## 6L.1

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

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HannahBui 2K
Posts: 101
Joined: Wed Sep 18, 2019 12:19 am

### 6L.1

I am having trouble figuring out the number of electrons being transferred (n) for this reaction: 2 Ce4+(aq) + 3 I−(aq) →2 Ce3+(aq) + I3−(aq). Can someone explain how I would figure this out?

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

### Re: 6L.1

Yes, so the value of n would be 2 in this case. As you can see, one half reaction is 2Ce4+ --> 2 Ce 3+. On the left side of the equation, 2 * 4 = 8 and the right side, 2*3 = 6. So, you have a net charge difference of 2. On the other hand, we have 3 I- --> I 3-. In this case, we have a -3 on the left side and a -1 on the right side. This, too, is a difference of 2. Therefore, the value of n, and the number of electrons transferred in the redox reaction, is 2.

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

### Re: 6L.1

In order to find the number of electrons transferred you need to find the two half reactions. Then you need to multiply the half reactions in order to have the same number of electrons transferred in each reaction, which allow the electrons to cancel in the overall reaction. This would give the overall number of electrons transferred.

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