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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tjh on March 07, 2009, 03:17:16 PM

Title: Inorganic lab question (copper)
Post by: tjh on March 07, 2009, 03:17:16 PM
Hi,

Im trying to explain the results of the following reaction of copper(II) which was performed in lab:

Solid potassium iodate followed by sodium hypochlorite solution were added to copper(II) sulfate solution and then the mixture was acidified with dilute HCl.

A white precipitate formed and there was vigorous production of gas.

Im thinking the precipitate may be CuI but Im not sure. Any help on what reactions are occuring would much appreciated.

Thanks.
Title: Re: Inorganic lab question (copper)
Post by: Arkcon on March 07, 2009, 03:47:21 PM
That's a fair guess.  Can you confirm it with some balanced reactions, and the known chemical properties of the product.  You will probably want to have the balanced reaction, come exam time.
Title: Re: Inorganic lab question (copper)
Post by: tjh on March 07, 2009, 04:46:45 PM
I have found 2 different equations for the formation of CuI (not sure which is correct):

2 Cu2+ + 5I-   --> 2 CuI  + I3-

or

Cu2+ + 2I-   --> CuI2
2 CuI2  --> 2 CuI + I2

But how does I- form from potassium iodate and sodium hypochlorite? And also what is the gas that is produced?

Thanks
Title: Re: Inorganic lab question (copper)
Post by: Borek on March 07, 2009, 05:57:00 PM
Both reactions are generally the same, they just differ in mechanism details, and you should know that

I- + I2  ::equil:: I3-

Otherwise the effect is identical.