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Chapter 19: Problem 56

A quantity of \(0.300 \mathrm{~g}\) of copper was deposited from a \(\mathrm{CuSO}_{4}\) solution by passing a current of \(3.00 \mathrm{~A}\) through the solution for 304 s. Calculate the value of the Faraday constant.

Short Answer

Expert verified
The Faraday constant is approximately 96500 C/mol.

Step by step solution

01

Calculate Moles of Copper Deposited

The molar mass of copper (Cu) is approximately \(63.55 \, \text{g/mol}\). We can calculate the moles of copper deposited using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). Thus, \(\text{moles of Cu} = \frac{0.300 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.00472 \, \text{mol}\).
02

Determine the Charge Required to Deposit Copper

For copper, \(\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}\), meaning 2 moles of electrons are needed to reduce 1 mole of copper ions. Thus, the charge required is \(0.00472 \, \text{mol} \times 2 = 0.00944 \, \text{mol of electrons}\).
03

Calculate the Total Charge Flow in Coulombs

The total charge \(Q\) that flowed through the solution is calculated using the formula \(Q = I \times t\), where \(I\) is the current in amperes and \(t\) is the time in seconds. Therefore, \(Q = 3.00 \, \text{A} \times 304 \, \text{s} = 912 \, \text{C}\).
04

Calculate the Faraday Constant

The Faraday constant \(F\) is the total charge required to deposit one mole of electrons. Thus, the Faraday constant is calculated as \(F = \frac{Q}{n}\), where \(Q\) is the total charge and \(n\) is the moles of electrons. Therefore, \(F = \frac{912 \, \text{C}}{0.00944 \, \text{mol}} \approx 96500 \, \text{C/mol}\).

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Key Concepts

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

Electrochemistry Basics
Electrochemistry is the study of chemical reactions that involve the movement of electrons, which leads to the conversion between chemical energy and electrical energy. It plays a crucial role in processes like electroplating, batteries, and fuel cells. In electrochemical reactions, there are two distinct sites where reactions occur: the anode, where oxidation happens, and the cathode, where reduction takes place.

Understanding these processes is vital for calculating how elements, like copper, can be deposited or removed from solutions. In electroplating, for example, ions from a solution are reduced and deposited onto a surface, forming a thin layer of metal.

The efficiency and rate of these reactions depend on several factors, including the type of metal, the current applied, and the duration of the reaction.
Copper Deposition
Copper deposition is a specific application of electrochemistry where copper ions (\(\text{Cu}^{2+}\)) in a solution gain electrons and are transformed into solid copper metal. This process is represented by the equation: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\).

During this reaction, electrons are transferred to the copper ions, reducing them to form solid copper. This process is crucial not only in electroplating but also in processes like copper refining.

The quality of the deposited copper layer depends on the precise control of several factors, including current density, temperature, and solution composition.
Understanding Coulombs
Coulombs measure the quantity of charge in electrical circuits. One coulomb is equivalent to the charge of approximately 6.242 x 10^18 electrons. In electrochemistry, coulombs help quantify the amount of electric charge used in reactions.

To find out how much charge is used during electroplating, the formula \(Q = I \times t\) is used, where \(Q\) is the charge in coulombs, \(I\) is the current in amperes, and \(t\) is the time in seconds.

For example, if you pass a current of \(3.00 \, \text{A}\) for \(304 \, \text{s}\), you calculate the total charge as: \(Q = 3.00 \, \text{A} \times 304 \, \text{s} = 912 \, \text{C}\).
Molar Mass Explained
Molar mass is a fundamental concept that refers to the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It allows chemists to convert between the mass of a substance and the moles of the substance, which is crucial for balancing chemical equations and calculating reactant and product masses.

For copper, the molar mass is approximately \(63.55 \, \text{g/mol}\). This means that one mole of copper atoms weighs 63.55 grams.

In the context of copper deposition, calculating the moles from a given mass helps determine how much copper can be deposited from a solution and how many electrons are involved in the process.
Basics of Mole Calculation
Mole calculations allow us to quantify the amount of substance used or produced in a chemical reaction, using the concept of moles, which are based on Avogadro's number (approximately 6.022 x 10^23).

To calculate moles, you use the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). For example, if you have \(0.300 \, \text{g}\) of copper, you use copper's molar mass (63.55 g/mol) to find the moles: \(\text{moles of Cu} = \frac{0.300 \, \text{g}}{63.55 \, \text{g/mol}} \approx 0.00472 \, \text{mol}\).

Knowing the number of moles is essential for determining how much of other substances will interact with it, and for calculating the Faraday constant, which represents the charge per mole of electrons.

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Most popular questions from this chapter

Given that: $$ \begin{array}{ll} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)$$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (b) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (c) \(\mathrm{CN}^{-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{CNO}^{-}+\mathrm{MnO}_{2}\) (in basic solution) (d) \(\mathrm{Br}_{2} \longrightarrow \mathrm{BrO}_{3}^{-}+\mathrm{Br}^{-}\) (in basic solution) (e) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (in acidic solution)

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(\mathrm{X}\) and \(\mathrm{Y}\) are: $$ \begin{aligned} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow & \mathrm{Y} & &\left|E^{\circ}\right|=0.34 \mathrm{~V} \\\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow & \mathrm{X} & &\left|E^{\circ}\right|=0.25 \mathrm{~V} \end{aligned}$$ where the \(\|\) notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(X\) and \(Y\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(\mathrm{X}\) to SHE. (a) Are the \(E^{\circ}\) values of the halfreactions positive or negative? (b) What is the standard emf of a cell made up of \(X\) and \(Y ?\)

The concentration of sulfuric acid in the lead-storage battery of an automobile over a period of time has decreased from 38.0 percent by mass (density \(=1.29 \mathrm{~g} / \mathrm{mL}\) ) to 26.0 percent by mass ( \(1.19 \mathrm{~g} / \mathrm{mL}\) ). Assume the volume of the acid remains constant at \(724 \mathrm{~mL}\). (a) Calculate the total charge in coulombs supplied by the battery. (b) How long (in hours) will it take to recharge the battery back to the original sulfuric acid concentration using a current of \(22.4 \mathrm{~A}\) ?

The nitrite ion \(\left(\mathrm{NO}_{2}^{-}\right)\) in soil is oxidized to the nitrate ion \(\left(\mathrm{NO}_{3}^{-}\right)\) by the bacterium Nitrobacter agilis in the presence of oxygen. The half-reactions are: \(\mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{NO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=0.42 \mathrm{~V}\) $$\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=1.23 \mathrm{~V}$$ Calculate the yield of ATP synthesis per mole of nitrite oxidized.
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