Problem 104 - Chemistry Principles and Practice

Question

Calculate the rate of oxygen gas production at standard temperature and pressure, in units of milliliters per minute \((\mathrm{mL} / \mathrm{min}),\) by the electrolysis of water at a \(0.250-\mathrm{A}\) current

Step-by-Step Solution

Answer

The rate of oxygen gas production is 0.867 mL/min.

Step 1: Write the half-reaction equation

In electrolysis, water is split into hydrogen and oxygen gas. The half-reaction for oxygen gas production at the anode is: \[ 2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- \] This reaction shows that 4 moles of electrons are required to produce 1 mole of \(O_2\).

Step 2: Calculate the moles of electrons

Determine the number of moles of electrons using the current and Faraday's constant. \[ \text{Moles of electrons} = \frac{I \times t}{F} \] where \(I = 0.250 \space A\), \(t\) is the time in seconds, and \(F = 96500 \space C/mol\). For simplicity, assume 1 minute: \[ \text{Moles of electrons per minute} = \frac{0.250 \times 60}{96500} \approx 1.55 \times 10^{-4} \space \text{mol} \]

Step 3: Relate moles of electrons to moles of oxygen

From the half-reaction equation, 4 moles of electrons produce 1 mole of \(O_2\). Therefore: \[ \text{Moles of } O_2 \per \text{minute} = \frac{1.55 \times 10^{-4}}{4} \approx 3.87 \times 10^{-5} \space \text{mol} \]

Step 4: Convert moles of oxygen to volume at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters (or 22400 milliliters). Therefore, the volume of \(O_2\) produced per minute is: \[ \text{Volume} = 3.87 \times 10^{-5} \times 22400 \approx 0.867 \space \mathrm{mL/min} \]

Key Concepts

Half-Reaction Equations

When we perform the electrolysis of water, we are essentially decomposing water molecules into hydrogen and oxygen gases. This happens in two separate reactions known as half-reactions, each occurring at different electrodes. In our case, at the anode, which is the positively charged electrode, water undergoes oxidation. The half-reaction that occurs is:\[2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-\]Here, water (\(H_2O\)) decomposes to produce oxygen gas (\(O_2\)), hydrogen ions (\(H^+\)), and electrons (\(e^-\)). This half-reaction equation tells us that to produce 1 mole of oxygen gas, 4 moles of electrons must be transferred. Breaking down such equations helps us understand how mass and charge are conserved, serving as a basis for calculating the quantity of gases produced in electrolysis.
Half-reaction equations are essential because they guide the stoichiometric calculation of electrolysis, which is crucial for accurately determining the amounts involved in the process.

Faraday's Constant

Faraday’s constant is a fundamental constant in electrochemistry that represents the charge of one mole of electrons. It is denoted by the letter \(F\) and its value is approximately \(96500\, C/mol\). This value is critical because it relates electric charge to chemical changes, letting us convert between them.

To calculate the moles of electrons involved in electrolysis, we utilize the formula:\[\text{Moles of electrons} = \frac{I \times t}{F}\]where \(I\) is the current in amperes, \(t\) is the time in seconds, and \(F\) is Faraday's constant. In the given problem, with a 0.250 A current running for a minute, we find:\[\text{Moles of electrons per minute} = \frac{0.250 \times 60}{96500} \approx 1.55 \times 10^{-4} \, \text{mol}\]Knowing the number of moles of electrons helps in determining how much product is formed or consumed in a reaction. Faraday's constant thus acts as a bridge between electrical circuits and chemical reactions.

Gas Volume at STP

Standard Temperature and Pressure (STP) is a reference point used in chemistry to define a set of conditions: a temperature of 0°C (273.15 K) and a pressure of 1 atm. At STP, one mole of any ideal gas occupies a volume of 22.4 liters. This simplifies the computation of gas volumes during reactions or processes such as electrolysis.

Using the relation that 1 mole of gas at STP equals 22.4 L, we can convert the moles of oxygen produced into volume. Given the moles of \(O_2\) is approximately \(3.87 \times 10^{-5} \, \text{mol/min}\), the volume in milliliters is calculated as:\[\text{Volume} = 3.87 \times 10^{-5} \times 22400 \approx 0.867 \, \text{mL/min}\]This means that under STP conditions, such a reaction will produce about 0.867 milliliters of oxygen gas per minute. By understanding gas volumes at STP, we can predict the outcomes of chemical reactions in terms of measurable quantities.