Chemistry · Redox Reactions

JEE Main 2026 — 6 April, Evening Shift — Question 62

500 mL of $0.2 \mathrm{M} \mathrm{MnO}_{4}^{-}$ 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 $\_\_\_\_$ .

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

$3 \mathrm{e}^{-}+\mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+4 \mathrm{OH}^{\ominus} 2 \mathrm{I}^{\ominus} \rightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-} \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{-2} \rightarrow 2 \mathrm{I}^{\ominus}+\mathrm{S}_{4} \mathrm{O}_{6}^{-2}$ gm equivalent of $\mathrm{MnO}_{4}^{-}$

=0.2 \times \frac{500}{1000} \times 3=0.3 \text { (Limiting reagent) }

gm equivalent of KI

=1.5 \times \frac{500}{1000} \times 1=0.75

gm equivalent of $\mathrm{MnO}_{4}^{-}$ $=$ gm equivalent of hypo $0.3=x \times \frac{300}{1000} \times 1$ $\mathrm{x}=1$

Answer key and solution verified before publishing.

Practise Redox Reactions

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Redox Reactions
Topic: n-Factor, Redox Titrations, Self Indicator & Miscellaneous Cases