020 6:16 pm

The given half reactions are:
Ag+ (aq) + e- ---> Ag(s) E*= 0.7996V
Ni 2+ (aq) + 2e- ---> E*= -0.25V
I solved for E*cell by using E*cell= E* right + E* left, which equaled to 1.0496V.
Then I went to solve Gibbs free energy by using delta G*= -nFE*cell:
-1 mol(96485 c/mol)(1.0497 V) and got -191270.656 c*V which I converted to -1.0127 X 10^5 c*V.
I didn't get the correct answer but is it because of the units? I thought c*V= J so that's what I put the units as.

MingdaH 3B
Posts: 133
Joined: Thu Jul 11, 2019 12:17 am

Re: Calculating Gibbs free energy of 1/2 rxn

Postby MingdaH 3B » Thu Mar 12, 2020 6:38 pm

Your units are fine, but I'm pretty sure n = 2, since there is a transfer of two electrons.

205192823
Posts: 100
Joined: Wed Sep 18, 2019 12:19 am

Re: Calculating Gibbs free energy of 1/2 rxn

Postby 205192823 » Thu Mar 12, 2020 6:56 pm

Okay that makes more sense, but how do you determine if I use that electron transfer or the other? I used 1 mol because I was looking at the Ag equation.

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

Re: Calculating Gibbs free energy of 1/2 rxn

Postby Jessica Chen 2C » Fri Mar 13, 2020 1:56 pm

Both equations have to have the same number of electrons being transferred, that's a part of balancing redox reactions. So you'd multiply the Ag half reaction by 2, and n=2 for both equations.


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