1. Copper sulfate solution and potassium iodide solution

2. Blue copper sulfate solution is poured into colourless potassium iodide solution.

3. The colourless solution changes to a cloudy brown colour – like milky coffee.

4. The cloudy substance is poured through filter paper.
The filtrate is a clear orange-brown colour.

5. What substance is in the filtrate?

6. The filter paper is stained with a very dark substance and a pale precipitate.
Can you identify the dark substance staining the filter paper?

7. Rinsing the filter paper with iodide solution and water removes the dark substance, leaving behind a white precipitate.

8. A small amount of the white precipitate is collected and added to water. It does not dissolve.

9. Silver nitrate solution is added.
A dark precipitate forms.
Can you identify the dark precipitate?

10. The liquid above the dark precipitate appears to be a very faint blue.
What do you think is colouring the liquid?

11. Upon addition of ammonia solution, this liquid turns royal blue.
Were you right?

12. Colourless cyclohexane (a non-polar solvent) is placed in a test tube.
A sample of the clear, orange-brown filtrate is taken.

13. When this filtrate is mixed with the cyclohexane, the organic (top) layer turns purple.
What substance turns cyclohexane purple?

14. Colourless iodide (I-)
has been oxidised to I2, which forms the orange-brown complex I3- in iodide,
and dissolves in non-polar solvents to form a purple solution. 2
I- → I2
+ 2e-
I2 isn’t very soluble, so in concentrated solution it precipitates out to form the dark solid we saw on the filter paper.

15. Identifying the white precipitate
The reaction between silver nitrate solution and the white precipitate produced silver metal (the black precipitate),
and Cu2+, which reacted with ammonia to form the royal blue complex.

16. Cu+(aq) + I-(aq) → CuI(s)
Silver metal comes from the reduction of Ag+ to Ag, so the white precipitate must have been oxidised to form Cu2+.
Possible oxidation states for copper are 0, +1 and +2.
Copper metal is brown, not white, so the white precipitate must contain Cu+. It is copper(I) iodide, CuI (sometimes written as Cu2I2).
The Cu2+ in copper sulfate is reduced by the I- to Cu+:
Cu2+(aq) + e- → Cu+(aq)
The Cu+ combines with I- to form insoluble copper(I) iodide which is white.
Cu+(aq) + I-(aq) → CuI(s)
So the overall reaction is:
2Cu2+(aq) + 4I- (aq) → 2CuI(s) + I2(s)