Copper sulfate solution and potassium iodide solution

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

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

The cloudy substance is poured through filter paper.

The filtrate is a clear orange-brown colour.

What substance is in the filtrate?

The filter paper is stained with a very dark substance and a pale precipitate.

Can you identify the dark substance staining the filter paper?

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

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

Silver nitrate solution is added.

A dark precipitate forms.

Can you identify the dark precipitate?

The liquid above the dark precipitate appears to be a very faint blue.

What do you think is colouring the liquid?

Upon addition of ammonia solution, this liquid turns royal blue.

Were you right?

Colourless cyclohexane (a non-polar solvent) is placed in a test tube.

A sample of the clear, orange-brown filtrate is taken.

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

What substance turns cyclohexane purple?

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.

I2 isn’t very soluble, so in concentrated solution it precipitates out to form the dark solid we saw on the filter paper.

2 I- → I2 + 2e-

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.

Identifying the white precipitate

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)