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### 13.3 a Multiplying by appropriate factors

Posted: **Mon Feb 02, 2015 2:43 pm**

by **Madison Davis 3F**

For 13.3 , The solutions manual says to multiply the oxidation and reduction half-reactions by appropriate factors that will result in the same # of electrons in both half reactions. I guess I'm just kind of confused with how to do this/ why you have to do this?

Thanks for the help!

### Re: 13.3 a Multiplying by appropriate factors

Posted: **Mon Feb 02, 2015 3:44 pm**

by **Chem_Mod**

Hello.

When you combine the half reactions to get the overall reactions, the electrons in the half reactions will have to cancel out, just like Hess's Law.

For example, if you have:

Reduction: 5e− + 8H+ + MnO4- →Mn2+ + 4H2O

Oxidation: 2I−→I2+2e−

You will need to multiply the bottom equation by a factor of 5 and the top by a factor of two. This way, there will be 10 electrons in each half reaction, and when you add the reactions you will end up with

10I−+16H++2MnO4-→5I2+2Mn2++8H2O without any electrons in the overall equation.

In the future please copy the question from the book onto your post so that those without a book on them can respond.

### Re: 13.3 a Multiplying by appropriate factors

Posted: **Tue Feb 03, 2015 12:03 am**

by **Alexandra Keir 3H**

Are there any instances where the electrons do not cancel out? Where we are given a reaction with an excess of electrons on either the reactant or product side (not in terms of half reactions)?

### Re: 13.3 a Multiplying by appropriate factors

Posted: **Tue Feb 10, 2015 12:19 am**

by **Justin Le 2I**

I don't think so because in redox reactions, the substance that is being oxidized is transferring electrons to the substance that is being reduced. There are no e- just lying around.