What is the oxidation state of $F e$ in the product formed when acidified potassium ferrocyanide is treated with $H_{2} O_{2}$ ?

+ 2

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

+ 6

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

+ 1

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

+ 3

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

Open in App

Solution

The correct option is B $+ 3$
$2 K_{4} [F e {(C N)}_{6}] + K_{2} {S O}_{4} + H_{2} O_{2} \to 2 K_{3} [F e {(C N)}_{6}] + K_{2} {S O}_{4} + 2 H_{2} O$

Suggest Corrections

Similar questions

What is the oxidation state of Fe in the product formed when acidified potassium ferrocyanide ( K4[Fe(CN6)] is treated with hydrogen peroxide ? How ?

I_{2}

The oxidation number of Fe in Potassium ferrocyanide is:

K I

Join BYJU'S Learning Program