Mr Cole Chemistry

Iodine Clock Questions

Iodine Clock Reactions

Rate Equations Worksheet

1. Hydrogen Peroxide & Iodide

Hydrogen peroxide will oxidise iodide ions to iodine in the presence of an acid. Iodine is reduced by thiosulfate ions (S2O32-), which are oxidised to form S4O62-.
a)
Write the following equations: – Half equation for the oxidation of iodide ions. – Half equation for the reduction of hydrogen peroxide. – The ionic equation for the reaction between iodide and hydrogen peroxide.

Oxidation: 2I–(aq) → I2 (aq) + 2e–

Reduction: 2H+(aq) + H2O2 (aq) + 2e– → 2H2O (l)

Overall Ionic: 2H+(aq) + H2O2 (aq) + 2I–(aq) → 2H2O (l) + I2 (aq)

b)
State the indicator and colour for testing for iodine.

Starch; turns blue-black (or black) in the presence of iodine.

c)
Explain, giving experimental details, how you could use a series of experiments to determine the order of this reaction with respect to iodide. In each experiment you should obtain a measure of the initial rate of reaction.

Step 1: Measure out 10 cm3 of sodium iodide, hydrogen peroxide, sulfuric acid and sodium thiosulfate into separate beakers. Measure out 2 cm3 of starch solution into a separate beaker. Ensure that there are more moles of iodide than thiosulfate.

Step 2: Mix all the solutions in a single container and start the timer immediately. (Note: you can premix some solutions, provided the final trigger reactant is added at time zero).

Step 3: Time how long it takes for the blue/black colour to appear. This is the end point.

Step 4: Repeat the experiment with different concentrations of sodium iodide (by varying volume and topping up with distilled water to keep total volume constant). All other concentrations and temperature must be constant.

Step 5: Calculate the rate of reaction using 1/time.

Step 6: Calculate the new concentration of iodide ions in the mixture for each run.

Step 7: Plot a graph of Rate vs [I–]. A straight line through the origin indicates first order; a curve suggests second order.

2. Iodide & Persulfate Clock

Iodide and persulfate will react in the following reaction:
2I− + S2O82− → I2 + 2SO42−
Iodine is reduced by thiosulfate ions (S2O32-), which are oxidised to form S4O62-. Sodium iodide, potassium persulfate and sodium thiosulfate are all reacted together in a container.
a)
State what you would observe and explain your answer.

There would be no visible change for a period of time, and then the solution would suddenly change from colourless to blue/black.

Explanation: The iodine produced in the main reaction reacts immediately with the thiosulfate ions present to reform iodide. The iodine concentration remains effectively zero until all the thiosulfate is consumed. Once the thiosulfate is used up, the iodine accumulates and reacts with the starch indicator.

b)
State which substance needs to be excess out of sodium iodide and potassium thiosulfate and explain your answer.

Sodium Iodide.

If there is more thiosulfate present than iodide (or more specifically, than the limiting reagent of the main reaction can produce iodine for), you would never run out of thiosulfate. Consequently, the iodine concentration would never rise above zero, and the blue-black colour would never appear.

c)

The following results were obtained for repeating the reaction multiple times. The concentration of all the stock solutions used was 0.5 mol dm-3. Construct a graph showing the effect of concentration on rate of reaction and determine the order of reaction with respect to iodide ions.

Volume / cm3
Exp 1 Exp 2 Exp 3 Exp 4
NaI15202530
S2O82-5555
S2O32-10101010
H2O151050
Starch2222
Time / s130856655

Step 1: Calculate Concentrations and Rates

[NaI] / mol dm-3 Rate (1/t) / s-1
0.1600.00769
0.2130.01176
0.2660.01515
0.3190.01818

Step 2: Plot Graph

Graph of Rate vs Concentration for Iodide

Conclusion:
The order of reaction is First Order. As the concentration of iodide ions increases, the rate of reaction increases proportionally (straight line through the origin).

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