\(Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O\)
The quantity of charge, in coulombs, needed to reduce 1 mole of \(Cr_2O_7^{2-}\) is
(Given \(\text{1 F = 96500 C mol}^{–1})\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(5.79 × 10^5\, \ C\)
One faraday (96500 C) is the charge on one mole of electrons. The half-reaction shows that 1 mol of \( \ce{Cr2O7^{2-}} \) takes up 6 mol of electrons (each Cr goes from +6 to +3):
\( \displaystyle \ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O} \)
\( \displaystyle Q = 6F = 6 \times 96500\ \mathrm{C} = 579000\ \mathrm{C} = 5.79 \times 10^5\ \mathrm{C} \)