Could someone please help with oxidation reduction problem??
MnO4(-) + H(+) =====> MnO2 + O2 +H20
Could someone please help me?? Please be as thorough as possible!!??
I ended up with 28H(+) + 7MnO4(-) =====> 4MnO2 + 3O2 + 14H2O which I know is incorrect. Have any idea where I messed up??
Answer:
First determine what is being oxidized (losing electrons) and what is being reduced (gaining electrons).
MnO4- the Mn+7 the O-2
MnO2 the Mn+4 the O-2
The Mn is being reduced, going from +7 to +4 (a gain of 3 electrons)
The O2 on the product side has lost 4 electrons (going from 2 atoms at -2 to O2 at O))
The electron gain and loss have to be balanced so for each Mn reduced, you need 3/4 O2 molecules.
1 Mn+7 ---> 3/4 O2
2 Mn+7 ---> 1.5 O2
3Mn+7 ---> 2 1/4 O2
4 Mn+7 ---> 3 O2 (finally, whole numbers!)
So balance what you know-
4 MnO4- ---> 3O2 + 4MnO2
16 O on this side and 8+6=14 on the other side
You can add H+ on the reactant side and H2O on the product side. To balance the O you will need 2 H2O. To balance the H you will need 4 H+.
4 MnO4 + 4H+ ---> 3 O2 + 4 MnO2 + 2 H2O
Yes, Mn is not balanced.
Here is what you are balancing:
MnO4(-) = MnO2 (balance half rxn in aqueous acid)
2 H2O = O2 + 4 H+ + 4 e- (oxygen in water is being oxidized)
Add together.
You got screwed up because some of the water is just being "used to balance things" and some is actually getting involved in the redox. That is a tricky situation. It helps to remember to write down the oxidation and the reduction steps explicitly.
And by the way Tony makes a good point about balancing the charge. Adding up the charges at the end is actually a really good way to check your work. And I agree with Tony that you should ultimately be able to solve these just by tinkering a little and thinking it through. But ... is that a method you want to trust on the test? I'd recommend sticking with one of the sure-fire methods of balancing until you're pretty spiffy at it.
2MnO4- + 4H+ >>2MnO2 + O2 + 2H2O
Hope it helps
So this one is just easier if you don't break it down into half reactions and just think it through
so there are only two charged species, the MnO4- and the H+ so obviously we have to keep the number of moles of these the same. The problem that we need to solve is the number of oxygen atoms - we'll only ever get an even number on the left side of the equation and we have an odd number on the right side of the equation. we make it an even number of oxygens by using 2 moles of water and subsequently adjusting H+ and MnO4- as well as MnO2. Then all we have to do is add 3 moles of O2
4 MnO4- + 4H+ ---> 4MnO2 + 3O2 + 2H2O
PS - the guy above me did a fine job of balancing the elements but failed to balance the charge.
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