Question

Calculate the number of moles of electrons required to convert \(1.1 \text{ mol } \text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\) in acidic medium.

• A. 7.1 mole
• B. 0.183 mole
• C. 6.6 mole
• D. 3.3 mole

Hint:

Remember: Dichromate → 6 electrons reduction.

Answer 1

The Correct Option is C

Step 1: Write half reaction. \[ \text{Cr}_2\text{O}_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

Step 2: Electrons required. \[ 1 \text{ mole } Cr_2O_7^{2-} \rightarrow 6 \text{ moles electrons} \]

Step 3: For 1.1 mole. \[ 1.1 \times 6 = 6.6 \text{ moles} \] Conclusion: \[ \text{Electrons required = 6.6 moles} \]