The following intramolecular redox reaction shows a loss and gain of how much electrons?
$(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O$

Six

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Seven

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Eight

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A Six
The oxidation number of Cr in ammonium dichromate is 6.
The oxidation number of Cr in chromium trioxide is 3.
Thus, the oxidation number of one Cr atom changes by 3.
Total change in the oxidation number for 2 Cr atoms is 6.
Hence, the redox change involves 6 electrons.

Suggest Corrections

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} ⟶ N_{2} + C r_{2} O_{3} + 2 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]

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

Join BYJU'S Learning Program