Ask Me Help Desk - Iodine clock chemical reaction

In our chem lab we're doing an iodine clock experiment. I need help with my equation KIO3 + NaHSO3 + starch --->? I'm pretty sure I2 is produced but I'm not sure of anything else. I need a molecular equation and a net ionic equation. Please help! Thanks

There are several variations of this reaction, but here's the most likely ex:

This can seem like a very complicated lab.

The goal is usually to try to look at how different things can affect the rate of the reaction... the most common being concentrations of reactants, temperature, and perhaps catalysts. You'll probably look at concentrations of reagents.

Now... when I ran this in my labs we were analyzing the following reaction:

(S2O8)2 - + 2I - --> I2 + 2(SO4)2 -

But the reaction is monitered by a follow-up reaction... that is

I2 + 2(S2O3)2- --> 2I- + (S4O6)2- (with a starch indicator present)

Basically... you are saying that the first reaction produces iodine (I2)... then the iodine immediately reacts with any thiosulfate (S2O3)2-, as long as it is present. Once the thiosulfate is gone, any new I2 produced will complex with the starch indicator... which shows a strong blue/black color.

So... this way you can mix the reagents, set a timer, and wait for the color to show. Then you mix different concentrations of reagents, set the timer, and again wait for the color to show.

You'll then look at the results. If you doubled the concentration of a reagent did the reaction rate change? Did it double? Triple? etc...

Quote:

Originally Posted by kp2171

there are several variations of this reaction, but heres the most likely ex:

this can seem like a very complicated lab.

the goal is usually to try to look at how different things can affect the rate of the reaction...the most common being concentrations of reactants, temperature, and perhaps catalysts. you'll probably look at concentrations of reagents.

now...when i ran this in my labs we were analyzing the following reaction:

(S2O8)2 - + 2I - --> I2 + 2(SO4)2 -

but the reaction is monitered by a followup reaction... that is

I2 + 2(S2O3)2- --> 2I- + (S4O6)2- (with a starch indicator present)

basically... you are saying that the first reaction produces iodine (I2)... then the iodine immediately reacts with any thiosulfate (S2O3)2-, as long as it is present. once the thiosulfate is gone, any new I2 produced will complex with the starch indicator.... which shows a strong blue/black color.

so.... this way you can mix the reagents, set a timer, and wait for the color to show. then you mix different concentrations of reagents, set the timer, and again wait for the color to show.

you'll then look at the results. if you doubled the concentration of a reagent did the reaction rate change? did it double? triple? etc...

All we have done in lab so far is mix I2 with the starch to see the reaction and then we mixed six different solutions with starch and got no reaction. Then we had several where we mixed two of the solutions with the starch and the only one that reacted was the KIO3 + NaHSO3 + starch. We're going to experiment with the concentrations of the solutions in the next lab. But since the solutions that reacted were KIO3 + NaHSO3 + starch, I have to come up with a net ionic equation and molecular equation for that formula. Someone once before on this website asked about balancing the formula
KIO3 + NaHSO3 ---> I2 + Na2SO4 + H2SO4 + K2SO4 + H20. So I'm wondering if this is the answer to my problem. But the equation is obviously not balanced. The balanced equation is 4 KIO3 + 10 NaHSO3 --> 2 I2 + 5 Na2SO4 + 3 H2SO4 + 2 K2SO4 + 2 H2O. Is this the answer to mine, but I just have to add starch to both sides? There didn't appear to be a precipitate in our mixture and how am I supposed to know that H2O is a product. I'm confused. I took the chemistry class six years ago and don't remember much and now I have to take this lab because I didn't take it before and it's been so long. Can you explain this to me?

Quote:

Originally Posted by Harmony138

KIO3 + NaHSO3 ---> I2 + Na2SO4 + H2SO4 + K2SO4 + H20.

The balanced equation is 4 KIO3 + 10 NaHSO3 --> 2 I2 + 5 Na2SO4 + 3 H2SO4 + 2 K2SO4 + 2 H2O.

Is this the answer to mine, but I just have to add starch to both sides?

There didn't appear to be a precipitate in our mixture and how am I supposed to know that H2O is a product.... Can you explain this to me?

well, as I mentioned before, the reactions here are a little difficult at first glance. You should know that nobody would anticipate the reactions... that is the students next to you don't intuitively know the iodate plus bisulfite gives iodide and sulfate.

first, the equation you posted is correct. At a cursory glance your balanced eq looks OK.

the starch issue can be solved often by writing it as "starch" on the left, and then "starch*I2 complex" on the right... though with the 2 coefficient in front of I2, if they want you to include starch, you'd have to say 2 starch ---> 2 complex... not sure if they'd care or not...

there is no precipitate. Not sure why you might think there would be one... but the complex that formed is the dark color you observed

you wouldn't intuitively know water was formed, and there's no visual cues. Again, the reaction steps are not intuitive. I'm surprised your lab manual didn't cover the reaction steps. I taught with a manual that we made, and we always included the reactions in the prelab discussion.

here... this breaks down the reation... its more than you need for the overall reaction, but it might help explain some of the things that are happening:

First, the iodate reacts with the bisulfite:

1) KIO3 + 3 NaHSO3 ---> KI + 3 NaHSO4

which can also be written as

IO3- + 3 HSO3- ---> I- + 3 SO42- + 3 H+

Then, some of the formed iodide reacts with more iodate to get iodine (I2)

2) 5 I- + IO3- + 6 H+ ---> 3 I2 + 3 H2O

But, the iodine will not react with starch as long as bisulfite is present. The reaction with starch is slower than the reaction with bisulfite. So the iodine will react with bisulfite to form more sulfate, until there is no more bisulfite.

3) I2 + HSO3- + H2O ---> 2 I- + SO42- + 3 H+

when there is no more bisulfite, then the I2 produced in step 2 will then complex with the starch.

4) I2 + starch ---> iodine-starch complex (blue-black)

over all, if you look back at the reactions, you can see that some of the I- produced in 1) is used in 2)... but then more I- is made again in 3). Also, iodate and bisulfate are consumed in 1) and 2) or 3)... so it gets used up. And then, as mentioned, I2 is both made and consumed in 2) and 3)... but more must get made than consumed in 3), otherwise you'd never get reaction with starch 4).

as I said. This is not a simple reaction. It has several rxns, all of which are not intuitive, and can be intimidating.

but look at the steps, see where things cancel out, and the overall rxn makes more sense.

Ah... a class where you teach yourself...

The "active learning through student-centered learning environments" craze... which when done right is just good teaching, and when done badly, is abandonment. Lovely.

I'm all for making student work some... it's the best way to really determine A students from B and from C... but this particular lab is complex enough at first that you need a little guidance, and there's more than one way to do the iodine clock, so its not like an internet search will give you a quick, easy answer necessarily.

Well... I don't do homework for people, or at least try not to, but if you have more questions like this feel free to post them. Most times ill at least get you headed in the right direction or ask you what you think the answer should be and then guide you. In this case I gave you more info straight up because the lab really isn't about digging up the reaction... its about understanding reaction rates by changing concentrations of reactants.