copper sulfate solution and potassium iodide solution
Post on 28-Jan-2016
Embed Size (px)
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)