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Chapter 21: Problem 101

Electrolysis of molten \(\mathrm{MgCl}_{2}\) is the final production step in the isolation of magnesium from seawater by the Dow process (Section 22.4). Assuming that \(45.6 \mathrm{~g}\) of Mg metal forms, (a) How many moles of electrons are required? (b) How many coulombs are required? (c) How many amps will produce this amount in \(3.50 \mathrm{~h} ?\)

Short Answer

Expert verified
3.752 moles of electrons, 362,030 coulombs, 28.73 amps.

Step by step solution

01

- Determine moles of Mg

First, find the moles of Mg formed using its molar mass. The molar mass of Mg is 24.305 g/mol. Therefore, the moles of Mg can be calculated as:\[ n_{Mg} = \frac{45.6 \text{ g}}{24.305 \text{ g/mol}} \approx 1.876 \text{ mol} \]
02

- Calculate moles of electrons required

Each mole of Mg requires 2 moles of electrons because of the reduction half-reaction:\[ \mathrm{Mg}^{2+} + 2e^{-} \to \mathrm{Mg} \]Thus, the moles of electrons required is:\[ n_{e^{-}} = 2 \times n_{Mg} = 2 \times 1.876 \approx 3.752 \text{ mol} \]
03

- Calculate coulombs required

Use the Faraday constant, which is approximately 96485 C/mol of electrons. The total charge required is:\[ Q = n_{e^{-}} \times F = 3.752 \times 96485 \approx 362,030 \text{ C} \]
04

- Determine the amperage required over 3.50 hours

Convert 3.50 hours to seconds:\[ 3.50 \, \text{hours} \times 3600 \text{ s/hour} = 12600 \text{ s} \]The current (in amps) can be found using the formula:\[ I = \frac{Q}{t} = \frac{362,030 \text{ C}}{12600 \text{ s}} \approx 28.73 \text{ A} \]

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Key Concepts

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

Electrolysis
Electrolysis involves using electrical energy to drive a non-spontaneous chemical reaction. In this problem, the electrolysis of molten magnesium chloride ( MgCl₂) is considered. During the process, MgCl₂ is broken down into its ions, Mg^{2+} and Cl^{-} . Applying an electric current allows for the reduction of Mg^{2+} ions to form magnesium metal. This process is critical in producing pure magnesium from its compounds found in nature, specifically in seawater.
Moles of Electrons
To understand how many moles of electrons are required for the reduction of MgCl₂, we need to first determine the moles of magnesium metal produced. Given in the problem is the mass of magnesium, which is 45.6 g. Using the molar mass of magnesium ( 24.305 \text{ g/mol} ), we can calculate moles of magnesium as follows:
\[ n_{Mg} = \frac{45.6 \text{ g}}{24.305 \text{ g/mol}} \ \text{ moles of Mg} \ \text{ moles of electrons }\]Next, we observe that each mole of magnesium requires 2 moles of electrons based on the reaction Mg^{2+} + 2e^{-} \to Mg . Therefore, to determine the moles of electrons required, we multiply the moles of magnesium by 2:
\[ n_{e^{-}} = 2 \times n_{Mg} \ \text{ moles of electrons }\]
Coulombs in Electrolysis
Coulombs measure the electric charge involved in the electrolysis process. The total charge needed depends on the moles of electrons transferred and is calculated using the Faraday constant ( 96485 \text{ C/mol} ). To find the total charge Q s required for the reduction of magnesium, we multiply the moles of electrons by the Faraday constant:
\[ Q = n_{e^{-}} \times F \ \text{ coulombs }\]This calculation gives us the total number of coulombs necessary to produce the given mass of magnesium via electrolysis.
Amperage Calculation
Amperage is a measure of the flow of electric charge per second, with the unit called an amp or ampere ( A ). In this scenario, we need to determine the amperage required to produce 45.6 g of magnesium in 3.50 hours.
First, convert hours to seconds:
\[ 1 \text{ hour} = 3600 \text{ seconds } \ \text{ total time in seconds: } 3.50 \text{ hours } \times 3600 \text{ s/hour } = 12600 \text{ s } \]Next, use the formula to find the amperage:
\[ I = \frac{Q}{t} \ \text{ amperage calculated using total charge and total time }\]This will give us the required current to achieve the electrolysis in the specified time frame.

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Most popular questions from this chapter

Electrodes used in electrocardiography are disposable, and many of them incorporate silver. The metal is deposited in a thin layer on a small plastic "button," and then some is converted to AgCl: $$\operatorname{Ag}(s)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \operatorname{AgCl}(s)+\mathrm{e}^{-}$$ (a) If the surface area of the button is \(2.0 \mathrm{~cm}^{2}\) and the thickness of the silver layer is \(7.5 \times 10^{-6} \mathrm{~m},\) calculate the volume (in \(\mathrm{cm}^{3}\) ) of \(\mathrm{Ag}\) used in one electrode. (b) The density of silver metal is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many grams of silver are used per electrode? (c) If \(\mathrm{Ag}\) is plated on the button from an \(\mathrm{Ag}^{+}\) solution with a current of \(12.0 \mathrm{~mA}\), how many minutes does the plating take? (d) If bulk silver costs \(\$ 28.93\) per troy ounce \((31.10 \mathrm{~g}),\) what is the cost (in cents) of the silver in one disposable electrode?

The \(\mathrm{MnO}_{2}\) used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is $$\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}$$ If a current of \(25.0 \mathrm{~A}\) is used, how many hours are needed to produce \(1.00 \mathrm{~kg}\) of \(\mathrm{MnO}_{2} ?\) At which electrode is the \(\mathrm{MnO}_{2}\) formed?

Balance the following skeleton reactions, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)\) [acidic] (b) \(\mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)[\mathrm{basic}]\) (c) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2}(g)\) [acidic]

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions: (a) \(\mathrm{Al}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cr}(s)\) (b) \(\mathrm{Cu}^{2+}(a q)+\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$\mathrm{Cu}(s)+\mathrm{SO}_{4}^{2-}(a q)+4 \mathrm{H}^{+}(a q)$$

Use the following half-reactions to write three spontaneous reactions, calculate \(E_{\text {cell }}^{\circ}\) for each reaction, and rank the strengths of the oxidizing and reducing agents: (1) \(\mathrm{Al}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(s) \quad E^{\circ}=-1.66 \mathrm{~V}\) (2) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NO}_{2}^{-}(a q) \quad E^{\circ}=0.867 \mathrm{~V}\) (3) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(I)+2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_{3}^{2-}(a q)+2 \mathrm{OH}^{-}(a q)\) \(E^{\circ}=0.93 \mathrm{~V}\)
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