Question

How many moles of acidified K₂ Cr_{2 ​}O₇ ​ is required to liberate 6 moles of I₂ ​ from an aqueous solution of I⁻ ?

• A. 0.5
• B. 2
• C. 1
• D. 0.25

Answer 1

The Correct Option is B

The reaction between acidified potassium dichromate (\( K_2Cr_2O_7 \)) and iodide ions (\( I^- \)) in acidic solution involves the reduction of \( Cr^{6+} \) to \( Cr^{3+} \), while \( I^- \) is oxidized to iodine (\( I_2 \)). The balanced equation for the reaction is: \[ \text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O \] From the equation, it is clear that 1 mole of \( K_2Cr_2O_7 \) liberates 3 moles of \( I_2 \). To liberate 6 moles of \( I_2 \), we need: \[ \frac{6 \, \text{moles of} \, I_2}{3 \, \text{moles of} \, I_2 \, \text{per mole of} \, K_2Cr_2O_7} = 2 \, \text{moles of} \, K_2Cr_2O_7 \] Thus, 2 moles of \( K_2Cr_2O_7 \) are required to liberate 6 moles of \( I_2 \).

The correct option is (B) : 2

Answer 2

The reaction involved in the liberation of iodine (I\(_2\)) from iodide (I\(^-\)) by acidified potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) can be represented as follows: \[ \text{6I}^-(aq) + \text{Cr}_2\text{O}_7^{2-}(aq) + 14H^+(aq) \rightarrow 3I_2(aq) + 2\text{Cr}^{3+}(aq) + 7H_2O(l) \] From the equation, we can see that 1 mole of K\(_2\)Cr\(_2\)O\(_7\) (which provides Cr\(_2\)O\(_7^{2-}\)) liberates 3 moles of iodine (I\(_2\)). To liberate 6 moles of I\(_2\), the number of moles of K\(_2\)Cr\(_2\)O\(_7\) required is: \[ \text{Moles of K}_2\text{Cr}_2\text{O}_7 = \frac{6 \text{ moles of I}_2}{3 \text{ moles of I}_2 \text{ per mole of K}_2\text{Cr}_2\text{O}_7} = 2 \text{ moles} \] Thus, 2 moles of acidified K\(_2\)Cr\(_2\)O\(_7\) are required to liberate 6 moles of iodine.