MydaMina
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I've been stuck on this question for a while and seem to not get my head around it...
Which substance is not produced in a redox reaction when solid sodium iodide reacts with concentrated sulfuric acid?
A) H2S
B) HI
C) SO2
D) I2
The equation is 2NaI + 2H2SO4 --> Na2SO4 + I2 + SO2 + 2H2O, but I can't seem to figure out how to get this equation, I got that it makes Na2SO4, but how does it form I2, SO2, and 2H2O? And how will I link it to the question?
The answer is B, can someone explain this in DETAIL, please?
Last edited by MydaMina; 2 years ago
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charco
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(Original post by MydaMina)
I've been stuck on this question for a while and seem to not get my head around it...
Which substance is not produced in a redox reaction when solid sodium iodide reacts with concentrated sulfuric acid?
A) H2S
B) HI
C) SO2
D) I2
The equation is 2NaI + 2H2SO4 --> Na2SO4 + I2 + SO2 + 2H2O, but I can't seem to figure out how to get this equation, I got that it makes Na2SO4, but how does it form I2, SO2, and 2H2O? And how will I link it to the question?
The answer is B, can someone explain this in DETAIL, please?
Hydrogen iodide has the hydroge in the same oxidation state (+1) as in sulfuric acid and the iodine the same (-1) as in sodium iodide, hence no oxidation or reduction.
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MydaMina
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Aha, that makes sense now, thank you! Can you also just explain the balanced equation? I really don't get how to figure it out if it was in an exam because I don't get the products formed..
(Original post by charco)
Hydrogen iodide has the hydroge in the same oxidation state (+1) as in sulfuric acid and the iodine the same (-1) as in sodium iodide, hence no oxidation or reduction.
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Rultan Safeed
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Sometimes, chemistry needs a bit, practice which helps you to memorise this equation.
I am doing my A2, and this also appears many times in A2
So i suggest you to make a chart of things you forget and important infront of your table to help you counter it.
I faced the same problem but overlooking it many times helps it.
Last edited by Rultan Safeed; 2 years ago
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MydaMina
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Yes, I knew I had to memorise this equation in some form, or even memorise the equations with the group 1 element halides with sulphuric acid.
Thank you for the tip!
(Original post by Rultan Safeed)
Sometimes, chemistry needs a bit, practice which helps you to memorise this equation.
I am doing my A2, and this also appears many times in A2
So i suggest you to make a chart of things you forget and important infront of your table to help you counter it.
I faced the same problem but overlooking it many times helps it.
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charco
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(Original post by MydaMina)
Yes, I knew I had to memorise this equation in some form, or even memorise the equations with the group 1 element halides with sulphuric acid.
Thank you for the tip!
The best way to "learn" redox equations is to learn to derive them from the two half equations. You will still need to know the products of the redox pairs, though.

sulfuric acid (when it behaves as an oxidising agent) gets reduced to sulfur dioxide. Now balance using only hydrogen ions, water molecules and electrons.

H2SO4 --> SO2

H2SO4 + 2H+ + 2e --> SO2 + 2H2O

iodide ions get oxidised to iodine

2I- --> I2 + 2e

Now, as the electrons are equal in both half-equations, you can simply add them together (when the electrons will cancel out)

H2SO4 + 2H+ + 2e --> SO2 + 2H2O
2I- --> I2 + 2e
-------------------------------------------------------------- add
H2SO4 + 2H+ + 2I- --> SO2 + 2H2O + I2

This is the balanced redox equation. If you wish to add in the spectator ions, then you have to balance with the appropriate ions.

2H2SO4 + 2KI --> SO2 + 2H2O + I2 + K2SO4
Last edited by charco; 2 years ago
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MydaMina
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I thought the equation was 8NaI(s) + 5H2SO4(l) ==> 4Na2SO4(s) + 4I2(g/s) + H2S(g) + 4H2O(l)?
Can sulfuric acid not get reduced to hydrogen sulphide in this stage?
(Original post by charco)
The best way to "learn" redox equations is to learn to derive them from the two half equations. You will still need to know the products of the redox pairs, though.

sulfuric acid (when it behaves as an oxidising agent) gets reduced to sulfur dioxide. Now balance using only hydrogen ions, water molecules and electrons.

H2SO4 --> SO2

H2SO4 + 2H+ + 2e --> SO2 + 2H2O

iodide ions get oxidised to iodine

2I- --> I2 + 2e

Now, as the electrons are equal in both half-equations, you can simply add them together (when the electrons will cancel out)

H2SO4 + 2H+ + 2e --> SO2 + 2H2O
2I- --> I2 + 2e
-------------------------------------------------------------- add
H2SO4 + 2H+ + 2I- --> SO2 + 2H2O + I2

This is the balanced redox equation. If you wish to add in the spectator ions, then you have to balance with the appropriate ions.

2H2SO4 + 2KI --> SO2 + 2H2O + I2 + K2SO4
Last edited by MydaMina; 2 years ago
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charco
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#8
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(Original post by MydaMina)
I thought the equation was 8NaI(s) + 5H2SO4(l) ==> 4Na2SO4(s) + 4I2(g/s) + H2S(g) + 4H2O(l)?
Can sulfuric acid not get reduced to hydrogen sulphide in this stage?
yes, it can get reduced further to hydrogen sulfide.

To generate this equation you go through the same process.
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MydaMina
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Aha, so it can get further reduced to H2S. Can I just use H2S as the example then or would I have to use SO2?
(Original post by charco)
yes, it can get reduced further to hydrogen sulfide.

To generate this equation you go through the same process.
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