X and Y are the number of electrons involved, respectively during the oxidation of I - to I 2 and S 2- to S by acidified K 2 Cr 2 O 7 . The value of X + Y is \ \ \ \ .

Chemistry · Redox Reactions

JEE Main 2026 — 24 January, Morning Shift — Question 66

X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$ to $\mathrm{I}_{2}$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ . The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$ .

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_{2}+7 \mathrm{H}_{2} \mathrm{O}$ no. of moles $\mathrm{e}^{-}$ involved $=\mathrm{x}=6$ $\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}+3 \mathrm{~S}^{2-}+14 \mathrm{H}^{+} \rightarrow+\mathrm{S}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$ No. of moles $\mathrm{e}^{-}$ involved $=\mathrm{y}=6$ Sum of $x+y=6+6=12$

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Chemistry
Chapter: Redox Reactions
Topic: Types of Redox Reactions & Balancing of Redox Reactions