6th edition 14.35

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Maddy Mackenzie
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6th edition 14.35

Postby Maddy Mackenzie » Wed Mar 06, 2019 5:32 pm

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
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Re: 6th edition 14.35

Postby Vy Lu 2B » Wed Mar 06, 2019 5:35 pm

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
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Re: 6th edition 14.35

Postby Xingzheng Sun 2K » Fri Mar 08, 2019 4:14 am

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

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Re: 6th edition 14.35

Postby JacobHershenhouse3G » Fri Mar 08, 2019 4:24 am

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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