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.
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!
To balance the hydrogen ions in the equation, you can use H+ ions. The number of H+ ions needed is determined by the imbalance in hydrogen atoms on both sides of the equation.
In this case, you have 4 hydrogen atoms on the right (from the 4H2O), but only 1 hydrogen atom on the left (from the CH3OH). To balance this, you need to add 3 H+ ions on the left side of the equation.
The balanced equation becomes:
5CH3OH + 1MnO4^- + 3H+ → Mn^2+ + 5CH2O + 4H2O
To balance the hydrogen atoms in the reaction, you can add H+ ions to the appropriate side of the equation. In this case, you need to balance the hydrogens on both sides of the equation.
Starting from:
5CH3OH + 1MnO4^- ----> Mn^2+ + 5CH2O + 4H2O
You can count the number of hydrogen atoms on each side. On the left side, you have 20 hydrogen atoms (5 hydrogens on each of the 5 CH3OH molecules), and on the right side, you have 8 hydrogen atoms (4 hydrogens in the 4 H2O molecules).
To balance the number of hydrogen atoms, you need to add 12 H+ ions to the left side of the equation:
5CH3OH + 1MnO4^- + 12H+ ----> Mn^2+ + 5CH2O + 4H2O
Now, we need to check the charges on both sides. On the left side, we have 5CH3OH (which has a charge of 0), 1MnO4^- (which has a charge of -1), and 12H+ (which have a charge of +1 each). The total charge on the left side is -1 + 12(+1) = +11.
On the right side, we have Mn^2+ (which has a charge of +2) and the neutral compounds 5CH2O and 4H2O. The total charge on the right side is +2.
Since the total charge needs to be the same on both sides of the equation, you need to balance the charges by adding electrons. In this case, you need to add 11 electrons to the right side of the equation:
5CH3OH + 1MnO4^- + 12H+ + 11e^- ----> Mn^2+ + 5CH2O + 4H2O
Now the equation is balanced both in terms of atoms and charges.