# What’s the oxidation state of carbon in methanal, H2CO?

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Can you please explain how I can determine the oxidation state of carbon in methanal? Thanks 🙂

• Use the oxidation states of the things you know, H is +1, O is -2 (except in O2 and peroxides). Remember, the total for all the atoms present must add to zero for a neutral compound. Also, don’t forget that if there are two of an element in the compound, they both count.

• Carbon Oxidation States

• Well, menthanal is a neutral atom, that is, is has no charge. So, the oxidation numbers should add up to 0. Also, we know that the oxidation number of H is always +1 except in metal hydrides and O is always -2 except in peroxides. Methanal isn’t a metal hydride and doesn’t contain a peroxide.

Calculations:

2 x +1 (there are 2 H atoms) + C + -2 (O) = 0

+2 + -2 + C = 0

C= 0

• C would have an oxidation number of zero (0) in this compound. While this may seem impossible, improbable, or “wrong” it does work when performing oxidation-reduction equation balancing. The system is based upon H being 1+ and O being 2- so in this compound C equates to zero… it works out in the end, so while unusual, this bookeeping system does indeed explain the redox balance.

• Acids: +3.

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