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?
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.
Oh no..it 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.
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.
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.
Sure, I can help you balance the rest of the equation. Balancing chemical equations involves ensuring that the number of atoms of each element is the same on both sides of the equation and that the charges are balanced as well.
Let's start by balancing the carbon (C) atoms. On the left side, we have 5 carbon atoms in CH3OH-, so we need to have 5 carbon atoms on the right side as well. We can achieve this by multiplying the CH2O species on the right side by 5.
5CH3OH- + 2MnO4- ----> Mn2+ + 5CH2O
Next, let's balance the hydrogen (H) atoms. On the left side, we have 15 hydrogen atoms (5 carbon atoms multiplied by 3 hydrogen atoms in CH3OH-), so we need to have 15 hydrogen atoms on the right side as well. We can achieve this by adding 15H+ to the right side.
5CH3OH- + 2MnO4- ----> Mn2+ + 5CH2O + 15H+
Now, let's balance the oxygen (O) atoms. On the left side, we have 24 oxygen atoms (5 CH3OH molecules multiplied by 4 oxygen atoms in each CH3OH- molecule), and on the right side, we have 13 oxygen atoms from the MnO4- and CH2O species. To balance the oxygens, we need to add water (H2O) molecules to the right side.
5CH3OH- + 2MnO4- ----> Mn2+ + 5CH2O + 15H2O
Finally, let's balance the charges. On the left side, we have 5 negative charges from CH3OH-. On the right side, we have 2+ charge from Mn2+, and each H2O molecule contributes 0 charge. Therefore, we need to add 5 positive charges to the right side. We can achieve this by adding 5H+ to the right side.
5CH3OH- + 2MnO4- + 8H+ ----> Mn2+ + 5CH2O + 15H2O
Now the equation is balanced, with an equal number of atoms for each element on both sides and balanced charges.