Currently studying for an lab exam. Just have a few question about a typical Iodine Clock reaction lab.
In an typical Iodine Clock reaction lab what would be the state of effect of an increase in the initial concentration of persulfate ion (S2O3 2-) and the increase in the initial concentration of iodide ion?
If the thiosulfate ion solution was omitted what would appear in the solution?
In a typical Iodine Clock reaction lab, the effect of an increase in the initial concentration of persulfate ion (S2O3 2-) and the increase in the initial concentration of iodide ion can be understood by considering the reaction mechanism.
The Iodine Clock reaction involves the following reactions:
1. S2O3 2- + I2 → S4O6 2- + 2I-
2. I2 + 2S2O3 2- → 2I- + S4O6 2-
The first reaction occurs at a slower rate, and the second reaction occurs at a faster rate. The overall reaction is the combination of these two reactions.
Increasing the initial concentration of persulfate ion (S2O3 2-) would increase the rate of the second reaction (as it is involved in the faster rate reaction), leading to a faster overall reaction rate. This would decrease the time taken for the appearance of the blue color, which is an indicator of the completion of the reaction.
Increasing the initial concentration of iodide ion would have a similar effect. Iodide ion (I-) is involved in the first reaction, and increasing its concentration would increase the rate of the first reaction, leading to a faster overall reaction rate and a shorter time for the appearance of the blue color.
If the thiosulfate ion solution was omitted, the reaction would be incomplete. Thiosulfate ion (S2O3 2-) is required to react with the excess iodine (I2) produced in the first reaction to convert it back to iodide ion (I-). This allows the iodine to be regenerated and react with more persulfate ion (S2O3 2-). Without the thiosulfate ion, the iodine would not be converted back to iodide, and it would continue to consume the reaction components. This would hinder the progress of the reaction and prevent the appearance of the blue color. Instead, the solution would remain brownish due to the presence of unreacted iodine.