## Problem 6L.1

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

Theodore_Herring_1A
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Problem 6L.1

In the problem 6L.1 (7th Edition),

Calculate the standard reaction Gibbs free energy for the following cell reactions:

(a) 2 Ce4+(aq) + 3 I−(aq) →2 Ce3+(aq) + I3−(aq), Ecell°=+1.08V,

What value n do I use for Ecell = -nFE?

marisaimbroane1J
Posts: 69
Joined: Fri Sep 28, 2018 12:23 am

### Re: Problem 6L.1

Your value for n will be the number of moles of e- in the balanced half reactions. So write out the two half reactions and then make sure the number of e- is the same for both of them. This number will be n in the equation.

David Effio 1H
Posts: 38
Joined: Wed Feb 14, 2018 3:01 am

### Re: Problem 6L.1

To find the value of n, the number of electrons, you should try to find what the half reactions will look like. Make sure the charges, reactants and products are all balanced first. You will then be able to plug in n for the formula.

Kirsty Star 2H
Posts: 48
Joined: Fri Sep 28, 2018 12:24 am

### Re: Problem 6L.1

Once you do what those people mentioned, I believe the balanced reaction comes out to 2 e-. So n = 2

Angela Grant 1D
Posts: 67
Joined: Fri Sep 28, 2018 12:25 am

### Re: Problem 6L.1

You'd use n = 2 electrons because when you split up the half-reactions you get:

Oxid.: Ce4+ + e- --> Ce3+
Red.: 3I- --> I(3)- + 2e-

Multiply the oxidation half-reaction by 2 to balance out the electrons and cancel them out, leaving you with two electrons getting cancelled.