The oxidation state of chromium in the final product formed in the reaction between KI and acidified K₂Cr₂O₇ solution is

Q. The oxidation state of chromium in the final product formed in the reaction between KI and acidified K₂Cr₂O₇ solution is :

A. +2

B. +3

C. +4

D. +6

Correct Answer: +3

Explanation

Potassium dichromate is a strong oxidizing agent in acidic medium. In acidified solution, dichromate ion undergoes reduction.

The reduction half reaction of dichromate ion is:

$$ \mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O} $$

Here, chromium is reduced from +6 oxidation state in $\mathrm{Cr_2O_7^{2-}}$ to +3 oxidation state in $\mathrm{Cr^{3+}}$.

Potassium iodide acts as a reducing agent and gets oxidized to iodine:

$$ \mathrm{2I^- \rightarrow I_2 + 2e^-} $$

Combining both half reactions, chromium is finally present as $\mathrm{Cr^{3+}}$ in the reaction mixture.

Therefore, the oxidation state of chromium in the final product is +3.