## Gibbs free energy and 6L.1

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Tracy Tolentino_2E
Posts: 140
Joined: Sat Sep 07, 2019 12:17 am

### Gibbs free energy and 6L.1

In the equation deltaG = -nFE, how do we figure out the n (if its moles of electrons).
For example, in part a 2Ce^+4(aq) + 3I^-(aq) + 2Ce^3+(aq) + I3^-(aq), Ecell° 5 11.08 V
How do I find the n in the equation?

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

### Re: Gibbs free energy and 6L.1

You have to first balance the half reactions so that there is conservation of charge. The molar coefficient of electrons is going to be your n value.

Grace Jansen 2A
Posts: 28
Joined: Sat Jul 20, 2019 12:16 am

### Re: Gibbs free energy and 6L.1

I'm confused how they got 2 moles for part a. I know Ce is being reduced, but how is I being oxidized? The charges are the same?

Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am

### Re: Gibbs free energy and 6L.1

For the oxidation, 3I- ->I3- +2e-. Since the charges need to balanced on both sides, 2e- need to be added (the charge of the left side is -3 and the charge of the right side is -1). Since there are 2 moles of e- transferred, n=2