The oxidation state of chromium in the final product formed in the reaction between KI and acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution is :

Question

Accepted Answer

A. 1. In acidified $\mathrm{K_2Cr_2O_7}$, chromium starts at the $+6$ oxidation state (in $\mathrm{Cr_2O_7^{2-}}$).

2. When KI is added, $\mathrm{I^-}$ is oxidized to $\mathrm{I_2}$, while dichromate is **reduced** via:
$$
\mathrm{Cr_2O_7^{2-}+14H^++6e^-\rightarrow2Cr^{3+}+7H_2O}
$$

3. So chromium in the final product is $\mathrm{Cr^{3+}}$, i.e., oxidation state $+3$.

Hence, the answer is **Option A**.

The oxidation state of chromium in the final product formed in the reaction between KI and acidified K2Cr2O7\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7K2​Cr2​O7​ solution is :

The oxidation state of chromium in the final product formed in the reaction between KI and acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution is :