Asked by Alex

CH3OH + MnO4- ----> Mn2+ + CH2O

Earlier I posted this question and then realized that CH3OH didn't have a charge. It changed my work because the oxidation state changed. Can you check if what I fixed is correct?

CH3OH- + MnO4- ----> Mn2+ + CH2O
1e-/C 5e-/Mn
x 5 x 1
= 5 = 5

5CH3OH + 1MnO4^- ----> Mn^2+ + CH2O

I now got a different coefficient for manganese. And now I need to balance it..


5CH3OH + 1MnO4^- ----> Mn^2+ + 5CH2O + 4H2O

I'm stuck. I know we have to add H+ ions to balance the H's but I don't know how much.
Answered by DrBob222
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.
Answered by Alex
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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