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. 480 faradays through an aqueous KF solution.

A) 6.48 g of OF2 at the anode.
B) 26.0 g of OF2 at the anode.
C) 6.48 g of OF2 at the cathode.
D) 26.0 g of OF2 at the cathode

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:

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

Answer 2

Using the data provided here, the mass of the compound produced is 6.48 g of OF2

What is electrolysis?

Electrolysis refers to the breaking u p of a molecule by the passage of direct current through it. The equation of the reaction is; H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻    E° = -2.15 V.

Now;

1 mole of OF2 is realeased by the passage of 4 F of electricity

x moles of OF2 is produced by the passage of  0.480F

x = 0.12 moles

Mass of OF2 = 0.12 moles * 54 g/mol = 6.48 g of OF2

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