I found the charges on all of the following atoms and found out that Mn was reduced and C was oxidized.

CH3OH- + MnO4- ----> Mn2+ + CH2O
2e-/C 5e-/Mn
x 5 x 2
= 10 = 10

5CH3OH- + 2MnO4- ----> Mn2+ + CH2O

Can anyone help me balance the rest?

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asked by Alex
  1. I don't believe the CH3OH has a charge on it.
    6H^+ + 5CH3OH + 2MnO4^- ==>2Mn^+2 + 5CH2O + 8H2O. Check it to make sure it balances BOTH with atoms and charge.

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  2. Oh doesn't. I misread it because of the print. Wouldn't that change the whole thing? I don't think it would have a coefficient of 5 anymore. I'm balancing redox equations. And now since the overall charge has changed, the charge on the carbon has changed, and so the oxidation has changed.

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    posted by Alex
  3. You are right and no, you're not right. Sound confusing?
    I balanced the equation as if the negative charge on the CH3OH was NOT there so my equation is correct, I think. You are right that a negative charge there, or is one WAS there and you took it off, it would change the oxidation states; however, C would still be oxidized and Mn still is reduced. Here is the way you do those redox equations.
    Mn goes from +7 on the left to +2 on the right. That is a gaiin of 5 electrons and it is reduced.
    C on the left, in CH3OH, has an oxidation state of -2. It goes to 0 on the right in CH2O. That is a loss of 2 elecrons which makes C oxidized. 10 is the lowest common multiple, so multiply Mn by 5 and CH3OH by 2 which gets 10 elecrons gained and 10 electrons lost. Now add water to the right side to balance the O atoms, then add H ion on the left to balance H atoms. Then check to make sure the charge balances. I just check it one more time and all atoms balance. All charges balance.

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  4. Check that. I made a big typo. The manganese half equation is multiplied by 2 and the CH3OH/CH2O half equation is multiplied by 5. The equation is wrote is balanced, both with atoms and with charge but I mistyped the coefficients in the explanation.

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