Yazomi
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This can be either A or C right- to me both of them shows the oxidation of sulfur so how can you tell?
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Igniform
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(Original post by Yazomi)
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This can be either A or C right- to me both of them shows the oxidation of sulfur so how can you tell?
No, it can't be C because H2S is fully reduced - it only has hydrogens. Any species that only has hydrogens is fully reduced.

For A you've written the oxidation numbers - Sulphur is being reduced from +6 to +4. So how can it be A?

For B, you've written the wrong number for Na2SO3 - there are 3 oxygen = -2 each, 2 sodium = +1 each, therefore S = +4. (for this reaction, there is no change in oxidation state).

Therefore the correct answer must be D.

I think you've got the idea of oxidation and reduction mixed up.
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Yazomi
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(Original post by Igniform)
No, it can't be C because H2S is fully reduced - it only has hydrogens. Any species that only has hydrogens is fully reduced.

For A you've written the oxidation numbers - Sulphur is being reduced from +6 to +4. So how can it be A?

For B, you've written the wrong number for Na2SO3 - there are 3 oxygen = -2 each, 2 sodium = +1 each, therefore S = +4. (for this reaction, there is no change in oxidation state).

Therefore the correct answer must be D.

I think you've got the idea of oxidation and reduction mixed up.
Ohhh right that makes sense yeah. I’m a bit confused when you said about c) any species that only had H is fully reduced- D) the sulfur here also only attached to Hydrogen- shouldn’t it also be fully reduced. Also for c) the sulphuric acid had Oxygen in it so the Sulfur isn’t just attached to Hydrogen? Therefore shouldn’t be fully reduced?
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Igniform
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(Original post by Yazomi)
Ohhh right that makes sense yeah. I’m a bit confused when you said about c) any species that only had H is fully reduced- D) the sulfur here also only attached to Hydrogen- shouldn’t it also be fully reduced. Also for c) the sulphuric acid had Oxygen in it so the Sulfur isn’t just attached to Hydrogen? Therefore shouldn’t be fully reduced?
In D, yes the sulphur is fully reduced in the hydrogen sulphide on the left hand side, and it's being oxidised to its elemental form (S0) because it's losing two hydrogens. (Generally, if it loses hydrogens, it is being oxidised)
In C, the sulphur in sulphuric acid is fully oxidised because it has +6 which is the highest oxidation state for sulphur.

You could have a look at the chemical structure of sulphuric acid to really understand it:
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The sulphur has two double bonded oxygens and two single bonded oxygens. It is fully oxidised because it is only bonded to oxygen. The 2 hydrogens in this structure don't matter because they are actually hydrogen ions which play no part in the redox reactions since they have no electrons. If you think about this reaction:
SO42- + 2H+ --> H2SO4. No redox reaction is taking place, the sulphate ion is equivalent to the sulphuric acid. You could replace the H+ with any positive ion.

However, in (C), the sulphuric acid is being reduced to hydrogen sulphide. These hydrogens are NOT just positive ions like in the above picture, they are hydrogen atoms so they reduce the sulphur because they donate their electrons to it. Hydrogen sulphide is therefore fully reduced.

Sorry if I've just confused you even more
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Yazomi
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(Original post by Igniform)
In D, yes the sulphur is fully reduced in the hydrogen sulphide on the left hand side, and it's being oxidised to its elemental form (S0) because it's losing two hydrogens. (Generally, if it loses hydrogens, it is being oxidised)
In C, the sulphur in sulphuric acid is fully oxidised because it has +6 which is the highest oxidation state for sulphur.

You could have a look at the chemical structure of sulphuric acid to really understand it:
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The sulphur has two double bonded oxygens and two single bonded oxygens. It is fully oxidised because it is only bonded to oxygen. The 2 hydrogens in this structure don't matter because they are actually hydrogen ions which play no part in the redox reactions since they have no electrons. If you think about this reaction:
SO42- + 2H+ --> H2SO4. No redox reaction is taking place, the sulphate ion is equivalent to the sulphuric acid. You could replace the H+ with any positive ion.

However, in (C), the sulphuric acid is being reduced to hydrogen sulphide. These hydrogens are NOT just positive ions like in the above picture, they are hydrogen atoms so they reduce the sulphur because they donate their electrons to it. Hydrogen sulphide is therefore fully reduced.

Sorry if I've just confused you even more
Ohhhhh right that kinda made more sense. In a sense if an a compound contains hydrogen, hydrogen would always be +1 right
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Igniform
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(Original post by Yazomi)
Ohhhhh right that kinda made more sense. In a sense if an a compound contains hydrogen, hydrogen would always be +1 right
Hydrogen is +1 in organic chemistry but not in metal hydrides, e.g. LiH, which has H = -1. So I was wrong to say hydrogen always means reduced, but for anything organic like sulphur or carbon compounds it does.
If you see NaH, MgH2, LiH... etc. you know that H = -1 because alkali metals and alkaline earth metals always have positive oxidation states. Lithium is the least electronegative element so it's always +1, like all the alkali metals. Alkaline earth metals are always +2.

There's a table here that has the rules/exceptions you need to remember for A level:
https://www.chemguide.co.uk/inorgani...idnstates.html
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Yazomi
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(Original post by Igniform)
Hydrogen is +1 in organic chemistry but not in metal hydrides, e.g. LiH, which has H = -1. So I was wrong to say hydrogen always means reduced, but for anything organic like sulphur or carbon compounds it does.
If you see NaH, MgH2, LiH... etc. you know that H = -1 because alkali metals and alkaline earth metals always have positive oxidation states. Lithium is the least electronegative element so it's always +1, like all the alkali metals. Alkaline earth metals are always +2.

There's a table here that has the rules/exceptions you need to remember for A level:
https://www.chemguide.co.uk/inorgani...idnstates.html
Ahaaaaa got it now, that’s really helpful thank you very much!!!!
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