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chemistry aqa aslevel inorganic question

A solution of sodium chlorate(l) was added to a colourless solution of potassium iodide. Suggest what is observed. Explain the reaction that leads to this observation.

answer: Goes brown (or shades of brown)
Due to iodine or I3-
Because I− oxidised
can someone pls explain this i don't get it
Original post by ricecakes1
A solution of sodium chlorate(l) was added to a colourless solution of potassium iodide. Suggest what is observed. Explain the reaction that leads to this observation.
answer: Goes brown (or shades of brown)
Due to iodine or I3-
Because I− oxidised
can someone pls explain this i don't get it

OK, I do OCR A instead of AQA, but I think I know enough to answer. If you have any questions, please feel free to ask.

So the overall equation is: NaClO3 + KI --> NaCl + KIO3
(NaClO3 being Sodium Chlorate (I), KI being Potassium Iodide, NaCl being Sodium Chloridee aka table salt, KIO3 being Potassium Iodate)

So we can derive the ionic equation:
ClO3- (aq) + K+ (aq) + I- (aq) --> Cl- (aq) + KIO3 (s)

Therefore we can use oxidation numbers to see if anything has been oxidised or reduced.
Cl: Oxidation number of +6 to -1, so Chlorine has been reduced.
O: Oxidation number of -2 to -2, so nothing has happened to Oxygen.
K: Oxidation number of +1 to +1. so nothing has happened to Potassium.
I: Oxidation number of -1 to +6, so Iodine has been oxidised.

Therefore, Iodine has been oxidised (we can use the Year 7/8 method to verify this in our heads - don't write this down in exams - where the Iodine ion, I-, gains Oxygen, to form an Iodate ion, IO3-)

As Iodine has been oxidised, it forms a solid, with the Oxygen and Potassium (due to how it has reacted), forming Potassium Iodate, a precipitate, which has a brown colour due to Iodine's nature of being brown in water (which the solution is).

Hope this helps, like I said, any questions, please feel free to ask
Original post by professorkuiki
OK, I do OCR A instead of AQA, but I think I know enough to answer. If you have any questions, please feel free to ask.
So the overall equation is: NaClO3 + KI --> NaCl + KIO3
(NaClO3 being Sodium Chlorate (I), KI being Potassium Iodide, NaCl being Sodium Chloridee aka table salt, KIO3 being Potassium Iodate)
So we can derive the ionic equation:
ClO3- (aq) + K+ (aq) + I- (aq) --> Cl- (aq) + KIO3 (s)
Therefore we can use oxidation numbers to see if anything has been oxidised or reduced.
Cl: Oxidation number of +6 to -1, so Chlorine has been reduced.
O: Oxidation number of -2 to -2, so nothing has happened to Oxygen.
K: Oxidation number of +1 to +1. so nothing has happened to Potassium.
I: Oxidation number of -1 to +6, so Iodine has been oxidised.
Therefore, Iodine has been oxidised (we can use the Year 7/8 method to verify this in our heads - don't write this down in exams - where the Iodine ion, I-, gains Oxygen, to form an Iodate ion, IO3-)
As Iodine has been oxidised, it forms a solid, with the Oxygen and Potassium (due to how it has reacted), forming Potassium Iodate, a precipitate, which has a brown colour due to Iodine's nature of being brown in water (which the solution is).
Hope this helps, like I said, any questions, please feel free to ask

That’s with sodium chlorate(V) - sodium chlorate(I) is NaClO. Your equations also show the formation of potassium iodate, which would form a colourless solution (remember group 1 salts usually aren’t coloured unless the anion contains a transition metal) - the product of interest is a brown solution.

What’s actually happening is the following reaction:

2I^- (aq) + ClO^- (aq) + H2O (l) -> I2 (aq) + Cl^- (aq) + 2OH^- (aq)

Your way of explaining that iodide is oxidised, however, is pretty good (other than the fact you’ve said the oxidation states of Cl in ClO3^- and iodine in IO3^- are +6) - all you need to do now is to apply it to the correct equation.

Of course, it isn’t necessarily obvious that you should form I2 and figuring out the equation isn’t straightforward. I think you just have to remember that when you mix an oxidising agent (such as ClO^-) and a halide salt, you make a halogen. You also need to remember the colours of aqueous solutions of the halogens (I2 is brown when dissolved in water). It’s a nice question that ties together a number of different areas rather well, I think.
(edited 1 month ago)
Original post by UtterlyUseless69
That’s with sodium chlorate(V) - sodium chlorate(I) is NaClO. Your equations also show the formation of potassium iodate, which would form a colourless solution (remember group 1 salts usually aren’t coloured unless the anion contains a transition metal) - the product of interest is a brown solution.
What’s actually happening is the following reaction:
2I^- (aq) + ClO^- (aq) + H2O (l) -> I2 (aq) + Cl^- (aq) + 2OH^- (aq)
Your way of explaining that iodide is oxidised, however, is pretty good (other than the fact you’ve said the oxidation states of Cl in ClO3^- and iodine in IO3^- are +6) - all you need to do now is to apply it to the correct equation.
Of course, it isn’t necessarily obvious that you should form I2 and figuring out the equation isn’t straightforward. I think you just have to remember that when you mix an oxidising agent (such as ClO^-) and a halide salt, you make a halogen. You also need to remember the colours of aqueous solutions of the halogens (I2 is brown when dissolved in water). It’s a nice question that ties together a number of different areas rather well, I think.

Oh sorry, yeah those first mistakes probably made my explanation wrong lol. Google couldn't show me the difference between NaClO and NaClO3, so I honestly thought it was NaClO3, thanks for the correction with that, and also the product. Thanks, I did try my best lol. @ricecakes1 the explanation above seems like the answer you were looking for. Sorry again 🙂
Original post by ProfessorKuiki
Oh sorry, yeah those first mistakes probably made my explanation wrong lol. Google couldn't show me the difference between NaClO and NaClO3, so I honestly thought it was NaClO3, thanks for the correction with that, and also the product. Thanks, I did try my best lol. @ricecakes1 the explanation above seems like the answer you were looking for. Sorry again 🙂

There’s no need to apologise - It was a good try and clearly you have got a good understanding of how to approach certain types of exam questions.

It’s probably worth noting that since you worked out that the oxidation state of Cl in ClO3^- was not +1, it couldn’t possibly have been chlorate(I). The roman numerals should indicate what oxidation state the central atom is in and so this is a useful means of making sure you are talking about the correct ions and compounds in your answers.
Reply 5
Original post by ProfessorKuiki
OK, I do OCR A instead of AQA, but I think I know enough to answer. If you have any questions, please feel free to ask.
So the overall equation is: NaClO3 + KI --> NaCl + KIO3
(NaClO3 being Sodium Chlorate (I), KI being Potassium Iodide, NaCl being Sodium Chloridee aka table salt, KIO3 being Potassium Iodate)
So we can derive the ionic equation:
ClO3- (aq) + K+ (aq) + I- (aq) --> Cl- (aq) + KIO3 (s)
Therefore we can use oxidation numbers to see if anything has been oxidised or reduced.
Cl: Oxidation number of +6 to -1, so Chlorine has been reduced.
O: Oxidation number of -2 to -2, so nothing has happened to Oxygen.
K: Oxidation number of +1 to +1. so nothing has happened to Potassium.
I: Oxidation number of -1 to +6, so Iodine has been oxidised.
Therefore, Iodine has been oxidised (we can use the Year 7/8 method to verify this in our heads - don't write this down in exams - where the Iodine ion, I-, gains Oxygen, to form an Iodate ion, IO3-)
As Iodine has been oxidised, it forms a solid, with the Oxygen and Potassium (due to how it has reacted), forming Potassium Iodate, a precipitate, which has a brown colour due to Iodine's nature of being brown in water (which the solution is).
Hope this helps, like I said, any questions, please feel free to ask

thank you!!
Reply 6
Original post by UtterlyUseless69
There’s no need to apologise - It was a good try and clearly you have got a good understanding of how to approach certain types of exam questions.
It’s probably worth noting that since you worked out that the oxidation state of Cl in ClO3^- was not +1, it couldn’t possibly have been chlorate(I). The roman numerals should indicate what oxidation state the central atom is in and so this is a useful means of making sure you are talking about the correct ions and compounds in your answers.

thank u! got it

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