## 14.9 6th Edition

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

Lauryn Shinno 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:24 am

### 14.9 6th Edition

The question asks you to calculate the standard reaction Gibbs free energy for a. 2Ce^4+(aq)+3I^-(aq)—>2Ce^3+(aq)+I3^-(aq). How do you find the moles of electrons transferred? Also, I understand that Ce^4+ is getting reduced, but how is 3I^- getting oxidized? It's oxidation number doesn't change.

Melissa Bu 1B
Posts: 36
Joined: Fri Sep 28, 2018 12:19 am

### Re: 14.9 6th Edition

I am also not sure about how I is being oxidized since its oxidation number is -1 on both sides of the equation.
As for the number of moles of electrons transferred, I believe we have to refer to the electrochemical series. The reduction equation for Ce4+ + e- --> Ce3+ in the table in Appendix 2B includes one electron. The equation given in the problem has 2Ce4+ and 2Ce3+, implying that 2 electrons were transferred. Therefore, n=2.

Katelyn Phan 2A
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

### Re: 14.9 6th Edition

You'll be able to find the number of moles transferred by balancing the equation through separating the half reactions.