copper sulfate solution and potassium iodide solution

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  • 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 isnt very soluble, so in concentrated solution it precipitates out to form the dark solid we saw on the filter paper.2I- 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)

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