JEE Main 2026 — Jan 24, Shift 1 — Chemistry solution

JEE Main2026Jan 24, Shift 1Chemistry

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 $K_{2} C r_{2} O_{7}$ . The value of $X + Y$ is ______ .

Your answer

Solution

$K_{2} C r_{2} O_{7}$ in acid reduces from $C r (V I)$ to $C r (I I I)$ , accepting $6 e^{-}$ per dichromate ion. Balanced equations:

Oxidation of iodide:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 C r^{3 +} + 3 I_{2} + 7 H_{2} O$

Electrons transferred per dichromate: $X = 6$ .

Oxidation of sulphide:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 3 S^{2 -} \to 2 C r^{3 +} + 3 S + 7 H_{2} O$

Electrons transferred per dichromate: $Y = 6$ .

$X + Y = 6 + 6 = 12$ .

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