# Half equations (Redox) AS ChemistryWatch

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#1
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?
Last edited by Crow_M; 1 year ago
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#2 0
1 year ago
#3
(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?
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#4
(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?
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1 year ago
#5
(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)
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#6
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?
Last edited by Crow_M; 1 year ago
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1 year ago
#7
the oxidation number of sulfur goes from +4 to +6 (oxidised )
and the oxidation number of O2 goes from 0 to -6 (reduced )

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?
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1 year ago
#8
(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
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#9
(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 )

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?
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1 year ago
#10
sorry, i accidently made a mistake its from 0 to -2 so still reduced
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#11
(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? 0
#12
(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? 0
1 year ago
#13
(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!
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#14
(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. .
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1 year ago
#15
(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.
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