Question

Calculate the theoretical quantity of chlorine obtainable by the electrolysis of \(2.0 \mathrm{~kg}\) of a \(20 \%\) sodium chloride solution. What other products would be obtained and what would be the weight of each?

Short answer

The theoretical quantity of chlorine gas obtainable from the electrolysis of \(2.0 \mathrm{~kg}\) of a \(20 \%\) sodium chloride solution is 242.97g. The other products in this process are sodium hydroxide (NaOH) and hydrogen gas (H2), but their weights cannot be calculated without more information about the specific conditions such as current and time of electrolysis.

Step-by-step solution

Understand the chemical reactions

During the electrolysis of sodium chloride solution, there are two main reactions that take place: 1. At the cathode: Na+ ions gain electrons to become Sodium atom - Na+ + e- → Na 2. At the anode: Chloride ions lose electrons to become Chlorine - 2Cl- → Cl2 + 2e- Additionally, the electrolysis of water happens simultaneously: 1. At the cathode: 2H2O + 2e- → H2 + 2OH- 2. At the Anode: 4OH- → O2 + 2H2O + 4e-

Calculate the amount of sodium chloride in solution

Given that 2.0 kg or 2000 g of the solution is 20% sodium chloride, this means there's 20/100 * 2000g = 400 grams of sodium chloride(NaCl) in the solution.

Calculate the moles of sodium chloride

To calculate the moles of sodium chloride in the solution, we use the formula \[ \text{{Moles}} = \frac{{\text{{mass}}}}{{\text{{molar mass}}}} \] For sodium chloride (NaCl), the molar mass is 58.44 g/mol (sum of Sodium(Na) and Chlorine(Cl) atomic masses). Therefore, we have: \[ \text{{Moles of NaCl}} = \frac{{400 g}}{{58.44 g/mol}} = 6.85 \text{{ moles}} \]

Calculate the theoretical quantity of chlorine

In electrolysis, from the balanced equation at the anode (2 Cl- → Cl2 + 2e-), 1 mole of chlorine (Cl2) is produced from 2 moles of sodium chloride. Therefore from 6.85 moles of NaCl, the moles of chlorine gas produced is 6.85/2 = 3.425 moles. We can convert moles of chlorine into grams using its molar mass (35.45*2 = 70.9 g/mol) as follows: \[ \text{{Mass of Cl2}} = 3.425 \text{{ moles}} \times 70.9 \text{{ g/mol}} = 242.97 g \]

Identify and calculate other products obtained

Sodium hydroxide (NaOH) and hydrogen gas (H2) are also produced. However, we don't have enough details to calculate the weight of NaOH and H2 obtained here. With more specific conditions such as current and time of electrolysis, the faraday's laws of electrolysis could be applied to do so. Accordingly, the theoretical quantity of chlorine gas obtainable is 242.97g, with sodium hydroxide and hydrogen gas as the other products.

Key concepts

Chlorine Gas Production

In the electrolysis of sodium chloride, chlorine gas is generated at the anode. The process starts with chloride ions (Cl\(^-\)) in the solution, which move towards the anode. Here, they lose electrons in a reaction characterized by oxidation. The equation for this reaction is:

• 2Cl\(^-\) → Cl\(_2\) + 2e\(^-\)

This means that for every two chloride ions, one molecule of chlorine gas (Cl\(_2\)) is produced. So, if you start with 6.85 moles of NaCl, you will obtain 3.425 moles of Cl\(_2\). By using chlorine's molar mass (70.9 g/mol), you can determine the mass of chlorine gas produced, which would be around 242.97 grams.
If ever asked about the practical applications, remember, this chlorine gas is often used for disinfecting water and in the production of various consumer products. Understanding this production is crucial in both industrial settings and for academic purposes.

Sodium Hydroxide Formation

Sodium hydroxide (NaOH) is a crucial byproduct in the electrolysis of sodium chloride. While chlorine gas forms at the anode, sodium ions (Na\(^+\)) from the dissolved NaCl migrate towards the cathode, where they have a different journey. Instead of directly forming NaOH, they play a role in creating conditions conducive to its formation. Here’s how:

• At the cathode, water undergoes reduction: 2H\(_2\)O + 2e\(^-\) → H\(_2\) + 2OH\(^-\)

This reaction not only releases hydrogen gas but also generates hydroxide ions (OH\(^-\)). These OH\(^-\) ions will combine with the nearby Na\(^+\) ions, forming NaOH in the solution. Sodium hydroxide is extensively used in cleaning products, paper manufacturing, and chemical manufacturing.
The electrolysis setup provides a scalable method to produce NaOH efficiently.

Hydrogen Gas Evolution

During electrolysis, hydrogen gas evolution occurs at the cathode. As electricity flows through the solution, water molecules (H\(_2\)O) are reduced to release hydrogen gas according to the reaction:

• 2H\(_2\)O + 2e\(^-\) → H\(_2\) + 2OH\(^-\)

In this process, hydrogen atoms are produced when excess electrons combine with the protons from water. These hydrogen atoms quickly form diatomic hydrogen gas (H\(_2\)). Over time, this accumulates and bubbles up from the solution, noticeable as hydrogen gas. This gas is incredibly useful in chemical industries, often applied in hydrogenation reactions, and even seen as a clean energy source for fuel cells in the future.
Understanding this evolution process is key in various fields, from industrial chemistry to renewable energy research.

Anode Reaction

The anode is where oxidation reactions occur in electrolysis. In this process involving sodium chloride, the anode performs a crucial function of oxidizing chloride ions:

• The reaction can be represented as: 2Cl\(^-\) → Cl\(_2\) + 2e\(^-\)

This shows that chloride ions lose electrons at the anode to form chlorine gas. It's notable that anode reactions involve a loss of electrons—thus, a redox process where oxidation predominates. The apparatus must provide a material that can resist these oxidation reactions, like graphite, which is commonly used as anode material.
In industrial settings, knowing the specifics of the anode reaction helps in designing more efficient production systems.

Cathode Reaction

At the cathode, reduction reactions take the lead. Within the electrolysis of sodium chloride, hydrogen gas formation dominates the cathode side, governed by the reduction of water molecules:

• 2H\(_2\)O + 2e\(^-\) → H\(_2\) + 2OH\(^-\)

This process results in both the evolution of hydrogen gas and the creation of hydroxide ions. While the hydrogen gas bubbles up, the OH\(^-\) ions remain in the solution, effectively participating in forming sodium hydroxide (NaOH).
Understanding these reactions is vital for students and professionals working in electrochemical processes, as the efficiency and outcome depend heavily on mastering the cathode reactions.