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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 \(3.50\) days? (b) How many minutes are needed to plate out \(10.00 \mathrm{~g} \mathrm{Mg}\) from molten \(\mathrm{MgCl}_{2}\), using \(3.50 \mathrm{~A}\) of current?

Short Answer

Expert verified
(a) The mass of magnesium formed by passing a current of 4.55 A through molten MgCl\(_2\) for 3.50 days is approximately 281.99 g. (b) Plating out 10.00 g of magnesium from molten MgCl\(_2\) using 3.50 A of current requires approximately 139.56 minutes.

Step by step solution

01

To convert the given time of 3.50 days into seconds, use the following unit conversion: \[t_{sec} = 3.50\, \mathrm{days} \times \frac{24\, \mathrm{hours}}{1\, \mathrm{day}} \times \frac{60\, \mathrm{minutes}}{1\, \mathrm{hour}} \times \frac{60\, \mathrm{seconds}}{1\, \mathrm{minute}}\]Calculate the magnesium mass

Use Faraday's law to calculate the mass of magnesium formed: \[m = \frac{I \times t_{sec} \times M}{n \times F}\]Where \(I = 4.55\, \mathrm{A}\), the time is converted to seconds, \(M = 24.31\, \mathrm{g/mol}\), \(n = 2\), and \(F = 96,485\, \mathrm{C/mol}\). #For Part (b)#Calculate the required charge
02

Use Faraday's law to calculate the required charge for depositing 10 g of Mg: \[Q = \frac{m \times n \times F}{M}\]Where \(m = 10.00\, \mathrm{g}\), \(n = 2\), \(F = 96,485\, \mathrm{C/mol}\), and \(M = 24.31\, \mathrm{g/mol}\). Calculate the time in minutes

Now, use the formula \(Q = I \times t\) and convert the time to minutes: \[t_{min} = \frac{Q}{I} \times \frac{1\, \mathrm{minute}}{60\, \mathrm{seconds}}\]Where \(Q\) is the required charge for depositing 10 g of Mg and \(I = 3.50\, \mathrm{A}\).

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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 laws of electrolysis are fundamental to understanding how electricity can cause chemical changes, such as the deposition or dissolution of elements during electrolysis. The first law states that the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity that passes through the cell. In simple terms, more electric charge means more product.

This concept is crucial in calculations involving electrochemical reactions, where Faraday's constant (\(F = 96,485 \, \text{C/mol}\)) represents the charge of one mole of electrons. The formula derived from Faraday's law, \(m = \frac{I \times t \times M}{n \times F}\), allows us to calculate the mass (\(m\)) formed based on the current (\(I\)), time (\(t\)), molar mass (\(M\)), and number of electrons exchanged (\(n\)).
  • For example, if you know the current and how long it flows, you can find out how much substance will be produced.
  • Understanding this principle is key to mastering electrochemical calculations as it links chemistry with the concept of electric charge.
Magnesium Production
Magnesium production through electrolysis involves extracting magnesium from its compound, usually magnesium chloride (\(\mathrm{MgCl}_{2}\)). The process is energy-intensive, as it requires molten salt for the electrolysis reaction. When a current is applied to the molten salt, magnesium ions move to the cathode, where they gain electrons and form metallic magnesium.

Why Use Electrolysis for Magnesium?
  • Magnesium is highly reactive; hence, obtaining it via simple chemical reactions isn't viable.
  • Electrolysis of molten salt is an efficient method to achieve the desired pure metal.
In our exercise, the electrolysis of \(\mathrm{MgCl}_{2}\) involves passing an electric current through the molten state, which breaks down the compound into magnesium and chlorine gas. The magnesium ions (\(\text{Mg}^{2+}\)) are reduced to magnesium (\(\text{Mg}\)) at the negative electrode while chlorine gas is formed at the positive electrode. Understanding this process is essential for grasping how electrolysis is utilized industrially to produce metals.
Electrochemical Calculations
Electrochemical calculations convert electrical measurements into chemical quantities like mass or volume. They are important in predicting the outcome of electrolysis reactions. We often use formulas involving current (\(I\)), time (\(t\)), molar mass (\(M\)), and Faraday’s constant (\(F\)) to perform these calculations.

Steps in Electrochemical Calculations:
  • Determine the total charge passed using current and time. Convert time to seconds if needed.
  • Use Faraday's law to link the electrical data with chemical changes. This includes calculating mass from charge or determining time required to produce a specific mass.
  • Conversion factors are crucial, like changing units from days to seconds or from grams to moles, ensuring accuracy.
By mastering these calculations, you can predict how much material will be produced in an electrolysis cell based on electrical inputs. This understanding is essential for industrial applications where precise amounts of elements are extracted or deposited.

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Most popular questions from this chapter

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\), and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}{ }^{2+}(a q)\). (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

The standard reduction potentials of the following halfreactions are given in Appendix E: $$ \begin{gathered} \mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{Ni}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ni}(s) \\ \mathrm{Cr}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(s) \end{gathered} $$ (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell emf, and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell emf, and calculate the value.

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q)-\cdots\) \(\mathrm{Fe}^{2+}(a q)+\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten \(\mathrm{NaCl}\) ?

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

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