Concentrated Sulfuric Acid reactions with Sodium Halides

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voltz
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I just did a past paper (AQA) and one of the questions asked for an equation for the reaction between Sodium Bromide and concentrated Sulfuric Acid.

The reaction that I wrote than did not include the redox products, as I have been taught it. I simply had Na2SO4 and HBr as the products.

This is the equation given on the mark scheme: Name:  Capture.PNG
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Is this always the overall equation you are expected to know? And could someone post the full equations for the reaction of Sodium Iodide with conc. H2SO4.
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username986184
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(Original post by voltz)
I just did a past paper (AQA) and one of the questions asked for an equation for the reaction between Sodium Bromide and concentrated Sulfuric Acid.

The reaction that I wrote than did not include the redox products, as I have been taught it. I simply had Na2SO4 and HBr as the products.

This is the equation given on the mark scheme: Name:  Capture.PNG
Views: 1393
Size:  3.5 KB

Is this always the overall equation you are expected to know? And could someone post the full equations for the reaction of Sodium Iodide with conc. H2SO4.
They're in the textbook.
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voltz
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(Original post by TeachChemistry)
They're in the textbook.
I only have the Maths skills textbook, and my notes dont actually have the full equations:/

Do you know where I can find the full equations? I have looked at chemguide but they're not on there
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username986184
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(Original post by voltz)
I only have the Maths skills textbook, and my notes dont actually have the full equations:/

Do you know where I can find the full equations? I have looked at chemguide but they're not on there
I just put your thread title into Google. The first hit was chemguide page containing exactly what you asked for.
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voltz
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(Original post by TeachChemistry)
I just put your thread title into Google. The first hit was chemguide page containing exactly what you asked for.
Im just wondering about the SO2 and S since they produced too?
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username986184
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(Original post by voltz)
Im just wondering about the SO2 and S since they produced too?
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Elemental S? Are you mixing this up with something else?
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voltz
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(Original post by TeachChemistry)
Elemental S? Are you mixing this up with something else?
Iodide reduces H2SO4 so SO2, S, and H2S if im not mistaken?
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username986184
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(Original post by voltz)
Iodide reduces H2SO4 so SO2, S, and H2S if im not mistaken?
6HI + H2SO4 --> S + 3I2 + 4H2O

Not sure if this is what you want. I just worked it out, didn't find it anywhere.
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Pigster
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(Original post by voltz)
Iodide reduces H2SO4 so SO2, S, and H2S if im not mistaken?
That may be true, but the OP concerned bromide, not iodide.

Bromide can only reduce sulfate to SO2.

Iodide is a better reducing agent, so sulfate can go to S and H2S.
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voltz
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(Original post by Pigster)
That may be true, but the OP concerned bromide, not iodide.

Bromide can only reduce sulfate to SO2.

Iodide is a better reducing agent, so sulfate can go to S and H2S.
Do you have the full equation for Iodide?
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username986184
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(Original post by voltz)
Do you have the full equation for Iodide?
I am still unclear about what you mean by the full equation.

SO42- is reduced to SO2 which is further reduced to S which is further reduced to H2S. There is no one specific equation that has all of these products in it.
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voltz
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(Original post by TeachChemistry)
I am still unclear about what you mean by the full equation.

SO42- is reduced to SO2 which is further reduced to S which is further reduced to H2S. There is no one specific equation that has all of these products in it.
The equation I attached at the start of the thread with bromide ions has the reduction products aswell, whereas I am used to simple showing Na2SO4 and HBr as the products, so am firstly unsure if you will always be expected to show the reduction products when asked for the reaction between a sodium halide and conc. sulfuric acid, and second was wondering if you know the full equation with NaI which has the redox products as well as Na2SO4 and HI.

Sorry for my lack of clarity, I hope this makes more sense
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username986184
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Want to have a go at this? I've run out of ideas.
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voltz
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(Original post by TeachChemistry)
Want to have a go at this? I've run out of ideas.
Appreciate the help
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Pigster
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(Original post by voltz)
so am firstly unsure if you will always be expected to show...
That depends on your exam board. AQA ? (assuming you're UK based).

If I'm right, they only expect ionic equations (unless things have majorly changed since I stopped teaching AQA).

(Original post by voltz)
was wondering if you know the full equation with NaI which has the redox products as well as Na2SO4 and HI.
The issue is that you cannot really show the three reduction products with iodide in one balanced equation as it would suggest the molar ratio of the three products.

You can show them as the full balanced equations below, but just about everyone uses the ionic (redox) equations. (I had to write these out as I couldn't find them to copy and paste - sorry if I've made any typos)

No redox: 2NaI + H2SO4 ->Na2SO4 + 2HI
1st reduction: 2H2SO4 + 2NaI -> Na2SO4 + SO2 + I2 + 2H2O
2nd reduction: 4H2SO4 + 6NaI -> 3Na2SO4 + S + 3I2 + 4H2O
3rd reduction: 5H2SO4 + 8NaI -> 4Na2SO4 + H2S + 4I2 + 5H2O

A rather full description of what is going on (and some other stuff): http://www.docbrown.info/page07/ASA2group7b.htm
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voltz
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(Original post by Pigster)
That depends on your exam board. AQA ? (assuming you're UK based).

If I'm right, they only expect ionic equations (unless things have majorly changed since I stopped teaching AQA).



The issue is that you cannot really show the three reduction products with iodide in one balanced equation as it would suggest the molar ratio of the three products.

You can show them as the full balanced equations below, but just about everyone uses the ionic (redox) equations. (I had to write these out as I couldn't find them to copy and paste - sorry if I've made any typos)

No redox: 2NaI + H2SO4 ->Na2SO4 + 2HI
1st reduction: 2H2SO4 + 2NaI -> Na2SO4 + SO2 + I2 + 2H2O
2nd reduction: 4H2SO4 + 6NaI -> 3Na2SO4 + S + 3I2 + 4H2O
3rd reduction: 5H2SO4 + 8NaI -> 4Na2SO4 + H2S + 4I2 + 5H2O

A rather full description of what is going on (and some other stuff): http://www.docbrown.info/page07/ASA2group7b.htm
Yes, AQA but the spec has changed recently.

I thought the same thing but the specimen paper is where I the answer included the redox products for the reaction with NaBr.

Would I just be best off writing the equation without the redox products unless asked for them?
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chem revision
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(Original post by voltz)
Yes, AQA but the spec has changed recently.

I thought the same thing but the specimen paper is where I the answer included the redox products for the reaction with NaBr.

Would I just be best off writing the equation without the redox products unless asked for them?
link for the specimen paper please
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Pigster
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(Original post by voltz)
I thought the same thing but the specimen paper is where I the answer included the redox products for the reaction with NaBr.
Your spec says: "The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid."

They want you to write equations that show the reduction products.

The trend they want you to show is that Cl- and F- do not reduce the S. Br- can reduce S to SO2 only and I- can reduce S to SO2, S and H2S. AND show these reactions as equations, either as overall equations OR as ionic equations.
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voltz
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(Original post by chem revision)
link for the specimen paper please
http://www.aqa.org.uk/subjects/scien...ment-resources

Its specimen paper 1 AS
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voltz
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(Original post by Pigster)
Your spec says: "The trend in reducing ability of the halide ions, including the reactions of solid sodium halides with concentrated sulfuric acid."

They want you to write equations that show the reduction products.

The trend they want you to show is that Cl- and F- do not reduce the S. Br- can reduce S to SO2 only and I- can reduce S to SO2, S and H2S. AND show these reactions as equations, either as overall equations OR as ionic equations.
Okay, that makes sense. Thank you
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