Question

Write the balanced chemical equation for the reaction between $K_{2}C{r}_2{O}_7$ and acidified KI solution.

• A. $C{r}_2{O}_7^{-2}+6I^{-}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$
• B. $2C{r}_2{O}_7^{-2}+6I^{-}+16H^{+}\to 4C{r}^{+3}+3I_2+8H_{2}O$
• C. $C{r}_2{O}_7^{-2}+8I^{-}+16H^{+}\to 2C{r}^{+3}+4I_2+8H_{2}O$
• D. None of these

Answer 1

The correct option is A $C{r}_2{O}_7^{-2}+6I^{-}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$

Option (A) is correct. Acidified potassium dichromate oxidizes potassium iodide to iodine. Potassium dichromate is reduced to chromic sulphate. Liberated iodine turns the solution brown. The net ionic equation is as shown.

$C{r}_2{O}_7^{-2}+6I^{-}+14H^{+}\to 2C{r}^{+3}+3I_2+7H_{2}O$