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A factory wants to produce \(1.00 \times 10^{3} \mathrm{kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{h}\) to accomplish this?

Short Answer

Expert verified
The current that must be applied for 4.00 hours to produce \(1.00 \times 10^3 \, \mathrm{kg}\) of barium from the electrolysis of molten barium chloride is approximately \(9.79 \times 10^4 \, \mathrm{A}\).

Step by step solution

01

Calculate the number of moles of barium to be produced

For barium, the molar mass (Ba) is approximately 137 g/mol. We're given the target production of barium is 1.00 x 10³ kg, so we need to first convert this mass to grams and then to moles: Mass of barium (g) = 1.00 x 10³ kg x (1000 g/kg) = 1.00 x 10⁶ g Now, convert the mass to moles: Moles = (Mass of barium) / (Molar mass of barium) Moles = \( \frac{1.00 \times 10^6 \, \mathrm{g}}{137 \, \mathrm{g/mol}} \approx 7.30 \times 10^3 \, \mathrm{mol} \)
02

Calculate the total charge required to produce the desired amount of barium

In electrolysis, two moles of electrons are required to produce one mole of barium (Ba²⁺ + 2e⁻ → Ba). From Faraday's law of electrolysis, 1 mole of electrons carries a charge, called a Faraday (F) which is approximately 96485 C/mol (Coulombs per mole). Total charge (C) = moles of barium x (moles of electrons per mole of barium) x Faraday's constant Total charge (C) = \( 7.30 \times 10^3 \, \mathrm{mol} \times 2 \, \mathrm{mol} \, \mathrm{e}^{-}/\mathrm{mol} \, \mathrm{Ba} \times 96485 \, \mathrm{C}/\mathrm{mol} \, \mathrm{e}^{-} \approx 1.41 \times 10^9 \, \mathrm{C} \)
03

Calculate the required current for the given time

We are given the total time required to produce 1.00 x 10³ kg of barium as 4.00 hours. To find the required current, first, we need to convert the time to seconds: Total time (s) = 4.00 hours x (3600 s/h) = 14400 s Now, from the formula: current (I) = total charge (C) / total time (s), we have: Current (I) = \( \frac{1.41 \times 10^9 \, \mathrm{C}}{14400 \, \mathrm{s}} \approx 9.79 \times 10^4 \, \mathrm{A} \) So, the current that must be applied for 4.00 hours to produce 1.00 x 10³ kg of barium from the electrolysis of molten barium chloride is approximately 9.79 x 10⁴ A.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law of Electrolysis
Electrolysis is a fascinating process where electricity is used to drive a chemical reaction. A key principle in understanding electrolysis is Faraday's Law. Essentially, Faraday's Law of Electrolysis tells us how much material will be deposited or dissolved at an electrode during electrolysis based on the quantity of electricity used. This law states that the amount of substance altered at an electrode during electrolysis is proportional to the quantity of electricity that passes through the system.
Faraday's Law is summarized by the formula: \[ m = \frac{Q}{Fz} \] where:
  • \( m \) is the mass of the substance produced at the electrode in grams,
  • \( Q \) is the total electric charge passed through the system in Coulombs (C),
  • \( F \) is Faraday's constant, approximately 96485 C/mol, which is the charge of one mole of electrons,
  • \( z \) is the number of electrons involved in the reaction for each ion of the substance.

Understanding Faraday’s Law is fundamental in determining how much of a compound can be produced during electrolysis, helping predict the quantities in large-scale manufacturing processes such as in factories.
Moles Calculation
One of the first steps in the electrolysis process involves calculating the number of moles of the substance to be produced or required for the reaction. The concept of a mole is crucial because it allows chemists to convert between mass and the number of particles in a substance.
To find the number of moles (\( n \)), you use the following formula: \[ n = \frac{m}{M} \] where:
  • \( m \) is the mass of the substance in grams,
  • \( M \) is the molar mass of the substance in grams per mole.

In our exercise, we need to produce barium where the target mass is given, and the molar mass of barium is known (~137 g/mol). Using these values, we can calculate that for 1.00 x 10³ kg (or 1.00 x 10⁶ g), roughly 7300 moles of barium are required.
This step is vital because knowing the moles helps determine how many electrons are involved in the electrochemical reaction, thus guiding the subsequent calculation of the current needed.
Current Calculation in Electrolysis
Once you have determined the total charge required from Faraday's law and the moles of the material to be produced, the next step is to calculate the current required for the electrolysis process. Current is the flow of electric charge, and for electrolysis, knowing the current helps control the rate at which the chemical reaction occurs.
The formula for calculating the current (\( I \)) required is: \[ I = \frac{Q}{t} \] where:
  • \( Q \) is the total charge in Coulombs (obtained from the product of moles of electrons, Faraday's constant, and moles of the material),
  • \( t \) is the total time in seconds for which the current is applied.

In the exercise, it was determined that a charge of approximately 1.41 x 10⁹ Coulombs is needed over 4 hours (14400 seconds).
By substituting these values, the required current is calculated to be approximately 9.79 x 10⁴ Amperes (A).
Understanding how to calculate current is crucial for planning and executing electrolysis in industrial settings, ensuring the efficient and safe operation of electrochemical processes.

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Most popular questions from this chapter

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

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

Combine the equations $$ \Delta G^{\circ}=-n F \mathscr{E}^{\circ} \quad \text { and } \quad \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} $$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

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