[Catherina]

The question is this:

Balance the following skeleton equations by adding H+ or OH- and H2O (if necessary) to the half reactions and then adding the half reactions to obtain a balanced net reaction equation.

CH3OH + MnO4(-)  -->  Mn(+)  + CH2O    (acidic)

The corrected answer I got back is:

5[CH3OH  -->  CH2O + 2H(+) + 2e-]

2[MnO4(-) + 8H(+) + 5e-  --> Mn(2+) + 4H20]

5CH3OH + 2MnO4(-) + 6H+  -->  5CH2O + 2Mn(2+) + 8H2O


What I'm not understanding is where the 5 electrons in the MnO4 reaction are coming from.  How does the Mn(2+) get the 2+ charge when in the initial equation it was just a + , not a 2+ ?  It's the 5 electrons that are throwing me off. I just can't see how it's only 5.. 

[Borek]

Mn⁺ must be a typo, no such animal (at least not in normal water solution). Mn²⁺ it is.

[Catherina]

Ok.  So if Mn is 2+ , I still don't see where the 5 electrons are coming from to balance the electrons out.  How does the 5 electrons fit into all this?

[Borek]

Show how you calculate number of electrons. Why do you think it is "only" 5 and should be more?

[Catherina]

MnO4(-) 

So O has an Ox. # of -2, x 4 = -8

Mn has an Ox. # of 7 (?) ..  7+ -8 = -1(charge of the MnO4(-) ion)

So I guess what I'm thinking it would be is the 8H+ + the -1 for the MnO4 ion, to balance, you would need 7 electrons?

I feel I'm missing a step somewhere.. but I can't seem to figure it out.

[DrCMS]

MnO4(-) 

So O has an Ox. # of -2, x 4 = -8

Mn has an Ox. # of 7 (?) ..  7+ -8 = -1(charge of the MnO4(-) ion)

You've got this bit correct.

So I guess what I'm thinking it would be is the 8H+ + the -1 for the MnO4 ion, to balance, you would need 7 electrons?

I feel I'm missing a step somewhere.. but I can't seem to figure it out.

Yes you're missing the other side of the equation.

On one side you have MnO₄^- + 8H⁺  and on the other  Mn²⁺ + 4H₂O

(Think of the MnO₄^- as Mn⁷⁺ + 4O^2- and 4H₂O as 8H⁺ + 4O^2-)

What do you need to equation to add to balance it?