Question

What is the quantity of charge, in Faraday units, required for the reduction of 3.5 moles of $Cr_2\text{O}_7^{2-} \text{ in acid medium?}$

• A. 6.0
• B. 10.5
• C. 21.0
• D. 3.0

Hint:

When dealing with redox reactions, remember that the number of moles of electrons required is proportional to the number of moles of reactant, and can be determined from the coefficients in the half-reaction.

Answer 1

The Correct Option is C

The reduction half-reaction for \( Cr_2O_7^{2-} \) in acid medium is given by: \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \] This equation shows that for each mole of \( Cr_2O_7^{2-} \), 6 moles of electrons are required for the reduction. Thus, for 3.5 moles of \( Cr_2O_7^{2-} \), the total number of moles of electrons required will be: \[ 6 \times 3.5 = 21 \text{ moles of electrons} \] Since 1 Faraday corresponds to 1 mole of electrons, the total charge required in Faraday units will be 21 Faradays. 
Thus, the correct answer is 21.0 Faradays.