Question
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
Iodide
Iodide
✓
Answer
K2Cr2O7 is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from +6 to +3. The oxidising action can be represented as follows:
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
Iodide: Iodide ion (J-) is oxidised to I by the acidfied solution of K2Cr2O7.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-\\ \overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$
$\text{Cr}_2\text{O}^{2-}_7+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}$
Iodide: Iodide ion (J-) is oxidised to I by the acidfied solution of K2Cr2O7.
Reaction:
$\text{Cr}_2\text{O}_7^{2-}+14\text{H}^++6\text{e}^-\rightarrow2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\text{I}^-\rightarrow3\text{I}_2+6\text{e}^-\\ \overline{\text{Cr}_2\text{O}^{2-}_7+14\text{H}^+6\text{I}^-\rightarrow3\text{I}_2+2\text{Cr}^{3+}+7\text{H}_2\text{O}}$
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