JEE Main 2026 Apr 6 Shift 2, Chemistry Q23: Advance Stoichiometry Based on Redox Reactions

JEE Main2026Apr 6, Shift 2Chemistry

500 mL of 0.2 M MnO $_{4}^{-}$ solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard $x$ M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of $x$ is _____.

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Solution

In basic medium MnO $_{4}^{-}$ is reduced to MnO $_{2}$ : 3-electron change per Mn.

Gram equivalents of MnO $_{4}^{-}$ = $0.2 \times 0.500 \times 3 = 0.30$ .

Gram equivalents of I $^{-}$ available = $1.5 \times 0.500 \times 1 = 0.75$ ; MnO $_{4}^{-}$ is the limiting reagent.

Equivalents of I $_{2}$ liberated $=$ equivalents of MnO $_{4}^{-}$ consumed $= 0.30$ .

I $_{2}$ + 2 S $_{2}$ O $_{3}^{2 -}$ $\to$ 2 I $^{-}$ + S $_{4}$ O $_{6}^{2 -}$ (each S $_{2}$ O $_{3}^{2 -}$ is a 1-equivalent reductant).

Equivalents of hypo used $= x \times 0.300 \times 1 = 0.30$ $\Rightarrow$ $x = 1$ .

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q23: Advance Stoichiometry Based on Redox Reactions

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