Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Catherina on March 16, 2008, 12:18:53 PM
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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.. ???
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Mn+ must be a typo, no such animal (at least not in normal water solution). Mn2+ it is.
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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?
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Show how you calculate number of electrons. Why do you think it is "only" 5 and should be more?
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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.
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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?