Question

The iodine clock experiment consists of the following three reactions, where Reaction 1 is slow relative to Reactions 2 and 3. Predict what would happen if Reaction 2 was slow relative to Reaction 1
Reaction 1:3I^- (aq) + S_2O_8^2- (aq) rightarrow I_3^-, (aq) + 2SO_4^2- (aq) slow
Reaction 2: l_3^-(aq) + 2S_2O_3^2- (aq) rightarrow 3I^- (aq) + S_4O_6^2- (aq) fast
Reaction 3: l_3^- (aq) + starch (aq) rightarrow 3I^---- starch (bluish black) fast
A. A color change would appear immediately because I_3^+, would be removed slowly from solution, and would therefore be able to react with starch.
B. The reaction would turn blue slower because I_3^-, is removed at a slower rate.
C. The rate of color change would indicate the rate of reaction 3 instead of reaction 1
D. Reaction 3 would not occur.

Answer 1

Answer: A color change would appear immediately because $I_3^+$, would be removed slowly from solution, and would therefore be able to react with starch, is the correct answer.

Explanation:

As the given reaction 2 is as follows.

$l_3^-(aq) + 2S_2O_3^2- (aq) \rightarrow 3I^- (aq) + S_4O_6^2-(aq)$

If this reaction is slow then the removal of $I_{3}^{-}$ will be slow in nature.

Since, reaction 3 is fast which is as follows.

$l_3^- (aq) + \text{starch}(aq) \rightarrow 3I^---- starch \text{(bluish black)}$

As reaction 3 is fast in nature then more amount of starch will react with $I_{3}^{-}$.

Therefore, we can conclude that a color change would appear immediately because $I_3^+$, would be removed slowly from solution, and would therefore be able to react with starch.

Answer 2

Iodine clock demonstration is used to display the chemical kinetics in a chemical reaction. It was discovered by Heinrich Landolt in the year 1886.

The correct answer is:

Option A. A color change would appear immediately because$I_3^+,$would be removed slowly from the solution, and would therefore be able to react with starch.

The reaction for the iodine clock experiment can be given as:

$\text {l}_3^-(aq) + 2S_2O_3^2- (aq) \rightarrow 3I^- (aq) + S_4O_6^{2-}$

Since, it is given that if the reaction is slow, then the iodine ions will be removed slowly.

Also, we know the third reaction is fast, in which

I₃⁻ + starch $\rightarrow$ 3 I ⁻

The reaction between iodine with starch will result in the fast removal of ions, giving bluish-black color.

Therefore, from the given reaction it can be concluded that color change would be seen immediately as the iodine ions will be removed slowly and will, therefore, react with the starch, imparting blue-black color.

Thus, option A is correct.

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