Question

One of the half-reactions for the electrolysis of water is: $$2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-}$$ If \(0.076 \mathrm{~L}\) of \(\mathrm{O}_{2}\) is collected at \(25^{\circ} \mathrm{C}\) and \(755 \mathrm{mmHg}\), how many faradays of electricity had to pass through the solution? \(?\)

Short answer

0.01234 Faradays

Step-by-step solution

Use Ideal Gas Law

First, we need to find the number of moles of oxygen gas. Use the ideal gas law equation: \[ PV = nRT \]where:- \( P \) is the pressure in atm (755 mmHg to atm conversion is required),- \( V \) is the volume in liters (0.076 L),- \( n \) is the number of moles,- \( R \) is the ideal gas constant \(0.0821 \text{ L atm K}^{-1} \text{mol}^{-1}\),- \( T \) is the temperature in Kelvin (Convert 25°C to Kelvin: \(25 + 273 = 298\text{ K}\)).Convert pressure from mmHg to atm:\[ P = \frac{755}{760} \text{ atm} \approx 0.993 \text{ atm} \].Substitute these values into the ideal gas law to find \( n \):\[ 0.993 \times 0.076 = n \times 0.0821 \times 298 \].\[ n \approx \frac{0.075468}{24.4758} \approx 0.003085 \text{ mol} \text{ of } \mathrm{O}_{2} \].

Relate Moles of Oxygen to Faradays

According to the half-reaction:\[ 2 \text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4 \text{H}^+(aq) + 4e^- \]1 mole of \( \text{O}_2 \) is produced with 4 moles of electrons. Therefore, the moles of electrons (\( n_{e^-} \)) required is:\[ n_{e^-} = 4 \times \text{ moles of } \text{O}_2 = 4 \times 0.003085 = 0.01234 \text{ mol} \text{ of electrons} \].1 Faraday corresponds to 1 mole of electrons. Hence, the number of Faradays required is:\ \ \ 0.01234 \text{ Faradays}.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to relate the properties of an ideal gas. It is represented as \( PV = nRT \). This law links pressure \( P \), volume \( V \), number of moles \( n \), the ideal gas constant \( R \), and temperature \( T \) in Kelvin. To effectively use the Ideal Gas Law, it is important to:

• Convert all units to the standard units such as pressure in atmospheres, volume in liters, and temperature in Kelvin.
• Know the value of the ideal gas constant \( R \), which is \( 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \).

In our exercise, we converted the pressure from mmHg to atm, the temperature from °C to Kelvin, and applied these values to determine the moles of oxygen gas. This calculation is crucial for understanding how gases behave under various physical conditions.

Faraday's Law

Faraday's Law is essential in electrochemistry and relates to the amount of substance altered or produced at the electrode during electrolysis to the total electric charge passed through the substance. It is based on the principle that 1 mole of electrons corresponds to 1 Faraday of charge. To apply Faraday's Law:

• Understand that 1 Faraday is equivalent to 96,485 coulombs, representing the charge of 1 mole of electrons.
• Use the relation between moles of electrons and moles of substance to determine how much charge is needed for a reaction.

In our context, we calculated the moles of electrons based on the half-reaction provided in the exercise. This allows us to determine how many Faradays, or moles of electrons, are required to produce a given amount of \( \text{O}_2 \). It leads us to the calculation of electrical energy utilized in the reaction.

Half-reaction in Electrolysis

Half-reactions in electrolysis describe the specific processes occurring at each electrode. For water electrolysis, the half-reaction at the anode involves the oxidation of water to produce oxygen gas and protons, along with electrons. The given half-reaction is:\[ 2 \text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4 \text{H}^+(aq) + 4e^- \]From this:

• Water molecules dissociate, contributing to the formation of oxygen gas through the loss of electrons.
• Each mole of \( \text{O}_2 \) production requires the transfer of 4 moles of electrons.

Understanding half-reactions is vital as it helps to decipher the electron flow, which is crucial for determining the moles of electrons needed, thereby allowing the calculation of charge through Faraday's Law.

Moles of Gas Calculation

Calculating the moles of gas is crucial when dealing with gases in chemical reactions, particularly in electrolysis to apply Faraday's law. The number of moles can be found using the Ideal Gas Law, linking the physical conditions of the gas to its quantity in moles. In our problem, this involved:

• Converting given conditions like pressure and temperature into suitable units.
• Substituting these into the ideal gas law equation to find the moles of \( \text{O}_2 \) produced.

Understanding how to calculate moles of a gas ensures the accuracy in further calculations, such as determining the charge in electrolysis. This foundational calculation is a stepping stone towards solving more complex chemical problems, following laws such as Faraday's.