In acidic medium, $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ shows oxidising action as represented in the half reaction $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{-} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_{2} \mathrm{O}$ $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and A are respectively are:

(D) 14, 6, 7 and $\mathrm{Cr}^{3+}$

(A) 8, 6, 4 and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

(B) 14, 7, 6 and $\mathrm{Cr}^{3+}$

(C) 8, 4, 6 and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Redox Reactions and Electrochemistry

In acidic medium,

K_{2} Cr_{2} O_{7}

shows oxidising action as represented in the half reaction

Cr_{2} O_{7}^{2 -} + XH^{+} + Ye^{-} \to 2 A + ZH_{2} O

X, Y, Z

and A are respectively are:

Answer: (D)

Balancing the half-reaction

Cr_{2} O_{7}^{2 -} + X H^{+} + Y e^{-} \to 2 A + Z H_{2} O

: chromium goes from

+ 6

+ 3

, so

A = Cr^{3 +}

, and each Cr gains 3 electrons, so

Y = 6

electrons for 2 Cr atoms. Balancing oxygen (7 O atoms need 7 H

_{2}

O, so

Z = 7

) and then hydrogen (14 H on the right from 7 H

_{2}

O, so

X = 14

Cr_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O

X, Y, Z, A = 14, 6, 7, Cr^{3 +}

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option