16.4 litre of a mixture of H2 and O2 were evolved by electrolysis of a quantity of water using 0.5 Ampere current. Find the volume of O2 liberated and the total Faraday consumed for the purpose

To solve the problem of determining the volume of oxygen (O2) liberated and the total Faraday consumed during the electrolysis of water, we need to understand the electrolysis process and the relationship between current, time, and the amount of gas produced. Let's break this down step by step.

Understanding Electrolysis of Water

During the electrolysis of water, water (H2O) is split into hydrogen (H2) and oxygen (O2) gases. The overall reaction can be represented as:

• 2 H2O(l) → 2 H2(g) + O2(g)

This means that for every 2 moles of water, 2 moles of hydrogen and 1 mole of oxygen are produced. In terms of volume, since gases at the same temperature and pressure occupy the same volume, we can say that 2 volumes of hydrogen are produced for every 1 volume of oxygen.

Calculating the Volume of Gases

Given that a total of 16.4 liters of gas mixture is produced, we can denote the volume of hydrogen as V(H2) and the volume of oxygen as V(O2). From the stoichiometry of the reaction, we know:

• V(H2) = 2 × V(O2)

Let’s denote the volume of oxygen as V(O2) and the volume of hydrogen as 2V(O2). Therefore, we can express the total volume of gases as:

• V(H2) + V(O2) = 16.4 liters

Substituting V(H2) with 2V(O2) gives us:

• 2V(O2) + V(O2) = 16.4 liters
• 3V(O2) = 16.4 liters

Now, solving for V(O2):

• V(O2) = 16.4 liters / 3 = 5.47 liters

Finding the Volume of Hydrogen

Now that we have the volume of oxygen, we can find the volume of hydrogen:

• V(H2) = 2 × V(O2) = 2 × 5.47 liters = 10.93 liters

Calculating Total Faraday Consumed

To find the total Faraday consumed, we need to know the charge required to produce the gases. The Faraday constant (F) is approximately 96500 coulombs per mole of electrons. In the electrolysis of water, 4 moles of electrons are required to produce 1 mole of O2.

From our reaction, we can see that:

• 1 mole of O2 requires 4 moles of electrons.

Using the ideal gas law, at standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the number of moles of O2 produced is:

• n(O2) = V(O2) / 22.4 = 5.47 liters / 22.4 liters/mole ≈ 0.244 moles

Now, the total moles of electrons required for this amount of O2 is:

• moles of electrons = 4 × n(O2) = 4 × 0.244 ≈ 0.976 moles

Now, we can calculate the total charge (Q) in coulombs:

• Q = moles of electrons × Faraday constant = 0.976 moles × 96500 C/mole ≈ 94188 coulombs

Relating Current, Time, and Charge

We can also relate the charge to current and time using the formula:

• Q = I × t

Where I is the current in amperes and t is the time in seconds. Given that the current is 0.5 A, we can find the time:

• t = Q / I = 94188 C / 0.5 A ≈ 188376 seconds

Summary of Results

To summarize:

• Volume of O2 liberated: 5.47 liters
• Total Faraday consumed: approximately 94188 coulombs

This analysis provides a comprehensive understanding of the electrolysis process and the calculations involved in determining the volumes of gases produced and the charge consumed. If you have any further questions or need clarification on any part, feel free to ask!