Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Write a balanced equation for the following reduction-oxidation reaction. SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+) can someone he...Asked by kevin
Write a balanced equation for the following reduction-oxidation reaction.
SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
can someone help me step by step??
SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
can someone help me step by step??
Answers
Answered by
DrBob222
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?
Answered by
kevin
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(+)
SO3(2-)-- SO4(2–)+ 2e(-)+?H(+)
Answered by
Damon
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)
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)
Answered by
kevin
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
Answered by
Damon
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)
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)
Answered by
DrBob222
Kevin--The numbers Damon provided are correct. Let me know what you don't understand about it and I can help you through it (as a chemist).
Answered by
DrBob222
Kevin, you may not have written the question the first time and perhaps it was some other person named Max, BUT both of you were/are using the same computer because the IP address is the same. But I still encourage you to tell me what you don't understand if you don't get how Damon balanced the equation for you.
Answered by
kevin
thanks damon
There are no AI answers yet. The ability to request AI answers is coming soon!