When an aqueous solution of zinc sulphate is subjected to electrolysis, \(280 \mathrm{ml}\) of oxygen gas at STP is liberated at anode. Calculate the quantity of electricity passed through the electrolyte. (a) \(0.05 \mathrm{~F}\) (b) \(0.5 \mathrm{~F}\) (c) \(0.005 \mathrm{~F}\) (d) \(5 \mathrm{~F}\)

Understanding the volume one mole of a gas occupies at Standard Temperature and Pressure (STP) is crucial for solving problems in chemistry, especially those involving gas calculations during electrolysis. The molar gas volume at STP is defined as the volume occupied by one mole of any gas at a standard temperature of 0°C (273.15 K) and a standard pressure of 1 atm (101.325 kPa). This volume is approximately 22.4 liters per mole.

When working with gas volumes in electrolysis calculations, it's often required to convert the volume of gas liberated during the reaction to moles, which can then be used to determine the quantity of electricity involved in the process. For instance, if we have 280 ml (or 0.28 liters) of oxygen gas at STP, we can calculate the number of moles by dividing the volume by the molar gas volume at STP:
\[ \text{Number of moles} = \frac{\text{Volume of gas}}{\text{Molar gas volume at STP}} \
\]With this calculation, we can proceed to the other steps of the electrolysis calculation. Remember, accuracy in this initial conversion from volume to moles is essential to ensure the correctness of subsequent calculations.

During electrolysis, the reactions that occur at the electrodes are crucial. The half-reaction at the anode involves oxidation where electrons are released. Particularly in aqueous solutions, the anode half-reaction often involves the liberation of oxygen from hydroxide ions:

\[ 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^- \
\]
In this half-reaction, four hydroxide ions (OH⁻) are oxidized to produce two water molecules (H₂O), one oxygen molecule (O₂), and four electrons (4e⁻). To fully understand the anodic reaction, it's necessary to notice the stoichiometric relationships—specifically, the ratio of electrons produced to the oxygen molecules formed. In this case, each mole of oxygen gas requires four moles of electrons to be released.

When performing calculations involving the half-reaction at the anode, knowing this stoichiometry helps determine the total charge or number of electrons involved in the process. Given the moles of oxygen gas produced, multiplying by the stoichiometric coefficient (in this case, 4) will yield the moles of electrons.

Faraday's constant is fundamental to understanding electrolysis and electrochemistry. It represents the amount of electric charge per mole of electrons. Numerically, Faraday's constant is approximately 96,485 coulombs per mole (C/mol). The value is significant because it allows the conversion between moles of electrons and the actual charge in units of coulombs or farads (since 1 farad = 1 coulomb per volt).

To use Faraday's constant in calculations, once the number of moles of electrons involved in the reaction is known, it is multiplied by Faraday's constant to determine the quantity of electricity (in coulombs or farads) that has passed through the electrolyte:
\[ \text{Quantity of electricity} = \text{Number of moles of electrons} \times \text{Faraday's constant} \
\]In the context of the provided exercise, after determining the moles of electrons required to liberate a certain volume of oxygen gas at the anode, multiplying this by Faraday's constant gives the total charge in farads needed for that electrolytic process. This is a pivotal step in electrolysis calculations as it bridges the gap between chemical reactions and the physical electricity involved.