I mixed 5ml H202 with 5ml KI with 5ml H2SO4 wit Na2So2O3 with 50ml. Now I am trying to figure out all the elementary steps for this reaction. A starch was added in this mixture to measure the reaction time with the thiosulfate and the I2. I was just wonder if anyone could give me all the elementary steps for this reaction?

Question

I mixed 5ml H202 with 5ml KI with 5ml H2SO4 wit Na2So2O3 with 50ml. Now I am trying to figure out all the elementary steps for this reaction. A starch was added in this mixture to measure the reaction time with the thiosulfate and the I2. I was just wonder if anyone could give me all the elementary steps for this reaction?

I came up with this but it is wrong or incomplete.

H2O2+2I^-+2H^+ ---> I2 + 2H20
I2+I^- ----> I3^-
I2+2S2O3^2- ---> 2I^- + S4O6^-2
STARCH + I2 ---- BLUE

Now what steps am I missing exactly?
Any help would be appreciated.

Answer 1

To determine the elementary steps for the reaction you described, let's break down the overall reaction and identify the intermediate species involved.

1. Hydrogen peroxide (H2O2) reacts with iodide ions (I-) and hydrogen ions (H+) to form iodine (I2) and water (H2O):
H2O2 + 2I- + 2H+ → I2 + 2H2O

2. Iodine (I2) reacts with additional iodide ions (I-) to form triiodide ions (I3-):
I2 + I- → I3-

3. Iodine (I2) reacts with thiosulfate ions (S2O3^2-) to form iodide ions (I-) and tetrathionate ions (S4O6^2-):
I2 + 2S2O3^2- → 2I- + S4O6^2-

4. Lastly, the blue color change is due to the formation of a complex between iodine (I2) and starch.

Therefore, the complete set of elementary steps for the reaction you described is as follows:

1. H2O2 + 2I- + 2H+ → I2 + 2H2O
2. I2 + I- → I3-
3. I2 + 2S2O3^2- → 2I- + S4O6^2-

The addition of starch forms a blue complex with the iodine. Please note that the reaction mechanism may involve additional steps or intermediates depending on the reaction conditions, such as pH or temperature.