Problem 65 - Chemistry and Chemical Reactivity

Question

A Electrolysis of molten \(\mathrm{NaCl}\) is done in cells operating at \(7.0 \mathrm{V}\) and \(4.0 \times 10^{4} \mathrm{A} .\) What mass of \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) can be produced in 1 day in such a cell? What is the energy consumption in kilowatt-hours? \(\left(1 \mathrm{kWh}=3.6 \times 10^{6} \mathrm{J} \text { and } 1 \mathrm{J}=1 \mathrm{C} \cdot \mathrm{V}\right)\)

Step-by-Step Solution

Answer

823.87 kg of Na and 1277.63 kg of Cl2 are produced, with 6720 kWh consumed.

Step 1: Calculate the Total Charge

First, find the total charge passed in one day. Current (I) is given by the formula \( I = \text{Charge} \div \text{Time} \). Rearranging for Charge, we get \( \text{Charge} = I \times t \).1 day = 24 hours = 24 \( \times \) 3600 seconds = 86400 seconds.Given current \( I = 4.0 \times 10^4 \) A:\[ \text{Charge} = 4.0 \times 10^4 \times 86400 = 3.456 \times 10^9 \text{ C}\]

Step 2: Calculate Moles of Electrons

Using Faraday's constant (1 mole of electrons = 96485 C):\[ \text{Moles of electrons} = \frac{3.456 \times 10^9 \text{ C}}{96485 \text{ C/mole}} = 35820.56 \text{ moles of electrons}\]

Step 3: Calculate Mass of Na Produced

The half-reaction for \text{Na} is \( \text{Na}^+ + e^- \rightarrow \text{Na(s)} \).- 1 mole of \( \text{Na} \) is produced per 1 mole of electrons.- Molar mass of \( \text{Na} \) is 23 g/mole.\[ \text{Mass of Na} = 35820.56 \text{ moles} \times 23 \text{ g/mole} = 823872.88 \text{ grams} \approx 823.87 \text{ kg}\]

Step 4: Calculate Moles of Cl2 Produced

The half-reaction for \text{Cl}_2 is \( 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \).- 1 mole of \( \text{Cl}_2 \) is produced per 2 moles of electrons.\[ \text{Moles of Cl}_2 = \frac{35820.56}{2} = 17910.28 \text{ moles} \]

Step 5: Calculate Mass of Cl2 Produced

Molar mass of \( \text{Cl}_2 \) is 71 g/mole.\[ \text{Mass of Cl}_2 = 17910.28 \text{ moles} \times 71 \text{ g/mole} = 1277630.88 \text{ grams} \approx 1277.63 \text{ kg}\]

Step 6: Calculate Energy Consumption

Power (P) is given by \( P = IV \), where \( V \) is 7.0 V and \( I \) is 4.0 \times 10^4 A.\[ P = 7.0 \times 4.0 \times 10^4 = 2.8 \times 10^5 \text{ W} \]Energy (E) is given by \( E = Pt \):\[ E = 2.8 \times 10^5 \times 86400 = 2.4192 \times 10^{10} \text{ J} \]Convert energy to kWh:\[ E = \frac{2.4192 \times 10^{10}}{3.6 \times 10^6} = 6720 \text{ kWh} \]

Key Concepts

Faraday's constant

Faraday's constant is an essential concept in the study of electrolysis and electrochemical reactions. It represents the charge carried by one mole of electrons. This constant is named after Michael Faraday, a pioneering scientist in the field of electromagnetism. The value of Faraday's constant is approximately 96485 coulombs per mole. This means that when a current passes through an electrolyte, that's the number of coulombs you need to deposit or release one mole of a substance.

Electrolysis, like in the example of producing sodium (\(\text{Na}\)) and chlorine gas (\(\text{Cl}_2\)), relies heavily on understanding this concept. By knowing the total charge that has passed through the circuit, we can determine how many moles of electrons have moved through the solution. This helps us calculate the amounts of the substances produced.

• 1 mole of electrons = 96485 C
• Used to calculate moles of any substance produced in electrolysis

Half-reaction

Half-reactions are a vital component of redox (reduction-oxidation) chemistry. They allow us to focus on one part of the redox process at a time, specifically the oxidation or reduction part. In the case of electrolysis of \(\text{NaCl}\), there are two main half-reactions:

For sodium (\(\text{Na}\)), the half-reaction is:\(\text{Na}^+ + e^- \rightarrow \text{Na(s)}\). This indicates that a single electron is needed to reduce one mole of sodium ions (\(\text{Na}^+\)) to one mole of sodium metal (\(\text{Na}\)).

For chlorine gas (\(\text{Cl}_2\)), the half-reaction is:\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\). Here, two moles of electrons are released when one mole of chlorine gas is formed from chloride ions. Knowing these half-reactions allows us to understand the stoichiometry involved in the production of each substance.

• Sodium half-reaction helps in calculating the sodium mass produced.
• Chlorine half-reaction is essential for accurately determining the production of chlorine gas.

Current (Amperes)

Current, measured in amperes (A), is the flow of electric charge. In electrolysis, the current dictates how many electrons are moving through the electrolytic solution per second. Larger currents allow for more electrons to pass, increasing the rate of electrochemical reactions and thus enhancing the production of substances.

In the problem, the current given is \(4.0 \times 10^4\) amperes. Knowing this, you can calculate the total charge that has flowed through the cell over a given period, like one day, using the equation:

\[\text{Charge} = \text{Current} \times \text{Time} \].

Since we know the time (24 hours or 86400 seconds), you can determine how much charge has been transferred. This charge is crucial in determining how many moles of electrons have participated in the half-reactions.

• Current is essential for calculating the total charge.
• Determines the rate of electrolysis in a cell.

Energy consumption (kWh)

Energy consumption is an important factor in electrolysis, particularly concerning efficiency and cost. In this case, energy is consumed in kilowatt-hours (kWh), a common unit of energy that power companies use to measure and bill electricity use.

To calculate energy consumption for the electrolysis cell operating at 7.0 volts and \(4.0 \times 10^4\) amperes, use the formulas:

• Find Power (\(P\)): \[P = IV\]
• Then find Energy (\(E\)) in Joules: \[E = P \times \text{Time}\]
• Convert Joules to kWh: \[E = \frac{E \text{ in Joules}}{3.6 \times 10^6}\]

The conversion is critical because households and businesses often reference energy use in kWh rather than in Joules for transparency with electricity expenses.

Molar mass

Molar mass is the mass of one mole of a given substance, expressed in grams per mole (g/mol). It's a fundamental concept in chemistry that helps translate the number of moles into a tangible mass that can be measured and understood.

For this electrolysis example, we calculate the mass of products formed, sodium (\(\text{Na}\)) and chlorine gas (\(\text{Cl}_2\)). Knowing the molar masses, which are 23 g/mol for sodium and 71 g/mol for chlorine gas, is necessary for these calculations.

Once the moles of these substances are determined through the electrolysis process, you multiply the number of moles by the molar mass to find the mass:

• Mass = Moles × Molar Mass
• Turns the abstract idea of moles into concrete measurements like thousands of grams or kilograms

Use of molar mass allows chemists to predict and measure the yield of reactions, making it invaluable for both theoretical and practical chemistry.