Write a balanced equation for the following reduction-oxidation reaction.
SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
can someone help me step by step??
I answered this earlier by saying S goes from oxidation state of +4 on the left to +6 on the right (loss of 2 electrons). Mn goes from an oxidation state of +7 on the left to +2 on the right (gain of 5 electrons). You want to make the electron loss = electron gain. What about this don't you understand or can you finish?
First of all I am not the same person as before...his name was Max, my name is Kevin so i do not know that you answer this question before. Secondly, I do not know what you mean about about loss of 2 electrons, how would i use hydrogen to balance the charges when I have:
SO3(2-)-- SO4(2–)+ 2e(-)+?H(+)
I did it brute force as follows (not a chemist)
O3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
first deal with S, Mn atoms
there must be the same number on each side
nSO3(2-) + mMnO4(-) --- nSO4(2–) + mMn(2+)
now try 2jH(+) on the left and jH2O on the right
nSO3(2-) + mMnO4(-) +2jH(+) --- nSO4(2–) + mMn(2+) + jH2O
O atoms
3 n + 4 m = 4 n + j
or
n+j = 4 m
+ charges
-2 n - 1 m + 2 j = -2 n + 2 m
so from the second one
3 m = 2 j
j = (3/2) m
from the first one
n + (3/2) m = 4 m
or
n = (5/2) m
Now go back using m = 2 to make the fraction go away
then j = 3
and n = 5
5 SO3(-2) + 2 MnO4(-1) + 6 H(+1) --> 3 H2O + 5 SO4(-2) + 2 Mn(+2)
uh i don't think that's how u do it because i think you need to add and remove electrons from two separate reactions first before multiplying it out...but thank you anyway
Now to do it right (for my own entertainment )
S(+4) ---> S(+6) + 2 e(-)
M(+7) + 5 e(-) ---> M(+2)
to balance the electrons multiply the first equation by 5 and the second by two.
5 S(+4) ---> 5 S(+6) + 10 e(-)
2 M(+7) + 10 e(-) ---> 2 M(+2)
There, that gives you the 5 and the 2 which is the hard part. (First time I have done one of these since about 1957)