The elements undergoing reduction and oxidation during the reaction,
$(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O$ are respectively:

N and Cr

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Cr and N

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N and O

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Cr and O

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Solution

The correct option is B Cr and N
$(N H_{4})_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 4 H_{2} O$
$2 N^{3 -} - 6 e^{-} ⟶ 2 N^{0}$ (Oxidation)
$2 C r^{V I} + 6 e^{-} ⟶ 2 C r^{I I I}$ (Reduction)

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Similar questions

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}

C r_{2} O_{3}

[N = 14, H = 1, O = 16, C r = 52]

C r O_{5} + S n C l_{2} \to C r O_{4}^{2 -} + S n C l_{4}

(N H_{4})_{2} C r_{2} O_{7} \to C r_{2} O_{3} + 4 H_{2} O + N_{2}

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