[aresofwar]

In the laboratory,copper(I) iodide is prepared by simply mixing an aqueous solution of sodium or potassium iodide and a soluble copper(II) salt such copper sulfate.

Here's my question: what is the role of sodium iodide in this reaction and why?

[Archer]

Half of the answer is in the question you have written!

[magician4]

another hint:

you might find it interesting to learn, that one of the steps for gaining purest iodine is the reaction of potassium iodide with copper(II)sulfate

regards

Ingo

[aresofwar]

Half of the answer is in the question you have written!

So sry I'm intellectually challenged compared to most people but I'm trying alr >_<

Is sodium iodide a reducing agent?

As for the other half of the answer, could you enlighten me? Or at least a hint on where to start...

[Archer]

Answer these questions in bullet points to help lay it out

1) What are you trying to make?

2) What are you trying to make it from?

3) What reagent are you using to make it?

[aresofwar]

1) copper 1 iodide

2) copper 2 salt

3) sodium iodide

anything i left out?

[Archer]

1) copper 1 iodide

2) copper 2 salt

3) sodium iodide

anything i left out?

Yes, you previously stated that you were making it from Copper II sulphate.

So the iodide is reducing the copper from Copper II to Copper I (as you quite rightly pointed out).

As Ingo points out this gives I₂ as the product.

So the other purpose of the sodium iodide is as a source of the Iodide counter ion of your product, making sodium sulphate.

Now you have all of the information try and write a balanced equation for the reaction.

[Archer]

http://en.wikipedia.org/wiki/Copper(I)_iodide

This has the equation which should help you to understand.

[aresofwar]

CuSO4 + 2NaI ---> Cu+ + Na2SO4 + I2
eliminating spectator ions:
Cu2+ +2I- ---> Cu+ + I2 can i write it this way?

CuI is poorly soluble in water (0.00042 g/L at 25 °C), but it dissolves in the presence of NaI or KI to give the linear anion [CuI2]−

could you explain this linear anion?

[kriggy]

It is probably coordination compound where one I^- is coordinated to CuI molecule. Linear means that the shape is like this: I-Cu-I

[Archer]

CuSO4 + 2NaI ---> Cu+ + Na2SO4 + I2
eliminating spectator ions:
Cu2+ +2I- ---> Cu+ + I2 can i write it this way?

This makes more sense if you do the ion exchange first.
i.e.
CuSO₄ + 2NaI  CuI₂ + Na₂SO₄.

Then the reduction

2CuI₂  2CuI + I₂

Then combine the two to give the balanced equation.

[aresofwar]

many thanks to the helpful posts by the community. i learnt quite a lot today! xD