Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Catherina on March 16, 2008, 12:18:53 PM

Title: Acidic skeleton equation.
Post by: Catherina on March 16, 2008, 12:18:53 PM
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..  ???
Title: Re: Acidic skeleton equation.
Post by: Borek on March 16, 2008, 01:20:50 PM
Mn+ must be a typo, no such animal (at least not in normal water solution). Mn2+ it is.
Title: Re: Acidic skeleton equation.
Post by: Catherina on March 17, 2008, 11:44:59 AM
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?
Title: Re: Acidic skeleton equation.
Post by: Borek on March 17, 2008, 12:01:29 PM
Show how you calculate number of electrons. Why do you think it is "only" 5 and should be more?
Title: Re: Acidic skeleton equation.
Post by: Catherina on March 17, 2008, 01:08:28 PM
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.
Title: Re: Acidic skeleton equation.
Post by: DrCMS on March 17, 2008, 01:27:41 PM
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+ + 4O2- and 4H2O as 8H+ + 4O2-)

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