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A factory wants to produce 1.00×103 kg barium from the electrolysis of molten barium chloride. What current must be applied for 4.00 h to accomplish this?

Short Answer

Expert verified
A current of 9.750 × 10⁴ A must be applied for 4.00 h to produce 1.00 × 10³ kg of barium from the electrolysis of molten barium chloride (BaCl2).

Step by step solution

01

Determine the molar mass of barium

From the periodic table, we know that the molar mass of barium (Ba) is 137.33 g/mol.
02

Convert the mass of barium into moles

Next, we will find the number of moles of barium (Ba) present in 1000 kg using the following formula: Number of moles (n) = mass (m) / molar mass (M) n(Ba) = 1.00×103kg137.33g/mol Before solving, we need to convert kg to g: 1.00 × 10³ kg = 1.00 × 10³ × 10³ g = 1.00 × 10⁶ g Now, we can calculate the number of moles: n(Ba) = 1.00×106g137.33g/mol = 7.280 × 10³ mol
03

Calculate the total charge needed

To find the total charge (Q) needed, we will use Faraday's law of electrolysis: Q = n × F × z where: - n: number of moles - F: Faraday's constant (96,485 C/mol) - z: Charge number (for Ba, z = 2 e-) Q = (7.280 × 10³ mol) × (96,485 C/mol) × (2 e-) = 1.404 × 10⁹ C
04

Convert the time to seconds

The given time is 4.00 h. We need to convert this time to seconds: 4.00 h = 4.00 × 60 min/h × 60 s/min = 1.440 × 10⁴ s
05

Calculate the current needed

Lastly, we calculate the current (I) needed using the formula: Current (I) = Charge(Q)time(t) I = 1.404×109C1.440×104s = 9.750 × 10⁴ A Thus, a current of 9.750 × 10⁴ A must be applied for 4.00 h to produce 1.00 × 10³ kg of barium from the electrolysis of molten barium chloride (BaCl2).

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Key Concepts

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

Faraday's Law
Faraday's Law of electrolysis is a fundamental principle in the field of electrochemistry that connects the process of electrical energy conversion to chemical reactions. At its core, Faraday's Law states that the amount of substance produced at each electrode during electrolysis is directly proportional to the total electric charge passed through the electrolyte. This means that if you increase the amount of electricity, more chemical product will be formed.
The law can be summarized in the equation:
Q=n×F×z
where:
  • Q is the total charge in coulombs (C),
  • n is the number of moles of electrons exchanged,
  • F is Faraday’s constant (96,485C/mol), which represents the charge of one mole of electrons,
  • z is the number of electrons involved in the electrochemical reaction.
This law is crucial when calculating the charge required to produce a specific mass of substance during electrolysis, as demonstrated in the exercise to produce barium from barium chloride.
Molar Mass
Molar mass is a key concept in chemistry and is defined as the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It acts as a bridge between the mass of a substance and the amount in moles, allowing us to convert between these two quantities.
To find the molar mass of an element like barium (Ba), one can refer to the periodic table. For barium, the molar mass is listed as 137.33 g/mol. This means that one mole of barium atoms weighs 137.33 grams. When determining how much barium can be produced, we must first convert the desired mass from kilograms to grams and then use the molar mass to find the number of moles.
In the exercise, by converting 1000 kg of barium to grams and using its molar mass, students calculate that they require approximately 7.280 × 10³ moles of barium to complete the electrolysis process.
Current Calculation
The calculation of current, required during electrolysis, involves the relationship between charge, time, and current. Current, denoted by I, is the rate at which charge flows through a conductor and is measured in amperes (A).
According to the formula:
I=Qt
where:
  • I is the current in amperes,
  • Q is the total charge in coulombs,
  • t is the time for which the current flows, measured in seconds.
In the exercise, the total charge needed to electrolyze the barium chloride is calculated to be 1.404 × 10⁹ C. Given that the reaction occurs over 4 hours (converted to seconds as 1.440 × 10⁴ s), the required current to complete the process is determined to be 9.750 × 10⁴ A. This calculation demonstrates how electrochemical processes can efficiently convert electrical energy to chemical energy, vital for industrial applications.
Electrochemistry
Electrochemistry is a fascinating branch of chemistry that explores the intersection between chemical reactions and electricity. It involves the study of how electric currents can drive chemical reactions, a process commonly referred to as electrolysis.
Electrochemical processes have practical applications, from metal refining and electroplating to powering batteries and fuel cells. During electrolysis, an electrolytic cell is used to pass an electric current through a liquid or molten ionic substance, causing the ions to move and undergo redox reactions.
In the context of the original problem, the electrolysis of molten barium chloride is used to produce barium metal. This process demonstrates the real-world application of electrochemistry in producing elements by breaking down compounds into their constituents using an electric current. Understanding electrochemistry helps in grasping how industries achieve large-scale chemical transformations and the role that electrical energy plays in these processes.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: Cu2+(aq)+2eCu(s)E=0.34V V2+(aq)+2eV(s)E=1.20V In this cell, the copper compartment contains a copper electrode and [Cu2+]=1.00M, and the vanadium compartment contains a vanadium electrode and V2+ at an unknown concentration. The compartment containing the vanadium (1.00L of solution) was titrated with 0.0800MH2EDTA2 , resulting in the reaction H2EDTA2(aq)+V2+(aq)VEDTA2(aq)+2H+(aq) K=? The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 mLH2EDTA2 solution added. At the stoichiometric point, Ecell was observed to be 1.98 V . The solution was buffered at a pH of 10.00 . a. CalculateEcell before the titration was carried out. b. Calculate the value of the equilibrium constant, K, for the titration reaction. c. Calculate Ecell at the halfway point in the titration.

Consider the following reduction potentials: Co3++3eCoE=1.26VCo2++2eCoE=0.28V a. When cobalt metal dissolves in 1.0M nitric acid, will Co3+ or Co2+ be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of HNO3 to get a different result in part a? Concentrated HNO3 is about 16 M .

Consider the following half-reactions: IrCl63+3eIr+6ClE=0.77V PtCl42+2ePt+4ClE=0.73V PdCl42+2ePd+4ClE=0.62V A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0M in chloride ion and 0.020M 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.)

A chemist wishes to determine the concentration of CrO42 electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 111) and a silver wire coated with Ag2CrO4 . The E value for the following half-reaction is 0.446 V relative to the standard hydrogen electrode: Ag2CrO4+2e2Ag+CrO42 a. Calculate E cell  and ΔG at 25C for the cell reaction when [CrO42]=1.00mol/L . b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at 25C) in which [CrO42]=1.00×105M, what is the expected cell potential? d. The measured cell potential at 25C is 0.504 V when the coated wire is dipped into a solution of unknown [CrO42]. What is [CrO42] for this solution? e. Using data from this problem and from Table 18.1, calculate the solubility product (Ksp) for Ag2CrO4.

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

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