Extra Credit 29
- Page ID
- 82787
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For the ΔG° values given here, determine the standard cell potential for the cell.
Q17.4.2
Solution:
1. To find the standard cell potential for the cell, you should first use the equation ΔG°=-nFE°cell since this equation interrelates the cell potential with the ΔG value for a cell.
2. You should rearrange the equation to isolate E°cell as you are given ΔG°, the number of moles, and Faraday's constant (96,486 C/mol e-). The rearranged equation should look like this:
\[E°_{cell}=\frac{-ΔG°}{nF}\]
a) 12kJ/mol, n=3
To solve for the standard cell potential, you must utilize the ΔG° value (12kJ/mol) and mole value you are given (n=3), and remember Faraday's constant.
\[E°_{cell}=\frac{-12kJ}{(3mol)(96,485\frac{J}{V\cdot mol})}\cdot \frac{10^{3}J}{1kJ}=\frac{-12\cancel{kJ}}{(3\cancel{mol})(96,485\frac{\cancel{J}}{V\cdot \cancel{mol}})}\cdot \frac{10^{3}\cancel{J}}{1\cancel{kJ}}=-0.0414V\]
b) -45kJ/mol, n=1
Like part a), you must solve for the standard cell potential by using the ΔG° value (-45kJ/mol) and mole value you are given (n=1), as well as remember Faraday's constant.
\[E°_{cell}=-\frac{-45kJ}{(1mol)(96,485\frac{J}{V\cdot mol})}\cdot \frac{10^{3}J}{1kJ}=\frac{45\cancel{kJ}}{(1\cancel{mol})(96,485\frac{\cancel{J}}{V\cdot \cancel{mol}})}\cdot \frac{10^{3}\cancel{J}}{1\cancel{kJ}}=0.466V\]
Q12.1.2
Ozone decomposes to oxygen according to the equation 2O3(g) --> 3O2(g). Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
Q12.1.2
Solution:
The balanced chemical equation shows that 2 moles of ozone (O3) must decompose for every 3 moles of oxygen (O2) produced. The molar ratios of O3 to O2 are 2:3. This means that the rate of change for [O3] and [O2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. It is important that [O3] will be negative since it is being used to be transformed into the product [O2]. The rate of the reaction of the decomposition of ozone into oxygen gas is represented as:
\[Rate=-\frac{Δ[O3]}{2ΔT}=\frac{Δ[O2]}{3ΔT}\]
Q12.4.20
For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene (CH2=CH-CH=CH2) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:
The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as 2.0 x 10-4 s-1 at 150°C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150°C with an initial pressure of 55 torr.
Q12.4.20
Solution:
1. For a first order reaction, the equation of the integrated rate law is:
\([A_{t}]=[A_{0}]e^{-kt}\)
\([A_{0}]\) is the initial concentration of the reactant, \[(A_{t}]\) is the concentration of the reactant left after a particular time t, k is the rate constant, and t represents time.
2. Since k and t are given, you can substitute those values into the integrated rate law equation for a first order reaction and solve for the initial pressure of cyclobutene:
\[[A_{30}]=(55 torr)e^{-(2.0x10^{-4}\frac{1}{sec})(30min\cdot\frac{60sec}{1 min})}\]
\[[A_{30}]=(55 torr)e^{-(2.0x10^{-4}\frac{1}{sec})(30\cancel{min}\cdot\frac{60 sec}{1\cancel{min}})}\]
\[[A_{30}]=(55 torr)e^{-(2.0x10^{-4}\frac{1}{\cancel{sec}})(1800\cancel{sec})}\]
\[[A_{30}]=(55 torr)e^{-0.36}\]
\[[A_{30}]=38.37 torr\]
3. To find the concentration of cyclobutene after 39 minutes, we must use the ideal gas equation since pressure (55 \(\cancel{torr}\cdot\frac{1atm}{760\cancel{torr}}=0.07237atm\)), volume (0.53 L), temperature (150°C or 423.15 K), and R 0.08206\(\frac{L*atm}{mol*K}\) are given.
\(PV=nRT\) or rearranging gives \(n=\frac{PV}{RT}\)
\(n=\frac{(55torr)(0.53L)}{(0.08206\frac{L*atm}{mol*K})(423.15K)}\)
\(n=\frac{(0.07237\cancel{atm})(0.53\cancel{L})}{(0.08206\frac{\cancel{L}*\cancel{atm}}{mol*\cancel{K}})(423.15\cancel{K})}=0.00110 moles\)
The initial moles of cyclobutene can be found by:
\([A_{0}]=\frac{n}{V}=\frac{0.00110 moles}{0.53 L}=0.00208 M\)
4. We found the initial concentration of cyclobutene; however, the question asks us to find the concentration after 30 minutes. To find this, we need to once again use the integrated rate law equation for a first order reaction.
\([A_{t}]=[A_{0}]e^{-kt}\)
\[[A_{30}]=(0.00208M)e^{-(2.0x10^{-4}\frac{1}{sec})(30min\cdot\frac{60sec}{1 min})}\]
\[[A_{30}]=(0.00208M)e^{-(2.0x10^{-4}\frac{1}{sec})(30\cancel{min}\cdot\frac{60 sec}{1\cancel{min}})}\]
\[[A_{30}]=(0.00208M)e^{-(2.0x10^{-4}\frac{1}{\cancel{sec}})(1800\cancel{sec})}\]
\[[A_{30}]=(0.00208M)e^{-0.36}= 0.00145M\]
Thus, the partial pressure of cyclobutene is 38.37 torr and its concentration after 30 minutes is 0.00145M.
Q21.3.4
Complete each of the following equations:
a. \(^{7}_{3}Li+?\rightarrow2^{4}_{2}He\)
b. \(^{14}_{6}C\rightarrow^{14}_{7}N+?\)
c. \(^{27}_{13}Al+^{4}_{2}He\rightarrow?+^{1}_{0}n\)
d. \(^{250}_{96}Cm\rightarrow?+^{98}_{38}Sr+4^{1}_{0}n\)
Q21.3.4
Solution:
1. To complete the equations below, start by determining the sum of the mass numbers on the right side of the equation. The sum numbers on the right side of the equation should equal the sum of the mass numbers on the left side of the equation. You may need to solve for the mass number (A).
2. Once you find the mass number (A), you may need to solve for the atomic number (Z) by setting the sum of the atomic numbers on the left side equal to the sum of the atomic numbers on the right side. After finding the atomic number, you can determine which element the nuclear reaction is missing.
This is the form of a nuclear particle involved in nuclear reactions: \(^{A}_{Z}X\)
A = Mass number, Z = Atomic Number, X = Element that corresponds to atomic number
a. \(^{7}_{3}Li+?\rightarrow2^{4}_{2}He\)
1. Calculating the Mass Number
Sum of the mass numbers on the left (reactants side): 7+x Sum of the mass numbers on the right (products side): 2⋅4= 8
Setting the mass numbers of left and right sides equal: 7+ x= 8
x= 8-7 =1 \(\therefore\ ^{1}_{?}X\)
2. Calculating the Atomic Number:
Sum of the atomic numbers on the left (reactants side): 3+x Sum of the atomic numbers on the right side (products side): 2⋅2= 4
Setting the mass number of left and right sides equal: 3+ x= 4
x= 4-3 =1 \( \therefore\ ^{1}_{1}X\)
Since you know the atomic number of the particle is 1, you can access a periodic table and find which element corresponds to this specific atomic number. In this case, the element is Hydrogen. Therefore, the final form of this nuclear particle is \(^{1}_{1}H\) and the full nuclear equation is
\[^{7}_{3}Li+^{1}_{1}H\rightarrow2^{4}_{2}He\]
b. \(^{14}_{6}C\rightarrow^{14}_{7}N+?\)
1. Calculating the Mass Number
Sum of the mass numbers on the left (reactants side): 14 Sum of the mass numbers on the right (products side): 14+x
Setting the mass numbers of left and right sides equal: 14 = 14+x
x= 14-14= 0 \(\therefore\ ^{0}_{?}X\)
2. Calculating the Atomic Number:
Sum of the atomic numbers on the left (reactants side): 6 Sum of the atomic numbers on the right side (products side): 7+x
Setting the mass number of left and right sides equal: 6 = 7+x
x= 6-7 = -1 \( \therefore\ ^{0}_{-1}X\)
In this case, the particle's mass number is 0 and its atomic number is -1, indicating that the missing particle is a β particle, which is an electron emitted by an unstable nucleus. A beta particle is written as \(^{0}_{-1}e^{-}\). Therefore, the final form of this nuclear particle is \(^{0}_{-1}e^{-}\) and the full nuclear equation is
\[^{14}_{6}C\rightarrow^{14}_{7}N+^{0}_{-1}e^{-}\]
c. \(^{27}_{13}Al+^{4}_{2}He\rightarrow?+^{1}_{0}n\)
1. Calculating the Mass Number
Sum of the mass numbers on the left (reactants side): 27+4=31 Sum of the mass numbers on the right (products side): x+1
Setting the mass numbers of left and right sides equal: 31= x+1
x= 31-1= 30 \(\therefore\ ^{30}_{?}X\)
2. Calculating the Atomic Number:
Sum of the atomic numbers on the left (reactants side): 13+2=15 Sum of the atomic numbers on the right side (products side): x+0= x
Setting the mass number of left and right sides equal: x= 15 \( \therefore\ ^{30}_{15}X\)
Since you know the atomic number of the particle is 15, you can access a periodic table and find which element corresponds to this specific atomic number. In this case, the element is Phosphorus. Therefore, the final form of this nuclear particle is \(^{30}_{15}P\) and the full nuclear equation is
\[^{27}_{13}Al+^{4}_{2}He\rightarrow^{30}_{15}P+^{1}_{0}n\]
d. \(^{250}_{96}Cm\rightarrow?+^{98}_{38}Sr+4^{1}_{0}n\)
1. Calculating the Mass Number
Sum of the mass numbers on the left (reactants side): 250 Sum of the mass numbers on the right (products side): x+98+4(1)= x+102
Setting the mass numbers of left and right sides equal: 250= x+102
x= 250-102= 148 \(\therefore\ ^{148}_{?}X\)
2. Calculating the Atomic Number:
Sum of the atomic numbers on the left (reactants side): 96 Sum of the atomic numbers on the right side (products side): x+38
Setting the mass number of left and right sides equal: 96= x+38
x=96-38= 58 \( \therefore\ ^{148}_{58}X\)
Since you know the atomic number of the particle is 58, you can access a periodic table and find which element corresponds to this specific atomic number. In this case, the element is Cerium. Therefore, the final form of this nuclear particle is \(^{148}_{58}Ce\) and the full nuclear equation is
\[^{250}_{96}Cm\rightarrow^{148}_{58}Ce+^{98}_{38}Sr+4^{1}_{0}n\]
Q20.1.1
Identify the oxidation state of the atoms in the following compounds:
a. PCl3
b. CO32-
c. H2S
d. S8
e. SCl2
f. Na2SO3
g. SO42-
Q20.1.1
Solution:
- Rules for Assigning Oxidation Numbers to Elements:
- Rule 1: The oxidization number of an element in its free (uncombined) state is zero. (ex: Al(s) or Zn(s))
- Rule 2: The oxidation number of a monatomic ion is the same as the charge on the ion. (ex: Na+ or S2-)
- Rule 3: The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic ion is equal to the charge on the ion.
- Rule 4: The oxidation number of an alkali metal (IA group) in a compound is +1, while the oxidation number of an alkaline earth metal (IIA group) in a compound is +2.
- Rule 5: The oxidation number of oxygen in a compound is usually -2; however, if the oxygen is in a peroxide it has an oxidation number of -1. If oxygen is bounded to fluorine, the oxidation number is +1.
- Rule 6: The oxidation state of hydrogen in a compound is usually +1. If the hydrogen is part of a binary metal hydride (compound of hydrogen and some metal), the oxidation state of hydrogen is -1.
- Rule 7: The oxidation number of fluorine is always -1.
a. PCl3 Chlorine is more electronegative than phosphorus, so its oxidation number is -1. The overall molecule is neutral so phosphorus would have a oxidation number of +3.
PCl3 \(\rightarrow\) x+3(-1) =0
x= +3
b. CO32- Oxygen is more electronegative than carbon, so its oxidation number is -2. The overall molecule has a charge of -2, so carbon should have an oxidation number of +4.
CO32- \(\rightarrow\) x+3(-2) =-2
x-6= -2
x=+4
c. H2S Sulfur is more electronegative than hydrogen, so its oxidation number is -2. They overall molecule is neutral, hydrogen should have an oxiation number of +1.
H2S \(\rightarrow\) 2(x)+(-2)=0
2x=2
x=+1
d. S8 In this case, sulfur is in an elemental form and therefore has an oxidation number of zero.
e. SCl2 Chlorine is more electronegative than sulfur, so its oxidation number is -1. The overall molecule is neutral, so sulfur should have an oxidation number of +2.
SCl2 \(\rightarrow\) x+2(-1)=0
x=+2
f. Na2SO3 Sodium is an alkali metal, so its oxidation number is +1. Oxygen is more electronegative than sulfur, so its oxidation number should be -2. The overall molecule is neutral, so sulfur should have an oxidation number of +4.
Na2SO3 \(\rightarrow\) 2(1)+x+3(-2)=0
x-4=0
x=+4
g. SO42- Oxygen is more electronegative than sulfur, so its oxidation number is -2. The overall molecule has a charge of -2, so sulfur should have an oxidation number of +6.
SO42- \(\rightarrow\) x+4(-2)=-2
x-8=-2
x=+6
Q20.4.19
Carbon is used to reduce iron ore to metallic iron. The overall reaction is as follows:
\[2Fe_{2}O_{3}\cdot xH_{2}O(s)+3C(s)\rightarrow4Fe(l)+3CO_{2}(g)+2xH_{2}O(g)\]
Write the two half-reactions for this overall reaction.
Q20.4.19
Solution:
1. Separate the half-reactions. The first step in determining the two half-reactions for the overall reaction is to identify which elements is oxidized and which element is reduced. The overall equation should be separated into two equations as shown below:
Reduction Half-Reaction: \[2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)\]
The oxidation number of iron on the left side of the equation should be +3 because oxygen is more electronegative than iron, making its oxidation number -2. Since the charge of the overall molecule is neutral, iron must be +3 to balance the charge of the oxygen atoms. The oxidation number of iron on the right side of the equation should be 0 since it is a free element (uncombined state). The iron in this half-reaction is reduced since the oxidation number of iron decreases from +3 in 2Fe2O3 (reactants side) to 0 in 4Fe(l) (products side).
Oxidation Half-Reaction: \[3C(s)\rightarrow3CO_{2}(g)\]
The oxidation number of carbon on the left side should be 0 since it is a free element (uncombined state). The oxidation number of carbon on the right side should be +4 since oxygen is more electronegative than carbon, making its oxidation number -2. The charge of the overall molecule is neutral, so carbon must be +4 to balance the charge of the oxygen atoms. The carbon in this half-reaction is oxidized as the oxidation number of carbon increases from 0 in 3C(s) (reactants side) to +4 in 3CO2(g) (products side).
2. Balance elements other than oxygen (O) and hydrogen (H). In this equation, all elements other than O and H are balanced.
\[2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)\]
\[3C(s)\rightarrow3CO_{2}(g)\]
3. Add H2O to balance oxygen. In the reduction half-reaction, 6 H2O molecules need to be added to the right side of the reaction to balance the oxygen molecules on the left side of the reaction. In the oxidation half-reaction, 6 H2O molecules must be added to the left side of the reaction to balance the oxygen molecules on the right side of the reaction.
Reduction Half-Reaction: \[2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)\]
Oxidation Half-Reaction: \[3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)\]
4. Balance hydrogen by adding protons (H+). 12 protons need to be added to the left side of the iron reaction to balance the 16 (2 per water molecule * 8 water molecules) hydrogens. 12 protons need to be added to the right side of the carbon reaction.
Reduction Half-Reaction: \[2Fe_{2}O_{3}\cdot xH_{2}O(s)+12H^{+}(aq)\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)\]
Oxidation Half-Reaction: \[3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)+12H^{+}(aq)\]
5. Balance the charge of each equation with electrons. The iron reaction has a +12 charge on the left side of the reaction due to the 12 protons, while the other side has a neutral (0) charge, so 12 electrons must be added to the left side of the reaction to balance the charge to 0. The carbon reaction has a +12 on the right side of the reaction due to the 12 protons, while the other side has a neutral (0) charge, so 12 electrons must be added to the right side of the reaction to balance the charge to 0.
This is the final form of the two half-reactions for the reduction of iron ore to metallic ore by carbon:
Reduction Half-Reaction: \[2Fe_{2}O_{3}\cdot xH_{2}O(s)+12H^{+}(aq)+12e^{-}\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)\]
Oxidation Half-Reaction: \[3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)+12H^{+}(aq)+12e^{-}\]
Q20.9.4
Two solutions, one containing Fe(NO3)2⋅6H2O and the other containing the same molar concentration of Fe(NO3)3⋅6H2O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer.
Q20.9.4
Solution:
1. To find which solution produces the most metal, you must first split the two compounds into their respective half reactions that form once they are electrolyzed.
For Fe(NO3)2⋅6H2O, the resulting half reaction would be: \(Fe^{2+}(aq)+2e^{-} \rightarrow Fe(s)\)
For Fe(NO3)3⋅6H2O, the resulting half reaction would be: \(Fe^{3+}(aq)+3e^{-} \rightarrow Fe(s)\)
2. The fact that both compounds were electrolyzed under identical conditions indicates that the same amount of electrons were transferred. This means that:
If a certain amount of electrons were transferred during the electrolysis of Fe(NO3)2·6H2O, then that specific amount of electrons divided by 2 will provide you with the moles of iron metal produced. The same is true for Fe(NO3)3·6H2O. If a certain amount of electrons were transferred during the electrolysis of Fe(NO3)3·6H2O, then that amount of electrons divided by 3 will provide you with the moles of iron metal produced. Thus, Fe(NO3)2·6H2O will produce the most metal.
Q20.8.3
Why is it important for automobile manufacturers to apply paint to the metal surface of a car? Why is this process particularly important for vehicles in northern climates, where salt is used on icy roads?
Q20.8.3
Solution:
Paint prevents contact between oxygen and water with the metal surface of the car, which prevents corrosion from occurring. It is important to remember that once paint is scratched on a painted iron surface, corrosion will occur and rust will quickly begin to form. This process is particularly important for vehicles in northern climates, where salt is used on icy roads, because salt will accelerate rust formation. Rust formation is an electrochemical process that accelerates the transfer of electrons from the anode to the cathode, ultimately allowing iron oxide compounds to form rust on vehicles.
The solutions to the questions are correct. Checked in phase 2.