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How do I find oxidation number?

Posted: Fri Feb 03, 2017 6:52 pm
by Andrew Uesugi 3I
This seems like I'm asking this question WAY too late in this, but how do I find the oxidation number of an element in a chemical equation?

For example: MnO4^- + H2C2O4 -> Mn^2+ + CO2

When focusing on MnO4^- -> Mn^2+, how come Mn in MnO4 has (-7) and O4 has -2, and Mn^2+ has +2?

I'm lost here.

Re: How do I find oxidation number?

Posted: Fri Feb 03, 2017 7:10 pm
by Jasmine_Esparza_2A
Hi I had a similar question. My PLF gave me an answer with specific steps.

1. The oxidation state of a lone element or free state is zero.
For example if you see a long element such as Cu, Zn, Ag. The O.S. will be zero. There are 8 diatomic elements to remember who's O.S. is zero as well: H2, O2, N2, Br2, S2, I2, F2, Cl2.

2. The O.S. is the same as the charge of the ion.
E.x. NH4+ the charge would be +1.

3. Group 1 elements have OS of +1
Group 2 elements have OS of +2
Group 16 have -2
group 17 have -1.

4. The sum of the O.S. in a neutral compound is zero. Using the OS of the parts you know can help you find an unknown OS.
ex. CH4 We know from #3 that H has OS of +1, since there are four H atoms, the OS of H4 is +4. If you want the overall charge of CH4 to be zero, then the OS of C should cancel the OS of H4. SO C should be -4 because -4 + 4 = 0.

5. H2O2 has a charge of -1.

Hope that helped!

Re: How do I find oxidation number?

Posted: Sun Feb 05, 2017 3:04 am
by Eric_pierce_3E
This was very helpful! Thanks