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Chemistry Paper 3: KI/ Starch question HELP!

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Hello all!

Guys, in some of the paper 3's,there is a question ( It's usually found as a section in the 3rd question), where they usually ask you to add POTASSIUM IODIDE, followed by a few drops of STARCH solution to the sample.

e.g:
http://www.xtremepapers.me/CIE/Internat ... _qp_31.pdf (PAGE 7)

http://www.xtremepapers.me/CIE/Internat ... _qp_32.pdf (PAGE 10)

Then they ask you about what type the reaction it is/ which was the oxidizing agent and so on..

Now i know that starch is used to test for the presence of iodine, but what i don't get is how do we come to the conclusion with the results we get!? :

How do we know that it's a redox reaction? How do we know that "compound X" was behaving as a reducing agent? ..etc.. They always ask you about oxidising/reducing agents, catalysts and reaction types in this section, so can someone PLEASE explain to me what is actually happening when we add those reagents (KI and starch) and how to come to the conclusion ( which is the e.g oxidising agent, etc.)!

Your help is greatly appreciated! Thank you and all the best! :)
 
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well oxidation no of I increases from -1 to 0 IF starch gives blue-black colur.....so means it has been oxidised or acting as a reducing agent
 
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Thanks, but can you please go into more detail? like what happens when we add the KI?
 
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Actually starch is an indicator of I3- ion, remember, we have to add iodine "solution" , not iodine itself. IODINE IS NOT SOLUBLE IN WATER
I2 + I-(FROM KI)= I3- (WHICH IS SOLUBLE)
So, as long as we get a blue black color, we can presume that I2 is present.

The original color of iodine solution is usually brown. The greater the concn of I3- (or I2) , the intense the solution. As the iodide ions are oxidised to iodine, the brown color starts appearing in the form of the initial yellow color

The ir shows that FA5 is Mn02
so:
Mn4+ + 2e-= Mn2+
2I- = I2 + 2e-

some of the iodine formed is either insouble (giving the blue/black color) or the iodine reacts with iodide to give a slightly golden solution(as mentioned above)
 
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