Yellow iodine appears after just a few minutes.
We next add starch and titrate with standardized thiosulfate solution.
Here we are getting ready for the reaction between
and KI(aq). (Tere page 13) The ozone generator utilizes Pt and
and 3 M H2SO4(aq)
in a Beral pipette. The stem of the pipet is drawn out to make a
delivery tube about 30 cm in length. Ozone generator is cooled in an
bath. Here you can see that we are testing for bubbles while
the air from the generator.
Here moist starch-iodide indicator paper detects
A view of the bubbles being generated.
We added a drop starch to the solution to detect
We then titrate with thiosulfate until the blue disappears. We
every 10 minutes and could do so without stopping the ozone
As you can see here, the ozone generator is still delivering ozone to
Tere has repeated the experiment for a period of 2
and the rate of ozone generation remains constant. As you can see
from the calculations below, we are making 785 nanomoles ozone per
and the gas produced is 0.25% O3
and (presumably) 99.75% O2:
6 H+ + IO3- + 6 S2O3-2 ---> I- + 3 S4O6-2 + 3 H2O
The titration (until the blue colour of I3-/starch disappeared) took 49.0 mL of the sodium thiosulfate solution. Given that 4.13 x 10-5 mol KIO3 was used and the above reaction requires 6 mol thiosulfate for each 1 mol iodate, the 49.0 mL of sodium thiosulfate solution contained 6 x 4.13 x 10-5 mol = 2.48 x 10-5 mol sodium thiosulfate. Thus, the concentration of the sodium thiosulfate solution is calculated to be 5.06 x 10-3 M.
We also prepared a dilute solution that was 5.06 x 10-4 M for use in some experiements.
Here are Tere's results with the Pb/C electrodes. (Tere page 16) Her rate of ozone generation is 14% larger than with the Pt electrode (897 nanomoles ozone per minute). Given that the rate of gas generation was 9.0 mL/minute, the percent ozone was about the same as with Pt. 0.24%.
The value at t = 90 s was non-linear because the blue turns to a funny green before it goes clear, so that value (0.099) was not included in the following graph:
With this we used the equation
determine that A = 0 at t = 89.4 s. Thus, the 1.57 x 10-6
mol PIT (6.29 x 10-5
mol/L X 0.025 L) reacts completely after 89.4 s. To relate this
the moles of ozone generated in this amount of time, we used the same
and same conditions to react ozone with excess KI for the same period
time (90 s) while titrating with thiosulfate. The titration was
with 11.8 mL of 5.06 x 10-4
M sodium thiosulfate solution, or 5.97 x 10-6
According the the reaction stoichiometry (earlier), 1 mol ozone
to 2 mol S2O3-2
that we have produced 2.99 x 10-6 mol O3
in 90 s. These numbers allow us to indirectly compare moles
ozone produced in 90 s with moles of PIT reacted in 90 s. The number of
moles of ozone is about twice the number of moles of PIT. (It
is known with indigo reactions that these are often nonstoichiometric.)
Results of 24 October 2002
The following results were obtained by Scot Eskestrand on 24 october 2002. (He has been doing this reaction for a few weeks and these are his best data.) The slope of the line gives
vol thiosulfate solution used = 0.6495t, where volume is in mL and t is in minutes
Converting volume into L gives:
vol thiosulfate solution used = 6.495 x 10-4t, where volume is in L and t is in minutes
Given that the molar concentration of thiosulfate is 5.06 x 10-3 mol/L, and that moles = molarity X volume, we get:
moles = molarity X volume
moles = (5.06 x 10-3 mol/L) X (6.495 x 10-4t) where t is in minutes.
Therefore, we are using 3.29 x 10-6 mol thiosulfate per minute. From the equations above (on my handwritten sheet), we know that 2 moles thiosulfate corresponds to 1 mol I2 and that 1 mole I2 corresponds to 1 mole O3, therefore: 2 moles thiosulfate corresponds to 1 mole O3
The amount of ozone generated is 1.65 x 10-6 mol O3 per minute. This value is about twice what Tere and I got last summer. The flow of total gas is also larger: 10.5 mL/minute. This corresponds to 4.29 x 10-4 mol gases/min. As a percentage, ozone represents:
Percent Ozone = 100% X (1.65 x 10-6 mol O3 per minute)/(4.29 x 10-4 mol gases/min)
= 0.38 % O3 (0.38 mol O3 per 100 mol total (H2 + O2 + O3))