14.9 6th Edition


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

14.9 6th Edition

Postby Lauryn Shinno 2H » Sat Feb 23, 2019 3:26 pm

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

Postby Melissa Bu 1B » Sat Feb 23, 2019 3:31 pm

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

Postby Katelyn Phan 2A » Sat Feb 23, 2019 7:44 pm

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


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