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Chapter 17: Problem 102

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 10鲁 kg, so we need to first convert this mass to grams and then to moles: Mass of barium (g) = 1.00 x 10鲁 kg x (1000 g/kg) = 1.00 x 10鈦� 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 (Ba虏鈦� + 2e鈦� 鈫� 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 10鲁 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 10鲁 kg of barium from the electrolysis of molten barium chloride is approximately 9.79 x 10鈦� 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鈥檚 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 10鲁 kg (or 1.00 x 10鈦� 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 10鈦� Coulombs is needed over 4 hours (14400 seconds).
By substituting these values, the required current is calculated to be approximately 9.79 x 10鈦� 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

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 A. How long will it take to plate out \(99 \%\) of the platinum in 0.50 L of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{K}\) is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83} $$ a. Calculate \(\mathscr{E}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

Consider a concentration cell that has both electrodes made of some metal M. Solution \(A\) in one compartment of the cell contains \(1.0 \mathrm{M} \mathrm{M}^{2+}\). Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathbf{M}^{2+}(a q)+\mathbf{S O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25 \(^{\circ}\) C. Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\)

Consider the following half-reactions: $$\begin{array}{ll} \operatorname{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \operatorname{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 \(M\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.
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