Question

The gas OF2 can be produced from the electrolysis of an aqueous solution of KF, as shown in the equation below.

Answer 1

Answer:

A) 6.48 g of OF₂ at the anode.

Explanation:

The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.

H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻    E° = -2.15 V

Oxidation takes place in the anode.

We can establish the following relations:

• 1 Faraday is the charge corresponding to 1 mole of e⁻.
• 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
• The molar mass of OF₂ is 54.0 g/mol.

The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:

$0.480F.\frac{1mole^{-} }{1F} .\frac{1molOF_{2}}{4mole^{-} } .\frac{54.0gOF_{2}}{1molOF_{2}} =6.48gOF_{2}$