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Laboratory work:
Determination of the rate constant for the oxidation
of iodine by hydrogen peroxide.
Hydrogen peroxide oxidizes iodine ions in an acid medium according to the
reaction:
2I
–
+ H
2
O
2
+ 2H
+
⇄
I
2
+ 2H
2
O (1)
This reaction can take place in the presence of sodium thiosulfate also, then
the resulting iodine is rapidly reduced by the thiosulfate ion:
I
2
+ 2S
2
O
3
2–
⇄
2I
–
+ S
4
O
6
2–
(2)
As a result, reaction (1) will proceed at a constant ion concentration I
–
. In the
presence of an excess of acid in the solution, the concentration of hydrogen ions
will also be practically constant. Consequently, the reaction rate (1) will depend
only on the concentration of hydrogen peroxide, and the kinetic equation for it
can be written as:
kС
dt
dС v
= =
, where C is [H
2
O
2
].
Thus, under these conditions reaction (1) proceeds as a first-order reaction.
Work way: pour 100 ml of a 0.4% solution of potassium iodide in a 200 ml
conical flask and add 5 ml of H
2
SO
4
with a concentration of 1 mol/l. A burette is
placed in a tripod and filled with a 0.05 M solution of sodium thiosulfate. Add
1 ml of Na
2
S
2
O
3
solution from the burette to the flask with potassium iodide. Add
5 drops of the starch solution. 10 ml H
2
O
2
0.025 M is poured into a measuring
tube and this solution is quickly poured into a flask. At the moment of mixing
the solutions, a stopwatch is activated.
Observe the state of the solution in the flask and record the first time of
appearance of the blue color. Then 1 ml of sodium thiosulfate solution is quickly
added from the burettewithout stopping time, mixed (the color disappears).
Waite for the next moment the appearance of the blue color. When it appears,
the time is again recorded. This operation is repeated 5 times. Then a few drops
of 0.05 M ammonium molybdate (catalyst) are added to the flask, and the iodine
released is titrated with a solution of sodium thiosulfate until the blue color
disappears. Record the total volume of sodium thiosulfate Na
2
S
2
O
3
(
V
) spent on
the experiment. The amount of Na
2
S
2
O
3
contained in this volume of solution is
equivalent to the total amount of hydrogen peroxide.
In subsequent moments, the amount of hydrogen peroxide will be
proportional to (
V‑n
), where
n
is the number of ml of Na
2
S
2
O
3
solution added to
the corresponding moment.