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Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the total energy requirement for this electrolysis if the applied emf is \(+5.00 \mathrm{~V} ?\)

Short Answer

Expert verified
Q = \(7.5 \times 10^{3}\) A × (48 × 60 × 60 s) = \(3.24 \times 10^{9}\) C #Step 2: Calculate moles of calcium produced# #tag_title# Determine the moles of calcium produced considering the cell efficiency #tag_content# Using Faraday's law, we can find the moles of calcium produced (n) by dividing the total charge passed (Q) by the product of Faraday's constant (F) and the number of electrons transferred (in this case, 2 for calcium): n = (Q x efficiency) / (F x number of electrons) Efficiency = 0.68 Faraday's constant (F) = \(9.65 \times 10^{4} \mathrm{C/mol}\) n = \((3.24 \times 10^{9} \mathrm{C} \times 0.68) / (9.65 \times 10^{4} \mathrm{C/mol} \times 2)\) = 12,170 mol Ca #Step 3: Calculate the mass of calcium produced# #tag_title# Calculate the mass of calcium produced from moles #tag_content# To convert moles of calcium to mass, we multiply the moles (n) by the molar mass of calcium (M): Mass of calcium = n x M M(Ca) = 40.08 g/mol Mass of calcium = 12,170 mol Ca × 40.08 g/mol = 487,600 g or 487.6 kg #Step 4: Calculate the total energy requirement# #tag_title# Determine the total energy requirement for the electrolysis process #tag_content# The total energy requirement for the electrolysis process can be found using Ohm's law: Energy = Voltage x Charge Energy = (Voltage) x (Current) x (Time) Voltage = +5.00 V Energy = (+5.00 V) x (\(7.5 \times 10^{3}\) A) × (48 × 60 × 60 s) = \(1.62 \times 10^{10}\) J #Answer# (a) The mass of calcium produced is 487.6 kg. (b) The total energy requirement for the electrolysis is \(1.62 \times 10^{10}\) J.

Step by step solution

01

Determine the total charge passed during the electrolysis

To find the total charge passed during the electrolysis, we need to multiply the given current (in Amperes) by the time the current is passed (in seconds). The formula is: Q = current x time Current = \(7.5 \times 10^{3}\) A Time = 48 hours = 48 x 60 x 60 seconds

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Key Concepts

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

Faraday's Laws of Electrolysis
One of the central theories in understanding electrolysis is Faraday's laws. These laws explain the relationship between the amount of electrical charge passed through a substance and the amount of material that is deposited, dissolved, or liberated at an electrode during the electrochemical reaction.

There are two laws of Faraday to consider:
  • Faraday's First Law: The mass (m) of the substance altered at an electrode during electrolysis is directly proportional to the amount of electric charge (Q) passed through the electrolyte. It tells us that mass is directly proportional to charge, which can be represented mathematically as:
    \[\begin{equation} m \text{is directly proportional to } Q. \text{Thus, } m = Z Q, \text{where } Z \text{is the electrochemical equivalent of the substance}. \text{\(Z = \frac {M} {n \times \text{F}}\)}, \text{\(M\)} \text{is the molar mass, } n \text{is the number of moles of electrons, and } \text{F}\ \text{is Faraday's constant}. \text{The units for } Z \text{are typically in grams per coulomb (g/C)}. \text{\(F\)} \text{is approximately } 96485 \frac{C}{mol}. \text{\(F\)} \text{represents} 96485\text{coulombs/mole, which is the charge of one mole of electrons.}
  • Faraday's Second Law: When the same electric charge passes through multiple electrolytes, the mass ofsubstances liberated at each electrode is directly proportional to their chemical equivalent (E).
Understanding Faraday's laws enables students to calculate the exact amount of a substance produced or consumed during electrolysis by calculating the charge and utilizing the chemical equivalents of the substances involved.
Electrochemical Cells
The basis of any electrolysis procedure is an electrochemical cell. These cells convert electrical energy into chemical energy and vice versa. An electrolytic cell, a type of electrochemical cell, is used in the process of electrolysis to drive a non-spontaneous chemical reaction.

In an electrolytic cell:
  • Electrical energy is applied to bring about a chemical change, which is what happens in the electrolysis of calcium chloride to produce calcium metal and chlorine gas.
  • The cell contains a cathode (negative electrode) where reduction occurs, and an anode (positive electrode) where oxidation happens. For example, in the case of calcium chloride, calcium ions are reduced to form calcium metal at the cathode, while chloride ions are oxidized to form chlorine gas at the anode.
  • An external voltage source is necessary to initiate and sustain the electrolysis process, as the reaction is non-spontaneous.

Efficiency of Electrochemical Cells

Efficiency is a key concept when it comes to electrolysis. It is important to note that not all the electrical energy is converted into chemical energy due to losses such as heat and the overpotential necessary to drive the reaction. The efficiency, often given as a percentage, tells us what portion of the electrical energy contributes to the intended chemical transformation.
Mole Concept
In chemistry, the mole concept is a fundamental aspect that links the mass of substances to the number of particles and the amount of substance. A mole is defined as exactly 6.022 x 10^23 (Avogadro's number) of particles, which can be atoms, molecules, ions, or electrons.

When calculating the mass of a product formed or reactant consumed in a chemical reaction, chemists often use the mole concept. For instance, in electrolysis:
  • You need to know the number of moles of electrons transferred in the reaction.
  • Relate the moles of electrons to the total charge passed (expressed in coulombs), since 1 mole of electrons carries a charge equivalent to Faraday's constant (\(96485 C/mol\)).
  • With Faraday's first law, we can relate the charge to the mass of substance deposited or dissolved.
Knowing the molar mass of the substance and using the mole concept, it's possible to calculate the actual mass of material produced in an electrolytic cell, in this case, calcium from calcium chloride, using the electrical charge passed during the electrolysis.

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Most popular questions from this chapter

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) \(\mathrm{NO}_{3}^{-}(a q)-\cdots \rightarrow \mathrm{NO}(g)\) (acidic solution) (d) \(\mathrm{O}_{2}(\mathrm{~g})-\longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (acidic solution) (e) \(\mathrm{Mn}^{2+}(a q)-\rightarrow \rightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (f) \(\mathrm{Cr}(\mathrm{OH})_{3}(s)-\cdots \mathrm{CrO}_{4}^{2-}(a q)\) (basic solution) (g) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution)

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