When $S O_{2}$ is passed through an acidic solution of potassium dichromate, chromium sulphate is formed. The change in oxidation number of chromium is:

+6 to +2

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

+6 to +3

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

+5 to +3

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

+6 to +4

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

Open in App

Solution

The correct option is B +6 to +3
$K_{2} C r_{2} O_{7} + 3 S O_{2} + H_{2} S O_{4} ⟶ K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O$
Here, the change in oxidation state is from +6 to +3.

Suggest Corrections

Similar questions

S O_{2}

S O_{2}

When SO₂ is passed through an acidified solution of K2Cr2O7, then chromium sulphate is formed. Change in the oxidation state of Cr is from.

The products formed when $S O_{2}$ reacts with acidified potassium dichromate solution is :

Join BYJU'S Learning Program