What quantity of electricity will liberate 0.125 mole of oxygen molecules during the electrolysis of dilute sodium chloride solution?

Analysis

To solve this problem, we need to understand the relationship between the amount of substance liberated during electrolysis and the quantity of electricity (charge) passed through the electrolyte. This is governed by Faraday’s Laws of Electrolysis.

Concept: Faraday’s Laws of Electrolysis

During the electrolysis of a dilute sodium chloride solution, the reaction at the anode involves the discharge of hydroxide ions ($OH^-$) from water to produce oxygen gas ($O_2$):

From this balanced equation, we can see that 4 moles of electrons are required to liberate 1 mole of oxygen gas.

The quantity of electricity ($Q$) is related to the number of moles of electrons ($n$) by the Faraday constant ($F$), where $F \approx 96,500 \text{ C/mol}$. The formula is:

Step-by-Step Solution

1. Determine the moles of electrons required: From the equation, 1 mole of $O_2$ requires 4 moles of electrons. Therefore, 0.125 moles of $O_2$ will require: $$n = 0.125 \times 4$$ $$n = 0.5 \text{ moles of electrons}$$

2. Calculate the quantity of electricity ($Q$): Using the Faraday constant ($F = 96,500 \text{ C/mol}$): $$Q = 0.5 \text{ mol} \times 96,500 \text{ C/mol}$$ $$Q = 48,250 \text{ C}$$

Conclusion

The quantity of electricity required to liberate 0.125 mole of oxygen molecules is 48,250 Coulombs. This matches option B.

Exam Tip

Always write out the half-equation for the gas being produced at the electrode. Remember that for oxygen ($O_2$), it is a 4-electron process, whereas for hydrogen ($H_2$), it is a 2-electron process. Knowing the stoichiometry of the half-equation is the key to solving any electrolysis calculation!