Question

The gas OF₂ can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below.
OF₂(g) + 2 H⁺(aq) + 4 e^- → H₂O(l) + 2 F^-(aq) E° = +2.15 V
Using the given standard reduction potential, calculate the amount of OF₂ that is produced, and the electrode at which the OF₂ is produced, upon the passage of 0.120 faradays through an aqueous KF solution.

The gas OF₂ can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below.
OF₂(g) + 2 H⁺(aq) + 4 e^- H₂O(l) + 2 F^-(aq) E° = +2.15 V
Using the given standard reduction potential, calculate the amount of OF₂ that is produced, and the electrode at which the OF₂ is produced, upon the passage of 0.120 faradays through an aqueous KF solution.

Options:

6.50 g of OF2 at the anode.

1.62 g of OF2 at the anode.

1.62 g of OF2 at the cathode.

6.50 g of OF2 at the cathode.

Answer 1

Date: ANS 1-62 gram Of at the Anade Page: Of2 c9) +2Ht + 4e 20 +2E E = 2.15 V Cadacter Reverse Equate H20 +2f D + 2HT+4e- OR H20 26 in the roduction e ofr Ga Loss o e Ofz +4e oxidatien occured At Anode 54 Jlmat matar mass 0 Of Q = Che No f- 9crao 1 mot ob ue rquicel or the eductien Q4F Q4x965ao = 4 F = 386 000 64 gom Of hoduced by OR 4f306000 c aduce Noo 0120 F educe =306000 C 4 foradays) S4gran Of2 x 0 120=1.62-9 S4 4 F 1.62 um Amale az at ANS 1