Indicate the reactant which undergoes both oxidation and reduction.

$2 K O H + B r_{2} \to K B r + K B r O + H_{2} O$

B r_{2}

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

K O H

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

Both of these

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 B $B r_{2}$
The oxidation number of Br in $B r_{2}$ increases from 0 to +1 in $K B r O$ . Thus, $B r_{2} S$ is oxidized.
The oxidation number of Br in $B r_{2}$ decreases from 0 to -1 in $K B r$ . Thus, $B r_{2}$ is reduced.

Suggest Corrections

Similar questions

C_{2} H_{4} + B r_{2} \to C_{2} H_{4} B r_{2}

4

B r_{2}

6

B r_{2}

B r_{2}

{B r}_{2}

K O H

(R - {N H}_{2})

B r_{2}

N H_{2}

S i + 2 K O H + H_{2} O \to K_{2} S i O_{3} + 2 H_{2}

Join BYJU'S Learning Program