Problem 88
Question
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?
Step-by-Step Solution
Verified Answer
(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.
1Step 1: 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
2Step 2: 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 \(I = 3.50\, \mathrm{A}\).
Key Concepts
Faraday's LawMagnesium ProductionElectrochemical Calculations
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 (\(n\)).
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?
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.
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:
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.
Other exercises in this chapter
Problem 86
(a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode
View solution Problem 87
(a) \(\mathrm{A} \mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after
View solution Problem 89
A voltaic cell is based on the reaction $$ \mathrm{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ Under standard condit
View solution Problem 90
Consider the voltaic cell illustrated in Figure \(20.5\), which is based on the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn
View solution