Question

What happens when KI solution is added to copper sulphate solution

Answer 1

KI + CuSO4 →K2SO4 + CuI2

People thought of the above reaction which is wrong, CuI2 decomposes immediately to CuI, the reaction is:

2CuSO4 + 4KI → 2CuI + I2 + 2K2SO4
This is not a simple reaction but a mixture of processes taking place in the same reaction vessel. If this is a test tube style reaction or it is happening in a beaker where chemicals are simply mixed together, then the quantities will not be exact. Left over reagents can account for residual colours.

Yellow / brown - Iodine

Blue - hydrated Cu^(2+)

Green - Both of the above (blue + yellow = green)

First: The reduction of Cu^(2+) to Cu^(+)

Cu^(2+) + 2 I^(-) → 2Cu^(+) + I2 = redox reaction

…Reduction of Cu^(2+) to Cu^(+).

…Oxidation of iodide to iodine.

Secondly: pptn of CuI

Cu^(+) + I^(-) → CuI (white ppt)

NB the ppt actually appears a yucky brown sludge because of the iodine. Iodine is sparingly (if at all) soluble in water but it does dissole in the presence of I^(-) ions to form a dark yellow / brown solution due to the formation of the comples ion. The is almost certainly some KI left to achieve this…

Thirdly: the dissolving of iodine:

I2 + I^(-) → I3^(-) in solution. Effectively this behaves as iodine solution and this is how iodine is dissolved for titration purposes.

Fourth: (Extension)

If thiosulfate solution is used, then the Iodine is re-oxidised back to Iodide and the yucky brown dissappears leaving the white ppt of CuI

I2 + 2 S2O3^(2-) → 2 I^(-) + S2O6^(2-)

…Reduction of iodine to iodide

…Oxidation of sulfate (VI) / thiosulfate to persulfate

All of the processes 1, 2 and 3 are taking place simultaneously. The thiosulfate is just an added part to help illustrate the chemistry and, if necessary, help to obtain a clean sample of CuI.

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Rahul Karda, former Student

Answered Feb 13

Continue Reading

Copper ions are reduced by the iodide ions to copper(I). The best way to look at this is via the half equations:

Cu2+ + 1e --> Cu+ 
2I- --> I2 + 2e

to balance these you double the first equation (to equalise the electrons) and add them together.

2Cu2+ + 2e --> 2Cu+ 
2I- --> I2 + 2e
------------------------------------------------------------- add
2Cu2+ + 2I- --> 2Cu+ + I2

Now add in the spectator ions:

2CuSO4 + 2KI --> Cu2SO4 + I2 + K2SO4

So the white precipitate is copper(I) sulphate and the brownish coloration is iodine solution...

Hope it helps :p

Upvotes are appreciated :)

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Anthony King, Head of Chemistry department (2008-present)

Answered May 9, 2017

It's a .redox reaction and not precipitation reaction or double decomposition reaction because it involves oxidising agent like cu2+ and 2I- .Cu2+ becomes reduced to CU+(blue to a white ppt whilst 2I- oxidises to I2 ie from colourless to yellow or brown.

2cuSO4+ 4KI———-^22CUI+ I2 +k2SO4

Cu(II) reacts with iodide to form CuI2, which immediately reacts to CuI and elemental iodine, thus resulting in a colourless precipitate and a brown solution (if excess iodide to form I3– is present).

Answer 2

CuSo4 + Kl ---- Cul2 + K2So4