Question

The oxidation state of 'Cr' in the final product formed by reaction between $\text{KI}$ and acidified $K_2Cr_2O_7$ is :

• A. $+2$
• B. $+6$
• C. $+4$
• D. $+3$

Hint:

Remember the standard redox reactions of dichromate: in acidic medium, $Cr_2O_7^{2-}$ is always reduced to $Cr^{3+}$, corresponding to a change in oxidation state from $+6$ to $+3$.

Answer 1

The Correct Option is D

Step 1: Acidified potassium dichromate ($K_2Cr_2O_7$) acts as a strong oxidizing agent.
Step 2: In acidic medium, iodide ions ($I^-$) are oxidized to iodine ($I_2$), while dichromate ions are reduced.
Step 3: Oxidation states: \[ Cr \text{ in } Cr_2O_7^{2-} = +6 \] Step 4: During the reaction, chromium gets reduced to $Cr^{3+}$.
Step 5: Balanced ionic equation: \[ Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O \] Step 6: Hence, chromium in the final product exists as $Cr^{3+}$.
Therefore, oxidation state of chromium is $+3$.