Complete and Balance the following equation using the half-reaction method;
(MnO4^-) + (CH3OH) --> (Mn^2+) + (HCO2H
an explanation/step by step would be nice because i keep getting wrong answer from the H+ & H2O (they are both suppose to be 12 according to the answer key but i can't get it)
I got 4MnO4^- +5CH3OH + 22H^+ ---> 4Mn^2+ +5HCO2H +11H2O
I Do
Here are tutorials on each phase of the redox process you may need. If you had shown your work I could hav found your problem easily. As it is I have no idea what you're doing wrong. The equation you posted isn't balanced, though, so you know it can't be right.
http://www.chemteam.info/Redox/Redox.html
The balanced equation you obtained, 4MnO4^- + 5CH3OH + 22H^+ ---> 4Mn^2+ + 5HCO2H + 11H2O, is almost correct. However, there seems to be a minor mistake in balancing the hydrogen (H+) and water (H2O) molecules.
To balance the hydrogen atoms, you need to make sure that the number of hydrogen atoms is equal on both sides of the equation. In this case, the hydrogen atoms are present in the MnO4^- and CH3OH molecules, as well as the H^+ and HCO2H products.
Let's go through the step-by-step process of balancing the equation using the half-reaction method:
1. Divide the equation into two half-reactions: the reduction half-reaction and the oxidation half-reaction.
Reduction half-reaction: MnO4^- --> Mn^2+
Oxidation half-reaction: CH3OH --> HCO2H
2. Balance the atoms other than hydrogen and oxygen in both half-reactions.
In the reduction half-reaction:
MnO4^- --> Mn^2+
There is one manganese (Mn) atom on both sides, so no balancing is necessary.
In the oxidation half-reaction:
CH3OH --> HCO2H
There is one carbon (C) atom on both sides, so no balancing is necessary.
There are four hydrogen (H) atoms on the reactant side, but only two on the product side. Therefore, we need to add two more hydrogen atoms to balance it.
The oxidation half-reaction becomes: CH3OH + 2H2O --> HCO2H + 4H+ + 4e^-
3. Balance the oxygen atoms by adding water molecules. Each added water molecule provides an oxygen atom to the equation.
In the reduction half-reaction:
MnO4^- + 8H+ + 5e^- --> Mn^2+ + 4H2O
By introducing 4 water molecules on the product side, we balance the oxygen atoms.
In the oxidation half-reaction:
CH3OH + 2H2O --> HCO2H + 4H+ + 4e^-
No further adjustment is needed as the oxygen atoms are already balanced.
4. Balance the charges by adding electrons (e^-). The number of electrons must be the same in both half-reactions.
In the reduction half-reaction:
MnO4^- + 8H+ + 5e^- --> Mn^2+ + 4H2O
There are 5 electrons on the reactant side, and no electrons on the product side.
In the oxidation half-reaction:
CH3OH + 2H2O --> HCO2H + 4H+ + 4e^-
There are 4 electrons on the product side, and no electrons on the reactant side.
To balance the electrons, we need to multiply the oxidation half-reaction by 5 and the reduction half-reaction by 4:
4CH3OH + 8H2O --> 4HCO2H + 16H+ + 16e^-
5MnO4^- + 40H+ + 25e^- --> 5Mn^2+ + 20H2O
5. Combine the two half-reactions.
Now that the electrons are balanced, we can combine the two half-reactions:
20H+ + 25MnO4^- + 16CH3OH + 64H2O --> 5Mn^2+ + 16HCO2H + 80H2O
Simplifying the equation, we get:
25MnO4^- + 16CH3OH + 20H2O --> 5Mn^2+ + 16HCO2H
So, the balanced equation using the half-reaction method for the given reaction is:
25MnO4^- + 16CH3OH + 20H2O --> 5Mn^2+ + 16HCO2H