Rate of Reaction (Iodine Clock) Experiment

May 2013
12
0
Hello :)

Can somebody please provide me with some assistance to this question? I have attached an image to verify what I mean, but for our Iodine Clock practical we have to design a procedure similar to Part A for Part B but Iodide ions are now the variable and Peroxydisulfate is the constant. What would the steps be? I assume it is very similar to Part A, but steps 2 and 3 are throwing me off...would Step 2 be add 10mL of Peroxydisulfate and Step 3 be to add 10mL of Thiosulfate (S2O3^2-) before adding Iodide ions in step 5 as that is the concentration variable?? I don't understand why adding Peroxydisulfate and Thiosulfate ions would even do anything before adding the Iodide ions anyway?

If somebody please could verify the Part B procedure that would be fantastic as I'm struggling for such an easy task!

Thanks
 

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Unknown008

CHF Hall of Fame
May 2010
1,658
37
Mauritius
Yes, you can add the iodine at step 5 instead of at step 2 and proceed similarly to Part A. This will work because S2O3^2- doesn't react with S2O8^2-. Otherwise, it wouldn't be working.

This is because the 'starting' reaction (the reaction with with everything starts) is between S2O8^2- and I^-.

Otherwise, to be safer or if you are not sure what reactants react with which, just carry out Part A, but with Step 2 as adding different concentrations of I^- and step 5 as adding a constant concentration of S2O8^2-.
 
May 2013
12
0
Thanks very much Unknown008, I knew I had the correct procedure but I just keep doubting myself as I wasn't sure what the point of adding S2O3^2- and S2O8^2- in separate steps was or if I was to leave a chemical out altogether...

Thanks again :D
 

Unknown008

CHF Hall of Fame
May 2010
1,658
37
Mauritius
You're welcome :)