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Chapter 20: Problem 92

Metallic magnesium can be made by the electrolysis of molten \(\mathrm{MgCl}_{2}\) (a) What mass of \(\mathrm{Mg}\) is formed by passing a current of 4.55 A through molten \(\mathrm{MgCl}_{2}\), for 4.50 days? (b) How many minutes are needed to plate out \(25.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\) using \(3.50 \mathrm{~A}\) of current?

Short Answer

Expert verified
For part (a), the mass of magnesium produced is 222.92 g. For part (b), it will take 947.77 minutes to plate out 25 g of magnesium using a current of 3.50 A.

Step by step solution

01

Identify relevant information

Given data for (a): - Current (I): 4.55 A - Time (t): 4.50 days - Molar mass of Mg: 24.31 g/mol - Faraday's constant (F): 96485 C/mol Given data for (b): - Mass of Mg: 25.00 g - Current (I): 3.50 A
02

Calculate the charge passed through the electrolyte for part (a)

First, convert time from days to seconds: Time (t) = 4.50 days * (24 hours/1 day) * (60 minutes/1 hour) * (60 seconds/1 minute) = 388800 seconds Next, calculate the charge (Q) using Q = It: Q = (4.55 A) * (388800 seconds) = 1769040 C
03

Calculate the moles of Mg produced for part (a)

Use Faraday's Law to find the moles of Mg produced (n): n = Q / (2 * F) n = 1769040 C / (2 * 96485 C/mol) = 9.17 mol
04

Calculate the mass of Mg produced for part (a)

Use the molar mass of Mg to find the mass of Mg produced: Mass of Mg = n * Molar Mass of Mg Mass of Mg = 9.17 mol * 24.31 g/mol = 222.92 g So, the mass of magnesium produced is 222.92 g.
05

Calculate the charge passed through the electrolyte for part (b)

We know the mass of Mg to be produced is 25 g. First, we'll find the moles of Mg: n = Mass of Mg / Molar Mass of Mg n = 25 g / 24.31 g/mol = 1.03 mol Now we'll use Faraday's Law to find the charge required: Q = n * (2 * F) Q = 1.03 mol * (2 * 96485 C/mol) = 199032.7 C
06

Calculate time needed to plate out Mg for part (b)

Use the relationship Q = It to find the time (t): t = Q / I t = 199032.7 C / 3.50 A = 56866.47 seconds Now convert the time to minutes: t = 56866.47 seconds * (1 minute / 60 seconds) = 947.77 minutes So, it will take 947.77 minutes to plate out 25 g of Mg using a current of 3.50 A.

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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 plays a crucial role in understanding the process of electrolysis. It essentially allows us to determine the amount of substance produced at an electrode during electrolysis. The essence of Faraday's Law is captured in the formula:\[ n = \frac{Q}{zF} \]where:
  • \( n \) is the amount of substance (in moles) produced,
  • \( Q \) is the total electric charge passed through the substance (in coulombs),
  • \( z \) is the number of electrons involved in the reaction (which is 2 for magnesium since it forms \( \text{Mg}^{2+} \)),
  • \( F \) is the Faraday's constant, approximately 96485 C/mol.
This component of the law is essential when calculating the mass of magnesium formed during electrolysis. By knowing the charge and using Faraday's Law, you can deduce the number of moles of magnesium produced. This approach is vital for accurately predicting the outcomes of electrochemical reactions.
Current and Charge Calculation
Calculating the charge that passes through the electrolyte is an important step in electrolysis problems. The charge \( Q \) can be obtained through the relationship:\[ Q = It \]where:
  • \( I \) is the current in amperes (A), and
  • \( t \) is time in seconds (s).
To convert time given in days to seconds, use:\[ t = \text{days} \times 24 \times 60 \times 60 \]As a result, if 4.55 A of current flows for 4.5 days, the total charge is calculated by substituting these values into the equation for \( Q \). It involves a straightforward multiplication to determine the total coulombs passing through the circuit.This computed charge is then used to find out how many moles of magnesium can be produced, making it a critical calculation for understanding the efficiency and completeness of the electrolysis process.
Moles and Molar Mass
The concept of moles and molar mass is fundamental to chemical calculations. In the context of electrolysis, we use these concepts to link the microscopic world of atoms to measurable macroscopic quantities.To find the moles of substance produced, the formula:\[ n = \frac{\text{Mass}}{\text{Molar Mass}} \]comes into play. Molar mass is the weight of one mole of a substance and is typically expressed in grams per mole (g/mol).For magnesium, given that its molar mass is 24.31 g/mol, we can calculate the number of moles from a given mass. For example, if you need to produce 25 g of Mg, the calculation would be:\[ n = \frac{25 \text{ g}}{24.31 \text{ g/mol}} \approx 1.03 \text{ mol} \]This calculated moles of magnesium are then essential for determining the charge required during electrolysis, thereby linking Faraday's Law, current, charge, and the physical quantity of magnesium required.

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Most popular questions from this chapter

Li-ion batteries used in automobiles typically use a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode in place of the \(\mathrm{LiCoO}_{2}\) cathode found in most Li-ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a \(\mathrm{LiCoO}_{2}\) cathode, approximately \(50 \%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a \(\operatorname{LiMn}_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

Complete and balance the following half-reactions. In each case, indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (acidic solution) (b) \(\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (d) \(\mathrm{S}(s)(a q) \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (acidic solution) (e) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{OH}^{-}(a q)\) (basic solution)

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

The purification process of silicon involves the reaction of silicon tetrachloride vapor \(\left(\mathrm{SiCl}_{4}(g)\right)\) with hydrogen to \(1250^{\circ} \mathrm{C}\) to form solid silicon and hydrogen chloride. \((\mathbf{a})\) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant?

(a) A \(\mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after 2.00 days? (b) What amperage is required to plate out \(0.250 \mathrm{~mol} \mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of \(8.00 \mathrm{~h}\) ?
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