Posted by Alex on Wednesday, February 20, 2008 at 6:40pm.
No. I don't remember the exact equation but I know the one I wrote for you that evening was correct. This one, the C has an oxidation state of -2 (4H=+4, 1 O = -2 so C must be -2) in CH3OH. It goes to zero on the other side with CH2O so it loses 2 electrons.
Mn changes from +7 to +2 so it is a gain of 5 electrons.
Multiply the C half equation by 5 and the Mn half equation by 2. Then add water to the right to balance the oxygen atoms, then H^+ on the left to balance the H atoms.
I get 6H^+ + 5CH3OH + 2MnO4^- ==> 2Mn^+2 + 5CH2O + 8H2O
Check my work.
Ahh! You're right. I multiplied the H with the O instead of adding and ended up losing a -1 charge. Thanks so much for the help!
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