MnO4^-(aq)+ 8H^+ --> Mn^2+(aq)+ 4H2O
H2C2O4(aq)+ 2H^+ --> CO2(g)+ 2H20
There are two ways to do it.
1. Look at the total charge on the left for the first equation. It is -1+8 = +7 and on the right it is +2. So you must add 5 electrons to the left [+7+(-5)=+2] to make it balance. In others words, add electrons to the appropriate side to make the charges balance.
2. The other way is to look at the oxidation state of Mn on the left. It is +7. On the right Mn is +2. Going from +7 to +2 means it must have gained 5e so you add 5e to the left side. The nice part about either of these tracks is that they always work.
Hey i have a question regarding disproportionate reactions:
Permanganate ions MnO4 reacts with oxalic acid, H2C2O4 in acidic aqueous solution, producing manganese(II) ions and carbon dioxide. The skeletal equation is
MnO4^-(aq)+ H2C2O4(aq)--> Mn^2+(aq)+ Co2 (g)
balance this reaction in acidic solution
So first i write down the 2 half reactions, and balance them
MnO4^-(aq)+ 8H^+ --> Mn^2+(aq)+ 4H2O
and
H2C2O4(aq)+ 2H^+ --> Co2(g)+ 2H20
I know i have to determine the electron charges, this is where i am sure what to do, i am getting confused as to how to add up the electron charges on each side , i know that i have to to that in both the half reaction and cancel them out and get the net ionic equation...can someone please help me understand the electron charges part:
I think MnO4 is oxidizing here, and H2C2O4 is reducing
2 answers
I forgot to mention that the H2C2O4 half equation is not balanced the way you have it.