\(Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O\)
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. \(5.79 × 10^5\, \ C\).
The balanced half-reaction provided is:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O \]
This reaction involves the reduction of \(Cr_2O_7^{2-}\) to \(2Cr^{3+}\) by gaining electrons.
The quantity of charge (in coulombs) needed to reduce 1 mole of \(Cr_2O_7^{2-}\) can be calculated by multiplying the number of moles by the Faraday constant (\(F\)).
From the balanced half-reaction, 6 moles of electrons are involved in the reduction of 1 mole of \(Cr_2O_7^{2-}\).
So, the quantity of charge (\(Q\)) is given by:
\( Q = n \times F \)
\( Q = 6 \, \text{moles} \times 96500 \, \text{C mol}^{-1} \)
\( Q = 579000 \, \text{C} \)
Therefore, the correct option is: (2) \(5.79 \times 10^5\, \text{C}\)