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NVAMediumJEE 2026Oxidation Number & Redox Reactions

JEE Chemistry 2026 Question with Solution

500 mL500 \, \text{mL} of 1.2 M1.2 \, \text{M} KI solution is mixed with 500 mL500 \, \text{mL} of 0.2 M0.2 \, \text{M} KMnO4\mathrm{KMnO_4} solution in basic medium. The liberated iodine is titrated with standard 0.1 M0.1 \, \text{M} Na2S2O3\mathrm{Na_2S_2O_3} solution in the presence of starch indicator till the blue colour disappears. The volume (in L) of Na2S2O3\mathrm{Na_2S_2O_3} consumed is _____ (nearest integer).

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: 500 mL500 \, \text{mL} of 1.2 M1.2 \, \text{M} KI, 500 mL500 \, \text{mL} of 0.2 M0.2 \, \text{M} KMnO4\mathrm{KMnO_4} in basic medium, and titration with 0.1 M0.1 \, \text{M} Na2S2O3\mathrm{Na_2S_2O_3}.

Find: The volume of Na2S2O3\mathrm{Na_2S_2O_3} consumed.

In basic medium, permanganate oxidises iodide to iodine:

2MnO4−+6I−+4H2O⟶2MnO2+3I2+8OH−2\mathrm{MnO_4^-} + 6\mathrm{I^-} + 4\mathrm{H_2O} \longrightarrow 2\mathrm{MnO_2} + 3\mathrm{I_2} + 8\mathrm{OH^-}

So, 22 mol KMnO4\mathrm{KMnO_4} produce 33 mol I2\mathrm{I_2}.

Moles of KI:

n(KI)=0.5×1.2=0.6 moln(\mathrm{KI}) = 0.5 \times 1.2 = 0.6 \, \text{mol}

Moles of KMnO4\mathrm{KMnO_4}:

n(KMnO4)=0.5×0.2=0.1 moln(\mathrm{KMnO_4}) = 0.5 \times 0.2 = 0.1 \, \text{mol}

From the balanced reaction, 11 mol KMnO4\mathrm{KMnO_4} requires 33 mol I−\mathrm{I^-}. Therefore, 0.10.1 mol KMnO4\mathrm{KMnO_4} requires:

0.1×3=0.3 mol0.1 \times 3 = 0.3 \, \text{mol}

Available iodide is 0.6 mol0.6 \, \text{mol}, so KMnO4\mathrm{KMnO_4} is the limiting reagent.

Moles of iodine liberated:

0.1×32=0.15 mol0.1 \times \frac{3}{2} = 0.15 \, \text{mol}

Now iodine is titrated with thiosulphate:

I2+2S2O32−→2I−+S4O62−\mathrm{I_2} + 2\mathrm{S_2O_3^{2-}} \rightarrow 2\mathrm{I^-} + \mathrm{S_4O_6^{2-}}

Thus, 11 mol I2\mathrm{I_2} reacts with 22 mol Na2S2O3\mathrm{Na_2S_2O_3}.

Moles of Na2S2O3\mathrm{Na_2S_2O_3} required:

2×0.15=0.30 mol2 \times 0.15 = 0.30 \, \text{mol}

Using M=nVM = \frac{n}{V},

V=nM=0.300.1=3.0 LV = \frac{n}{M} = \frac{0.30}{0.1} = 3.0 \, \text{L}

Therefore, the volume of Na2S2O3\mathrm{Na_2S_2O_3} consumed is 3 L3 \, \text{L}.

Mole Ratio Shortcut

Given: 0.1 mol0.1 \, \text{mol} of KMnO4\mathrm{KMnO_4} and excess iodide.

Find: Volume of 0.1 M0.1 \, \text{M} Na2S2O3\mathrm{Na_2S_2O_3} needed.

Use the two stoichiometric links directly:

  • 2KMnO4→3I22\mathrm{KMnO_4} \rightarrow 3\mathrm{I_2}
  • 1I2→2Na2S2O31\mathrm{I_2} \rightarrow 2\mathrm{Na_2S_2O_3}

Combining them:

2KMnO4→6Na2S2O32\mathrm{KMnO_4} \rightarrow 6\mathrm{Na_2S_2O_3}

So,

1KMnO4→3Na2S2O31\mathrm{KMnO_4} \rightarrow 3\mathrm{Na_2S_2O_3}

Since iodide is in excess, 0.1 mol0.1 \, \text{mol} KMnO4\mathrm{KMnO_4} gives:

0.1 \times 3 = 0.30 \, \text{mol Na_2S_2O_3}

Now,

V=0.300.1=3.0 LV = \frac{0.30}{0.1} = 3.0 \, \text{L}

Therefore, the correct numerical answer is 33.

Common mistakes

  • Using KI as the limiting reagent is incorrect because the balanced reaction requires only 0.3 mol0.3 \, \text{mol} iodide for 0.1 mol0.1 \, \text{mol} KMnO4\mathrm{KMnO_4}, while 0.6 mol0.6 \, \text{mol} is available. First compare required and available moles before deciding the limiting reagent.

  • Taking the KMnO4:I2\mathrm{KMnO_4} : \mathrm{I_2} ratio as 1:11:1 is wrong. In basic medium, 2KMnO42\mathrm{KMnO_4} produce 3I23\mathrm{I_2}. Always use the balanced redox equation in the specified medium.

  • Using a 1:11:1 ratio between I2\mathrm{I_2} and Na2S2O3\mathrm{Na_2S_2O_3} gives an incorrect titre. The correct titration reaction is I2+2S2O32−→2I−+S4O62−\mathrm{I_2} + 2\mathrm{S_2O_3^{2-}} \rightarrow 2\mathrm{I^-} + \mathrm{S_4O_6^{2-}}, so iodine requires twice as many moles of thiosulphate.

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