## 6th edition 14.35

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

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### 6th edition 14.35

For this question, why does the half reaction In 3+ + 2e- --> In 2+ when there is only a change of 1 e- between In 3+ and In2+? Even in the back of the book the reduction half reaction only has 1 e- so I'm wondering why they have it this way in the solutions manual.

Vy Lu 2B
Posts: 59
Joined: Fri Sep 28, 2018 12:24 am

### Re: 6th edition 14.35

This might have been an error on the solution manual's part; however, would this reduction be a result of combining two different reduction phrases together? I believe I saw one question that had that similar issue.

Xingzheng Sun 2K
Posts: 62
Joined: Fri Sep 28, 2018 12:29 am

### Re: 6th edition 14.35

I think the correct equation for this is In3+ (aq) + e- -----> In2+ (aq). This is from my solution manual.

JacobHershenhouse3G
Posts: 32
Joined: Thu Jan 10, 2019 12:17 am

### Re: 6th edition 14.35

It seems that you have the right idea and the book may have a typo. the oxidation state on a lone ion is the charge so it makes sense that In3+ is reduced to In2+ with the gain of an electron.

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