Formation of Mn(OH)3 from Mn(OH)2 and Oxygen

[SUchica10]

When hydroxide ion is added to Mn(II) in aqueous solution, an insoluble white precipitate of manganous hydroxide, Mn(OH)2, is formed. If there is oxygen present, this material reacts with it to form manganic hydroxide, Mn(OH)3


Part one was easy because it wasnt a redox it just needed to be balanced so I got:

+
2OH + Mn --> Mn(OH)2

The second part I am having trouble with...

+2 -2+1 0 +3 -2+1
Mn(OH)2 + O2 ---> Mn(OH)3

+2 -2+1 +3 -2+1
Oxidation - Mn(OH)2 ---> Mn(OH)3 + 1e-

0 -2+1
Reduction - O2 + 2e- ---> Mn(OH)3
(I don't understand this bc there's no Mn on the left side)

alright this I don't know if anyone can help me with because its really hard to type out and I don't think this is going to look like the way i set it up but if anyone can try to explain this to me through here it would be greatly appreciated.

 

[Decan]

For starters, what do you have to do? Do you have to balance the redox rxn, or simply figure out the oxidation state?

 

[f(x)]

After a simple search i got this


4Mn(OH)2 + O2 + 2H2O ==> 4Mn(OH)3