# Half equations (Redox) AS Chemistry Watch

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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?

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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#3

(Original post by

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

**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

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(Original post by

Wait but isn't it 2SO2 + O2 --> 2SO3?

**Sapphire_Rose**)Wait but isn't it 2SO2 + O2 --> 2SO3?

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#5

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

**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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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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#7

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

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?
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#8

(Original post by

Wouldn't O2 be -2 as it has an oxidation number of -2 in SO2? That's what's causing me some confusion

**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

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(Original post by

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

**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?And on the left when side Sulfur is +4 so should O be -2 as well?

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(Original post by

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

**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

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?

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(Original post by

sorry, i accidently made a mistake its from 0 to -2 so still reduced

**ablog**)sorry, i accidently made a mistake its from 0 to -2 so still reduced

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#13

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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?

**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?

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

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(Original post by

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!

**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!

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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#15

(Original post by

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. .

**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. .

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