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 0.240 faradays through an aqueous KF solution.

Question

Quizplus · Accepted Answer

The standard reduction potential for the electrolysis of KF is positive, indicating that the reaction is spontaneous and will proceed as written. The anode is the electrode at which oxidation occurs, and the cathode is the electrode at which reduction occurs. In this case, OF₂ is produced at the anode, and H₂O is produced at the cathode. The amount of OF₂ produced can be calculated using the equation:```n = Q / (nF)```where:* n is the number of moles of OF₂ produced* Q is the charge passed through the solution (in coulombs)* n is the number of electrons transferred in the reaction (4 in this case)* F is the Faraday constant (96,485 C/mol)Substituting the given values into the equation, we get:```n = 0.240 C / (4 mol e- / 96,485 C/mol) = 6.23 x 10^-6 mol```The molar mass of OF₂ is 54.04 g/mol, so the mass of OF₂ produced is:```mass = n * molar mass = 6.23 x 10^-6 mol * 54.04 g/mol = 3.36 x 10^-4 g```Therefore, the correct answer is A) 3.24 g of OF₂ at the anode.

The Gas OF2 Can Be Produced from the Electrolysis of an Aqueous

The Gas OF₂ Can Be Produced from the Electrolysis of an Aqueous