Experiment 2.2 (printed pp. 12-14): effect of iodide concentration on the rate of 2I⁻ + H2O2 + 2H⁺ → I2 + 2H2O at room temperature. A fixed, small amount of Na2S2O3 and starch are present from the start; thiosulphate reduces the liberated I2 back to I⁻ as fast as it forms (I2 + 2S2O3²⁻ → S4O6²⁻ + 2I⁻), and once it is used up I2 builds up and gives the blue starch complex. The time for the blue colour to APPEAR is recorded (clock reaction); it is reproducible. Runs: 25 mL of A (25 mL 3% H2O2 + 25 mL 2.5 M H2SO4 + 5 mL starch + 195 mL water) + 25 mL of B/C/D (10 mL 0.04 M Na2S2O3 + 10/20/30 mL 0.1 M KI + water to 100 mL); only [I⁻] is varied. The manual states no rate law.
-- NCERT Chemistry Lab Manual Class 12, Unit 2, p. 12Kinetic Iodide H₂O₂
8 MCQs3 revision cards9-step worked example
Source: NCERT Practical and Analytical ChemistryOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Kinetic Iodide H₂O₂, explained for NEET
Where this lesson goes: the question the experiment answers → the two reactions → how the clock works → what goes into each flask → why only iodide changes → reading the times → precautions → eight questions and one full calculation.
The question: does the rate of the reaction between iodide ions and hydrogen peroxide change when you change the concentration of iodide, and by how much? NCERT sets this up as Experiment 2.2, "To study the effect of variation in concentration of iodide ions on the rate of reaction of iodide ions with hydrogen peroxide at room temperature" (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12).
The trap first: this experiment is not a titration. Nothing is run from a burette, no end-point is judged and no aliquots are withdrawn. You mix two solutions, start a stopwatch, and stop it when a blue colour appears. The only reading is a time. A student who remembers "starch, thiosulphate, iodine" and writes about a titre has described a different experiment.
The two reactions (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12):
2I⁻ + H₂O₂ + 2H⁺ → I₂ + 2H₂O (the reaction being timed)
I₂ + 2S₂O₃²⁻ → S₄O₆²⁻ + 2I⁻ (thiosulphate removes iodine as fast as it forms)
How the clock works, step by step:
- Hydrogen peroxide starts oxidising iodide to iodine the moment the solutions are mixed.
- A small, fixed amount of sodium thiosulphate is already in the flask. It reduces each bit of iodine back to iodide "as fast as it is formed" (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12).
- So while any thiosulphate is left, free iodine cannot build up, and the starch that is present from the start stays colourless.
- When the last thiosulphate is used up, iodine builds up quickly and forms the intense blue complex with starch.
- The blue therefore marks the moment a fixed amount of iodine has been made. NCERT notes this time is reproducible and that, because the appearance of colour is timed, the reaction is called a clock reaction (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12).
What goes into the flasks (NCERT Chemistry Lab Manual Class 12, Unit 2, page 13):
| Flask | Contents | Total |
|---|---|---|
| A | 25 mL 3% H₂O₂ + 25 mL 2.5 M H₂SO₄ + 5 mL freshly prepared starch + 195 mL water | 250 mL |
| B | 10 mL 0.04 M Na₂S₂O₃ + 10 mL 0.1 M KI + 80 mL water | 100 mL |
| C | 10 mL 0.04 M Na₂S₂O₃ + 20 mL 0.1 M KI + 70 mL water | 100 mL |
| D | 10 mL 0.04 M Na₂S₂O₃ + 30 mL 0.1 M KI + 60 mL water | 100 mL |
Each run pours 25 mL of A into flask E, then adds 25 mL of B (or C, or D) with stirring; the stopwatch starts when half of the second solution has gone in, and stops when the blue appears. Each run is repeated and the times averaged (NCERT Chemistry Lab Manual Class 12, Unit 2, pages 13–14).
What that means inside flask E (50 mL; worked out from the volumes above):
- Thiosulphate: 25 mL of B, C or D carries 0.10 mmol, the same in every run (2.0 × 10⁻³ mol L⁻¹ in E).
- Iodide: 0.25, 0.50 and 0.75 mmol for B, C and D, that is 5.0 × 10⁻³, 1.0 × 10⁻² and 1.5 × 10⁻² mol L⁻¹, in the ratio 1 : 2 : 3. This is the one thing the experiment changes.
- Iodine that must form before the blue appears: 0.10 mmol thiosulphate ÷ 2 = 0.050 mmol, which is 1.0 × 10⁻³ mol L⁻¹, the same in every run.
Why only iodide changes during the timing:
- Hydrogen peroxide: flask A holds about 22 mmol in 250 mL (taking 3% as 3 g per 100 mL), so the 25 mL of A used in a run carries about 2.2 mmol. Only 0.050 mmol reacts before the blue, about 2%, so its concentration hardly moves.
- Acid: flask A holds 62.5 mmol H₂SO₄, so 25 mL of it carries 6.25 mmol, far more than the 0.10 mmol H⁺ used.
- Iodide: every I₂ that thiosulphate reduces gives back its two I⁻, so the iodide concentration stays at its starting value until the thiosulphate runs out.
So each time measures an initial rate at a known iodide concentration, with everything else held steady. That is the initial-rate method: change one reactant's concentration, keep the others constant, and compare rates (NCERT Class 12 Chemistry Chapter 3, page 67).
Reading the times: every run makes the same 1.0 × 10⁻³ mol L⁻¹ of iodine before the blue, so
rate = 1.0 × 10⁻³ mol L⁻¹ ÷ t, and rate ∝ 1/t.
A shorter time means a faster reaction. If the rate law is rate = k[I⁻]ˣ (the other reactants being in excess), then t_B / t_C = 2ˣ and t_B / t_D = 3ˣ. The order x is whatever the times say. It is found by experiment, not read off the balanced equation, and the order with respect to a reactant is its exponent in the rate law (NCERT Class 12 Chemistry Chapter 3, page 68).
Precautions NCERT lists (NCERT Chemistry Lab Manual Class 12, Unit 2, page 14): keep the thiosulphate concentration below the potassium iodide concentration; use freshly prepared starch; use fresh hydrogen peroxide and potassium iodide; measure with the same cylinders in both sets, cleaning a cylinder before measuring a different solution; record the time the moment the blue appears.
Bridge to the questions: the first two questions fix what is measured and what thiosulphate does. The next three do the flask arithmetic. The last three read an order and a rate off the times and connect the experiment to oxidation numbers.
Can you answer these Kinetic Iodide H₂O₂ MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In NCERT's experiment on the reaction of iodide ions with hydrogen peroxide, what is recorded for each mixture?
Show answer and why every option is right or wrong
Answer: B. The stopwatch starts when half of the second solution has been added and stops when the blue colour appears; the average time is recorded for each mixture (NCERT Chemistry Lab Manual Class 12, Unit 2, pages 13–14).
Why A is wrong: A is wrong because nothing is titrated in this experiment: the thiosulphate is added in a fixed amount before the reaction starts, not run from a burette (trap: describing an iodometric titration instead of the clock reaction).
Why C is wrong: C is wrong because the colour appears at the end of the timed interval; it does not start blue and fade. Blue appears only once the thiosulphate is used up.
Why D is wrong: D is wrong because no iodine is weighed; the amount of iodine made before the blue is fixed by the thiosulphate, and it is the time that varies.
MCQ 2Easy RecallPractice
What is the job of the sodium thiosulphate placed in flasks B, C and D?
Show answer and why every option is right or wrong
Answer: C. Thiosulphate reduces the liberated iodine back to iodide as fast as it is formed until all of it is oxidised to tetrathionate; only then does iodine build up and give the blue colour with starch (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12).
Why A is wrong: A is wrong because starch is the indicator; it is added to flask A and forms the blue complex with iodine.
Why B is wrong: B is wrong because hydrogen peroxide is the oxidant; thiosulphate is a reductant, converted to tetrathionate (S₄O₆²⁻).
Why D is wrong: D is wrong because the H⁺ comes from the 2.5 M sulphuric acid in flask A.
MCQ 3Direct ApplicationPractice
Flask C contains 10 mL of 0.04 M Na₂S₂O₃, 20 mL of 0.1 M KI and 70 mL of water. After 25 mL of flask C is mixed with 25 mL of flask A, the iodide concentration in the reaction flask is:
Show answer and why every option is right or wrong
Answer: A. In flask C, 20 mL of 0.1 M KI is diluted to 100 mL: 0.020 mol L⁻¹. 25 mL of it carries 0.50 mmol, and mixing with 25 mL of flask A makes 50 mL, so [I⁻] = 0.50 mmol ÷ 50 mL = 1.0 × 10⁻² mol L⁻¹. The volumes are NCERT's (NCERT Chemistry Lab Manual Class 12, Unit 2, page 13).
Why B is wrong: B is wrong because 2.0 × 10⁻² mol L⁻¹ is the concentration inside flask C; it forgets that mixing with an equal volume of flask A halves it.
Why C is wrong: C is wrong because 5.0 × 10⁻³ mol L⁻¹ is the iodide concentration of the flask B run (10 mL of KI), not flask C.
Why D is wrong: D is wrong because 0.10 mol L⁻¹ is the stock KI solution before any dilution.
MCQ 4Direct ApplicationPractice
In every run the reaction flask contains 0.10 mmol of thiosulphate. How much iodine must hydrogen peroxide produce before the blue colour can appear?
Show answer and why every option is right or wrong
Answer: D. I₂ + 2S₂O₃²⁻ → S₄O₆²⁻ + 2I⁻, so 1 mol of iodine uses 2 mol of thiosulphate. 0.10 mmol thiosulphate removes 0.050 mmol iodine; only the iodine made after that stays free and turns starch blue (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12).
Why A is wrong: A is wrong because it takes the iodine–thiosulphate ratio as 1 : 1; the equation needs two thiosulphate ions per I₂.
Why B is wrong: B is wrong because it multiplies by 2 instead of dividing: each I₂ consumes two S₂O₃²⁻, so the iodine is half the thiosulphate, not double.
Why C is wrong: C is wrong because it applies the factor of 2 twice (once for the two iodide ions in the first equation, which do not enter this step).
MCQ 5Direct ApplicationPractice
With H₂O₂ and acid in large excess, the blue colour appears after 120 s in the flask B run and after 60 s in the flask C run, where [I⁻] is twice that of flask B. The order of the reaction with respect to iodide is:
Show answer and why every option is right or wrong
Answer: B. Both runs make the same amount of iodine before the blue, so rate ∝ 1/t. Halving the time doubles the rate. With rate = k[I⁻]ˣ, 2ˣ = 120/60 = 2, so x = 1. The order is the exponent in the rate law, found by experiment (NCERT Class 12 Chemistry Chapter 3, page 68).
Why A is wrong: A is wrong because zero order would give the same time in both runs; here the time changed.
Why C is wrong: C is wrong because second order would make the rate four times faster when [I⁻] doubles, so the time would fall to 30 s, not 60 s.
Why D is wrong: D is wrong because half order would raise the rate by √2 ≈ 1.41, giving about 85 s, not 60 s.
MCQ 6Concept TrapPractice
During the timed interval of each run, the iodide ion concentration stays essentially at its starting value. Why?
Show answer and why every option is right or wrong
Answer: A. Iodide is oxidised all through the run, but thiosulphate reduces each I₂ back to 2I⁻ as fast as it forms, so the iodide concentration is restored until the thiosulphate runs out (NCERT Chemistry Lab Manual Class 12, Unit 2, page 12). That is why each time measures the rate at a known [I⁻].
Why B is wrong: B is wrong because the reverse holds: about 2.2 mmol H₂O₂ reaches the flask, against 0.25 to 0.75 mmol iodide. It is hydrogen peroxide that is in excess.
Why C is wrong: C is wrong because the oxidation of iodide by hydrogen peroxide runs from the moment of mixing; thiosulphate only removes the iodine it makes.
Why D is wrong: D is wrong because starch forms its blue complex with iodine (I₂), not with iodide, and it does not stop oxidation.
MCQ 7CalculationPractice
In the flask B run the reaction volume is 50 mL, 0.10 mmol of thiosulphate is present, and the blue appears after 125 s. The average rate of formation of iodine over this interval is:
Show answer and why every option is right or wrong
Answer: C. Iodine formed = 0.10 mmol ÷ 2 = 0.050 mmol (I₂ + 2S₂O₃²⁻, NCERT Chemistry Lab Manual Class 12, Unit 2, page 12). In 50 mL that is 1.0 × 10⁻³ mol L⁻¹. Rate = 1.0 × 10⁻³ ÷ 125 = 8.0 × 10⁻⁶ mol L⁻¹ s⁻¹.
Why A is wrong: A is wrong because it divides the thiosulphate concentration (2.0 × 10⁻³ mol L⁻¹) by the time, skipping the 2 : 1 ratio between thiosulphate and iodine.
Why B is wrong: B is wrong because it divides the iodide concentration (5.0 × 10⁻³ mol L⁻¹) by the time; iodide is recycled and is not the quantity the blue colour marks.
Why D is wrong: D is wrong because it halves the answer a second time, as if two thiosulphate ions were needed per iodine atom rather than per I₂ molecule.
MCQ 8Easy RecallPractice
Thiosulphate ends up as the tetrathionate ion, S₄O₆²⁻. The oxidation number of sulphur in tetrathionate is:
Show answer and why every option is right or wrong
Answer: D. 4x + 6(−2) = −2, so x = +2.5. NCERT gives Na₂S₄O₆ as an example of a fractional oxidation number, an average over sulphur atoms in different states (NCERT Class 11 Chemistry Chapter 7, page 245); the lab manual asks exactly this as a discussion question (NCERT Chemistry Lab Manual Class 12, Unit 2, page 14).
Why A is wrong: A is wrong because +2 is sulphur's oxidation number in thiosulphate, S₂O₃²⁻, before the reaction.
Why B is wrong: B is wrong because +4 is sulphur's oxidation number in sulphite or SO₂, which does not take part in this experiment.
Why C is wrong: C is wrong because +6 is sulphur's oxidation number in sulphate, from the sulphuric acid, which is unchanged.
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Kinetic Iodide H₂O₂: quick recall before you leave
How do you solve a Kinetic Iodide H₂O₂ question? A worked example
- 1
Given
NCERT's three mixtures (25 mL of flask A + 25 mL of flask B, C or D; 50 mL in all). The volumes are from NCERT Chemistry Lab Manual Class 12, Unit 2, page 13. The times are illustrative average readings a student might record, not NCERT data: B 124 s, C 63 s, D 41 s.
- 2
Required
The rate of iodine formation in each run, and the order of the reaction with respect to iodide.
- 3
Concept
Each run contains 0.10 mmol thiosulphate, so each must make 0.050 mmol I₂ (1.0 × 10⁻³ mol L⁻¹ in 50 mL) before the blue appears. Hydrogen peroxide and acid are in large excess and iodide is recycled, so each time measures the rate at a fixed [I⁻] of 5.0 × 10⁻³, 1.0 × 10⁻² and 1.5 × 10⁻² mol L⁻¹.
- 4
Formula
rate = Δ[I₂] ÷ t; and for two runs, t₁ ÷ t₂ = (c₂ ÷ c₁)ˣ
- 5
Substitution
rate(B) = 1.0 × 10⁻³ ÷ 124; rate(C) = 1.0 × 10⁻³ ÷ 63; rate(D) = 1.0 × 10⁻³ ÷ 41
B against C: 124 ÷ 63 = 2ˣ; B against D: 124 ÷ 41 = 3ˣ - 6
Calculation
rate(B) = 8.1 × 10⁻⁶, rate(C) = 1.6 × 10⁻⁵, rate(D) = 2.4 × 10⁻⁵ mol L⁻¹ s⁻¹
124 ÷ 63 = 1.97, so x = log 1.97 ÷ log 2 = 0.294 ÷ 0.301 = 0.98
124 ÷ 41 = 3.02, so x = log 3.02 ÷ log 3 = 0.480 ÷ 0.477 = 1.01
Note on exact constants: the 2 in the thiosulphate–iodine ratio and the concentration ratios 2 and 3 are exact (from the equation and from 10, 20 and 30 mL of the same KI solution) and do not limit significant figures. - 7
Final answer
Rates 8.1 × 10⁻⁶, 1.6 × 10⁻⁵ and 2.4 × 10⁻⁵ mol L⁻¹ s⁻¹ (2 significant figures, set by the times). Both comparisons give x ≈ 1, so the reaction is first order in iodide under these conditions: doubling [I⁻] doubles the rate, tripling it triples the rate.
- 8
Common trap
Reading a longer time as a faster reaction, or comparing the times directly with the concentrations (124 : 63 : 41 against 1 : 2 : 3) and concluding "no pattern". Convert each time to a rate first: 1/124 : 1/63 : 1/41 ≈ 1 : 2 : 3, which matches the iodide ratio.
- 9
Similar NEET-style question
"In a clock reaction the blue colour appears after 90 s at an iodide concentration c and after 45 s at 2c, all else unchanged. What is the order with respect to iodide, and after how long would the colour appear at 3c?"
(Answer: 90 ÷ 45 = 2 = 2ˣ, so x = 1; at 3c the rate is three times that at c, so t = 90 ÷ 3 = 30 s.)
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What to remember before solving Kinetic Iodide H₂O₂ questions
More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Kinetic Iodide H₂O₂ questions from past NEET papers
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Sources
NCERT refs: Class 12 Chemistry Chapter 3, p.67 | Class 12 Chemistry Chapter 3, p.68
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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