[Seany (OP)]

Hi this is Eunice, Sean's sister, My AS level Chemistry is next week and im stuck, help me out please

2004 Jan, Module 2
3e) describe two observations that you would make when concentrated sulphuric acid is added to solid sodium iodide. Write an equation for a reaction that occurs in which iodide ions are oxidised by the sulphuric acid.

NaI + H2SO4 --> HI + NaHSO4

2HI + H2SO4 --> I2 + SO2 + 2H2O

6HI + H2SO4 --> 3I2 + S + 4H2O

8HI + H2SO4 --> 4I2 + H2S + 4H2O

I understand the above, but

The answer is:

2NaI + 2H2SO4 --> HI + Na2SO4 + I2 + SO2 + 2H2O

I’m not sure how you get Na2SO4

Thanks

[Cut Chemist]

Jeeze, that's an intense test question!!!

are you sure you didn't leave one of the reactions out??

Since H2SO4 is a diprotic acid so it will react twice.  So

NaI + H2SO4 --> HI + NaHSO4

and then

NaI + NaHSO4 --> HI + Na2SO4

The rest of the problem is all about balancing redox reactions
(see wiki if confused...)

-figure out half reactions for each sub-reaction
-find out which species are oxidized and which are reduced
-balance with H+ and then with electrons
-cancel out the electrons when you add the equations together

and after all of that work you should get the bottom equation.

[Cut Chemist]

After thinking about it for a minute,

I don't think you need all of that redox balancing

with that one extra equation
you can just add and subtract equations to reach the one at the bottom.  

That makes it much, much easier.

[Seany (OP)]

Ermm.. Sorry I'm lost!!

Could you write out the WHOLE equation please?? I know I'm asking alot.. But.. []

[Cut Chemist]

     NaI + H2SO4 --> HI + NaHSO4
+    NaI + NaHSO4 --> HI + Na2SO4
-----------------------------------
2NaI + H2SO4 + NaHSO4 --> 2HI + Na2SO4 + NaHSO4 

(NaHSO4 is on both sides of the reaction so it goes away...leaving this)

2NaI + H2SO4 --> 2HI + Na2SO4

then,

     2NaI + H2SO4 --> 2HI + Na2SO4
+    2HI + H2SO4 --> I2 + SO2 + 2H2O
-------------------------------------

2NaI + 2H2SO4 + 2HI --> I2 + SO2 + 2H2O + Na2SO4 + 2HI

(The 2HI's on both sides also go away...leaving this)


2NaI + 2H2SO4 --> Na2SO4 + I2 + SO2 + 2H2O
This might be the answer

But definately not this...
2NaI + 2H2SO4 --> HI + Na2SO4 + I2 + SO2 + 2H2O
The equation is not balanced
(There are too many I's on the product side)

[Seany (OP)]

Thanks! []

[lightarrow]

3e) describe two observations that you would make when concentrated sulphuric acid is added to solid sodium iodide.

One observation could be that, at the beginning, when NaI is in excess with respect to H₂SO₄, only HI should form from the reaction, and when the amount of H₂SO₄ increases, adding it drop by drop, HI then should react again with it forming I₂, SO₂ and H₂O, so you should observe a change in colour of the mix, from white to yellow-brown, during the addition of the acid.
I ask for confirm of this.

[DrDick]

That sounds about right.

You need relatively high concentrations of H₂SO₄ and I^– before you start getting a redox reaction.  Under standard conditions (1 mol/L each of sulfuric acid and iodide), the redox reaction is not spontaneous.