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Redox Reactions · Question 1

JEE Main 2026 (Online) 6th April Evening ShiftNumericalMedium📈 Topic in 13% papers+4
1

500 mL of 0.2 M MnO4- 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 .

Answer

1

Solution

Basic medium: MnO4- + 2H2O + 3e- MnO2 + 4OH-; 2I- I2 + 2e-

n(MnO4-) = 0.5 × 0.2 = 0.1 mol ⇒ electrons accepted = 0.3 mol

n(I-) = 0.5 × 1.5 = 0.75 mol (excess, needs only 0.3 mol)

⇒ n(I2) = 0.32 = 0.15 mol

I2 + 2S2O32- 2I- + S4O62- ⇒ n( S2O32-) = 2 × 0.15 = 0.3 mol

⇒ x = 0.3 mol0.300 L = 1 M

Answer: 1

JEE Main 2026 (Online) 6th April Evening Shift

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In order to oxidise a mixture of 1 mole each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required is

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4

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7

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