Question

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of 2.50 \(\mathrm{A}\) in 15.0 \(\mathrm{min}\) ?

Short answer

During the electrolysis of water with a current of 2.50 A in 15.0 min, approximately 0.261 L of hydrogen gas (H₂) and 0.131 L of oxygen gas (O₂) are produced at standard temperature and pressure (STP).

Step-by-step solution

Convert time to seconds

To solve this problem, we need to have the time in seconds. So, we'll convert the 15.0 min to seconds: \[15.0 \ \text{min} \times \frac{60 \ \text{s}}{1 \ \text{min}} = 900 \ \text{s}\]

Calculate the charge passed during electrolysis

We need to calculate the charge passed during the electrolysis process. We can use the formula: \[Q = It\] where Q is the charge, I is the current, and t is the time. So the charge passed is: \[Q = (2.50 \ \text{A})(900 \ \text{s}) = 2250 \ \text{C}\]

Applying Faraday's Laws of Electrolysis

We have to determine the amount of gas formed during electrolysis using Faraday's laws. The reaction that takes place is: \[2H_2O(l) \longrightarrow 2H_2(g) + O_2(g)\] From Faraday's law, we know that the charge needed to produce one mole of a substance is given by: \[Q = nfF\] where Q is the charge, n is the number of moles, f is the number of electrons involved in the reaction, and F is Faraday's constant (\(96500\ \text{C mol}^{-1}\)). For the electrolysis of water: - 2 electrons are required to produce 1 mole of hydrogen gas (H₂) - 4 electrons are required to produce 1 mole of oxygen gas (O₂)

Calculate the moles of H₂ and O₂ produced

Now, we can calculate the moles of H₂ and O₂ generated during electrolysis. For H₂: \[n_{H_2} = \frac{Q}{(2 \times F)} = \frac{2250 \ \text{C}}{(2 \times 96500 \ \text{C mol}^{-1})} = 0.011672 \ \text{mol}\] For O₂: \[n_{O_2} = \frac{n_{H_2}}{2} = \frac{0.011672}{2} = 0.005836 \ \text{mol}\]

Calculate the volume of H₂ and O₂ produced at STP

Now that we have the moles of gases produced, we can calculate the volumes of H₂ and O₂ at STP. The volume of one mole of any gas at STP is 22.4 L/mol. So, the volumes of gases produced are: For H₂: \[V_{H_2} = n_{H_2} \times V_{molecular} = 0.011672 \ \text{mol} \times 22.4 \ \text{L/mol} \approx 0.261 \ \text{L}\] For O₂: \[V_{O_2} = n_{O_2} \times V_{molecular} = 0.005836 \ \text{mol} \times 22.4 \ \text{L/mol} \approx 0.131 \ \text{L}\] Thus, from the electrolysis of water with a current of 2.50 A in 15.0 min, approximately 0.261 L of hydrogen gas and 0.131 L of oxygen gas are produced at STP.

Key concepts

Faraday's Laws of Electrolysis

Faraday's laws are fundamental principles that relate electricity to chemical change, key for understanding electrolysis.

• The **first law** states that the amount of chemical change is proportional to the quantity of electricity that passes through the electrolyte. This means if you know the charge, you can determine how much of a substance can be converted or produced.
• The **second law** states that the amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights. This helps to determine the number of moles formed from electrolysis, given the type of ions involved.

In electrolysis, electricity causes a chemical reaction which results in the production of gases when water is split. This is why it's essential to understand how Faraday's laws apply to the process, as they help predict the amount of gas generated by a certain charge. In simple terms, the laws provide a bridge between the electrical input (current and time) and the chemical output (amount of substance).

Stoichiometry

Stoichiometry is the area of chemistry that involves the calculation of reactants and products in chemical reactions. It helps in quantifying the exact amounts of substances needed and produced in reactions.
In the electrolysis of water, the stoichiometric equation is: \[2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\]
This tells us that two moles of water (H₂O) produce two moles of hydrogen gas (H₂) and one mole of oxygen gas (O₂).
Using stoichiometry:

• **Mole ratio for hydrogen and oxygen**: In the formula, for every two moles of water that decompose, two moles of hydrogen and one mole of oxygen are produced.
• **Application with Faraday's laws**: Once you know how many moles of electrons are involved, stoichiometry allows you to convert that to moles of gases produced.

Stoichiometry is crucial in predicting volumes of gases by converting the moles of gas calculated during electrolysis into liters at certain conditions.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure, commonly referred to as STP, is a set of conditions for the measurement of gases. It provides a standard because changes in temperature or pressure can significantly alter the volume of gases.
The conditions for STP are:

• **Temperature at STP** is 273.15 K (0°C).
• **Pressure at STP** is 1 atm (101.3 kPa).
• **Volume of one mole of gas** at STP is 22.4 liters.

This standardization allows chemists to make accurate calculations and comparisons.

In the context of electrolysis, knowing that each mole of gas occupies 22.4 liters at STP, we can determine the actual volume of gases produced. This is particularly useful for comparing gases generated under different conditions back to a standard and expected volume.

Gas Production Calculations

In electrolysis, you can calculate the volumes of hydrogen and oxygen gases produced using a few steps based on Faraday's laws and stoichiometry. Here's how you can do it:

• **Determine the charge passed**: Multiply the current (in amperes) by the time (in seconds).
• **Relate charge to moles of gas**: Use Faraday's laws. You need the charge to understand how many moles of gas are formed. Use the relation \( Q = nfF \), involving Faraday's constant (\( 96500 \, \text{C mol}^{-1} \)).
• **Use stoichiometry**: Convert moles of electrons to moles of gas using the stoichiometric ratios from the equation \(2H_2O \rightarrow 2H_2 + O_2\).
• **Calculate gas volume at STP**: Multiply the moles of gas by 22.4 L/mol to find the volume produced at STP.

This approach allows for explicit volume determination from known electrical conditions, providing useful insight into both the chemistry involved and the practical applications of electrolysis.