# Half equations (Redox) AS Chemistry

Hi, so I'm trying to improve my speed with balancing redox equations so I can balance them with numbers or with redox, however I'm a bit stuck with this one...

SO2 02 -> SO3

Before anyone say you can do it with a simple method, I know. I know that the answer would be 2SO2 O2 -> 2SO3

However, can someone please show me what's going on with half equations, please?

Thanks

EDIT: I also know that Sulfate oxidises from 4 to 6, but what am I supposed to do?
(edited 5 years ago)
Original post by Crow_M
Hi, so I'm trying to improve my speed with balancing redox equations so I can balance them with numbers or with redox, however I'm a bit stuck with this one...

SO2 + 02 -> SO3

Before anyone say you can do it with a simple method, I know. I know that the answer would be 2SO2 + 3O2 -> 2SO3

However, can someone please show me what's going on with half equations, please?

Thanks

Wait but isn't it 2SO2 + O2 --> 2SO3?
Original post by Sapphire_Rose
Wait but isn't it 2SO2 + O2 --> 2SO3?

Oops I mixed up two questions, yeah, that's the right answer, but how would we show that as a redox equation?
Original post by Crow_M
Oops I mixed up two questions, yeah, that's the right answer, but how would we show that as a redox equation?

I would assume you show the oxidation states so S from 4 to 6 (oxidation) and O2 from 0 to -6 (reduction)
Wouldn't O2 be -2 as it has an oxidation number of -2 in SO2? That's what's causing me some confusion, it should be -2 on the RHS as well because of Sulfur being 6 and therefore should be balanced, no?
(edited 5 years ago)
the oxidation number of sulfur goes from +4 to +6 (oxidised )
and the oxidation number of O2 goes from 0 to -6 (reduced )

therefore redox equationSubmit reply
the (Original post by Crow_M)Oops I mixed up two questions, yeah, that's the right answer, but how would we show that as a redox equation?
Original post by Crow_M
Wouldn't O2 be -2 as it has an oxidation number of -2 in SO2? That's what's causing me some confusion

Yes, the oxidation of state of O2 is -2 but there are 3x O2 molecules, so 1x O2 is -2 which means 3x O2 molecules are -6, but you could still write that oxygen has an oxidation of 2- now as it would still show it has been reduced
Original post by ablog
the oxidation number of sulfur goes from +4 to +6 (oxidised )
and the oxidation number of O2 goes from 0 to -6 (reduced )

therefore redox equationSubmit reply
the (Original post by Crow_M)Oops I mixed up two questions, yeah, that's the right answer, but how would we show that as a redox equation?

Shouldn't O be -2 on the right side so that it balances with the +6 of Sulfur?
And on the left when side Sulfur is +4 so should O be -2 as well?
sorry, i accidently made a mistake its from 0 to -2 so still reduced
Original post by rehman15
Yes, the oxidation of state of O2 is -2 but there are 3x O2 molecules, so 1x O2 is -2 which means 3x O2 molecules are -6, but you could still write that oxygen has an oxidation of 2- now as it would still show it has been reduced

So, in SO2, S has an oxidation state of +4, O (or 1/2 of O2) has an oxidation state of -2. Which makes O2 have an oxidation state of -4 in SO2. But on O2 it's oxidation state is zero as it's an uncombined element, so which number should I take?

In SO3, Sulfur has an oxidation state of +6 whereas each O has an oxidation state of -2 so O3 has an oxidation state of -6, so as it is -2 in both sides, so how would I write this?
Original post by ablog
sorry, i accidently made a mistake its from 0 to -2 so still reduced

Oh that makes more sense now, my only question now is why is it 0 and not -2? Like, in SO2 it is -2, but in O2 is 0, so why wouldn't take 0 and not -2?
Original post by Crow_M
So, in SO2, S has an oxidation state of +4, O (or 1/2 of O2) has an oxidation state of -2. Which makes O2 have an oxidation state of -4 in SO2. But on O2 it's oxidation state is zero as it's an uncombined element, so which number should I take?

In SO3, Sulfur has an oxidation state of +6 whereas each O has an oxidation state of -2 so O3 has an oxidation state of -6, so as it is -2 in both sides, so how would I write this?

Ok, maybe i've made it a lil confusing, on the left put the oxidation state of sulfur as +4 and the oxidation state of O2 on its won 0. then on the right side, put the oxidation state of sulfur as +6 and the oxidation state of oxygen as -2.

I'm really crap at explaining so sorry if i confused you!
Original post by rehman15
Ok, maybe i've made it a lil confusing, on the left put the oxidation state of sulfur as +4 and the oxidation state of O2 on its won 0. then on the right side, put the oxidation state of sulfur as +6 and the oxidation state of oxygen as -2.

I'm really crap at explaining so sorry if i confused you!

Oh that makes much more sense now, thanks!

Quick question, why is the oxidation 0 on the left and not -2? I'm trying to understand it so that I can make half equations much more easily, so that's why I chose this example. .
Original post by Crow_M
Oh that makes much more sense now, thanks!

Quick question, why is the oxidation 0 on the left and not -2? I'm trying to understand it so that I can make half equations much more easily, so that's why I chose this example. .

For this reaction to work as a redox reaction, the oxidation of oxygen must decrease since the oxidation state of sulfur increases. So if you used the oxygen in the 2SO2 molecule, the oxidation state for oxygen would go from -2 to -2 in SO3 which doesn't show redox, but because we used the oxidation of oxygen on its own which is 0, it shows the redox reaction happening as it now goes from 0 to -2.
Original post by ablog
the oxidation number of sulfur goes from +4 to +6 (oxidised )
and the oxidation number of O2 goes from 0 to -6 (reduced )

therefore redox equationSubmit reply
the (Original post by Crow_M)Oops I mixed up two questions, yeah, that's the right answer, but how would we show that as a redox equation?

oxidation number stays as -2 not -6