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Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgC}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of \(97,000 \mathrm{~A}\) for a period of \(24 \mathrm{~h}\) ?

Short Answer

Expert verified
(a) An aqueous solution of \(\mathrm{MgCl}_{2}\) is not used for electrolysis because water molecules in the solution tend to get reduced at the cathode instead of magnesium ions, resulting in the production of hydrogen gas rather than magnesium metal. (b) The mass of magnesium formed by passing a current of 97,000 A for 24 hours with a 96% efficiency is approximately 1,009,000 grams, or 1,009 kg.

Step by step solution

01

Part A - Using Aqueous Solution

In the electrolysis process, we want to extract magnesium metal from \(\mathrm{MgCl}_{2}\). However, if an aqueous solution of \(\mathrm{MgCl}_{2}\) is used instead of a molten solution, the water molecules in the solution will tend to get reduced at the cathode instead of the magnesium ions. This occurs because the reduction potentials of water and magnesium ions are such that favor the reduction of water. Thus, hydrogen gas will be produced, rather than magnesium metal.
02

Part B - Total Charge

To calculate the total charge passed through the cells, we'll use the formula: Charge (Q) = Current (I) × Time (t) Given the current, I = 97,000 A, and the time, t = 24 hours, we first need to convert hours to seconds (1 hour = 3600 seconds). t = 24 hours × 3600 s/h = 86,400 s Now, let's calculate the total charge (Q): Q = 97,000 A × 86,400 s = 8,380,800,000 C (coulombs)
03

Part B - Moles of Magnesium

Now, let's use Faraday's Law of electrolysis to find how many moles of magnesium are formed. Faraday's constant, denoted as "F," is approximately 96,485 C/mol. n = Q / (z × F) Here, n is the number of moles, z is the number of electrons transferred during the reaction (for magnesium, z = 2, as it is divalent), and F is Faraday's constant. n = 8,380,800,000 C / (2 × 96,485 C/mol) = 43,385 mol
04

Part B - Mass of Magnesium Produced

To find the mass of magnesium produced, we'll multiply the number of moles formed by the molar mass of magnesium (Mg). The molar mass of magnesium is approximately 24.3 g/mol. Taking into account the efficiency, we also multiply by 0.96: Mass of Mg = 43,385 mol × 24.3 g/mol × 0.96 = 1,009,000 g Hence, the mass of magnesium formed by passing a current of 97,000 A for 24 hours when the process is 96% efficient is approximately 1,009,000 grams, or 1,009 kg.

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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
Faraday's Law of Electrolysis is a fundamental principle that quantifies the relationship between the electric charge passed through an electrolyte and the amount of substance deposited at an electrode. According to this law, the amount of chemical change or the mass of elements deposited in electrolysis is directly proportional to the total electric charge passed through the substance. The law can be broken down into two separate parts: First, the amount of a substance produced at an electrode during electrolysis is proportional to the amount of charge passed. Second, the amount of different substances liberated by the same quantity of electricity passing through the electrolyte is proportional to their chemical equivalent weights.

In a practical scenario, like the extraction of magnesium from magnesium chloride, Faraday's Law helps us calculate how many moles of magnesium will be produced by a certain amount of charge. Remember, the charge is the product of the current and the time during which the current flows, measured in coulombs (C), where 1 C equals the charge passed by a current of one ampere in one second. Therefore, knowing the molar mass of magnesium, we can convert moles to grams to find the actual mass of magnesium deposited.
Reduction Potentials
Reduction potentials are a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each reduction reaction has an associated standard reduction potential, which is measured in volts. In an electrolytic cell such as one that performs the electrolysis of magnesium chloride, the substance with the higher reduction potential will be reduced preferentially.

For example, water has a higher reduction potential than magnesium ions, and hence, if electrolysis is performed in an aqueous solution, water will be reduced to hydrogen gas, not magnesium metal. This tendency is quantified by the reduction potential of each species involved in the reaction. It is crucial to select the correct electrolyte and conditions to ensure that the desired reaction occurs; in this case, molten \(\mathrm{MgCl}_{2}\) is used over an aqueous solution to ensure magnesium metal is deposited.
Calculating Moles in Electrolysis
Calculating the moles of a substance produced during electrolysis involves applying Faraday's Law, which relies on the total charge passed through the electrolyte and the number of electrons involved in the redox reaction. To calculate the moles (\( n \) of element produced, you divide the total charge (\( Q \) in coulombs) by the product of the charge number of electrons (\( z \) which is typically 2 for divalent metal ions like magnesium) and Faraday's constant (\( F \) approximately 96,485 C/mol).

\( n = \frac{Q}{z × F} \)
Once the charge has been calculated, applying the formula gives you the number of moles of substance that have been deposited at the electrode. This step is vital to determine the quantity of the product obtained from the electrolysis process.
Molar Mass
Molar mass is defined as the mass of one mole of a substance, expressed in grams per mole (\( g/mol \)). It is an essential concept in chemistry that links the microscopic world of atoms and molecules to the macroscopic world we can measure and observe. In electrolysis, understanding molar mass allows you to convert the number of moles of a substance produced to the mass of that substance.

For the element magnesium, the molar mass is approximately 24.3 \(\text{g/mol}\). When you multiply the number of moles of magnesium obtained from the electrolysis by its molar mass, you find out the mass in grams of magnesium that has been deposited. This calculation is crucial when determining the yield of electrolytic processes in industrial applications, such as producing metallic magnesium from its chloride salt.

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Most popular questions from this chapter

A voltaic cell utilizes the following reaction: $$ 4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M}^{[}\left[\mathrm{Fe}^{3+}\right]=\) \(0.010 \mathrm{M}, P_{\mathrm{o}_{2}}=0.50 \mathrm{~atm}\), and the \(\mathrm{pH}\) of the solution in the cathode half-cell is \(3.50\) ?

Consider a redox reaction for which \(E^{b}\) is a negative number. (a) What is the sign of \(\Delta G^{\text {e }}\) for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than 1? (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings? [Section 20.5]

At \(298 \mathrm{~K}\) a cell reaction has a standard cell potential of \(+0.17 \mathrm{~V}\). The equilibrium constant for the reaction is \(5.5 \times 10^{5}\). What is the value of \(n\) for the reaction?

The \(K_{s p}\) value for \(\mathrm{PbS}(s)\) is \(8.0 \times 10^{-23}\), By using this value together with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction $$ \mathrm{PbS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{S}^{2-}(a q) $$

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: (a) Fe(s) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Ca}(s)\) or \(\mathrm{Al}(s)\) (c) \(\mathrm{H}_{2}\) (g, acidic solution) or \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (d) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{lO}_{3}^{-}(a q)\) 20.44 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\operatorname{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{2}(\mathrm{~g})\) 20.45 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an exidant or a reductant: (a) \(\mathrm{Cl}_{2}\) (g), (b) \(\mathrm{MnO}_{4}^{-}\)(aq, acidic solution), (c) \(\mathrm{Ba}\) (s), (d) \(\mathrm{Zn}(\) s). 20.46 Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(\mathrm{aq})\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{CO}_{3}^{-}(\mathrm{aq})\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ? \(20.47\) (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{C}_{2}, \mathrm{O}_{2}\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{AL}\). 20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in acidic solution: \(\mathrm{Br}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Zn}, \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} ?\) (b) Which of the following is the strongest reducing agent, and which is the weaket in acidic solution: \(\mathrm{F}^{-}, \mathrm{Zn}, \mathrm{N}_{1}{ }^{+}\), \(\mathrm{I}_{\mathrm{n}} \mathrm{NO}\) ?

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