Extra Credit 43
- Page ID
- 82957
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Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel.
A17.6.2
In order for a metal to be used as a sacrificial anode in cathodic protection, the metal should be more readily oxidized that the parent metal it is protecting. To determine this, we look at a standard reduction potential table and find the values for each metal.
Ag = 0.80 V
Au = 1.50 V
Mg = -2.37 V
Ni = -0.25 V
Zn = -0.76 V
As the standard reduction potential becomes more negative, the strength as a reducing agent increases. Because the standard reduction potential of Fe is -0.447 V, we would need to use a metal or metals whose reduction potential is greater than that value, in the negative direction - those metals would lie lower on the standard reduction potential table than Fe.
From using the table and the values we found, we can see that the two metals that satisfy the need are Mg and Zn, therefore those two metals could be used a sacrificial anodes in the cathodic protection of an underground steel storage tank.
Q17.6.2
Aluminum (E∘Al3+/Al=−2.07V)] is more easily oxidized than iron (E∘Fe3+/Fe=−0.477V), and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation.
A17.6.2
Untreated aluminum has a better corrosion resistance than untreated iron because aluminum will spontaneously form a thin oxide layer that protects itself from further oxidation. Aluminum, which is more easily oxidized, will oxidize more quickly than iron, but after forming the initial first layer of aluminum oxide, further corrosion is blocked by that layer. Aluminum oxide is impermeable and varies from oxides of other metals in that in binds very strongly to the parent metal. When iron oxidizes, it does not form a protective layer, thus the metal will continue to corrode.
Q12.3.6
Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction NO+O3⟶NO2+O2 is first order with respect to both NO and O3 with a rate constant of 2.20 × 107 L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 × 10−6 M and [O3] = 5.9 × 10−7 M?
A12.3.6
NO + O3 → NO2 + O2
rate = R = k[NO][O3]
k is the rate constant for the reaction, and is given to be 2.20x107 M/s, so the rate equation would be,
R = (2.20x107 M/s)[NO][O3]
The rate can also be defined in terms of the individual species of the reaction.
-[NO]/dt = -[O3]/dt= [NO2]/dt= [O2]/dt
To find the rate of disappearance of NO, you would use the first equation. You could then set this equation equal to the overall rate reaction to get the following,
-[NO]/dt = (2.20x107 M/s)[NO][O3]
To find the instantaneous rate of disappearance of NO, you would plug in the given values for the concentrations of NO and O3.
-[NO]/dt= (2.20x107 M/s)(3.3x10-6 M)(5.9x10-7 M)
= - 4.284x10-5 M/s
We have found a negative value because the NO is being used up, so its concentration is decreasing over time as the reaction proceeds. We would adjust this to find that the rate is equal to 4.284x10-5 M/s.
Q12.5.15
The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
C12H22O11+ H2O ⟶ C6H12O6 + C6H12O6
follows a first-order rate equation for the disappearance of sucrose: Rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
- In neutral solution, k = 2.1 × 10−11 s−1 at 27 °C and 8.5 × 10−11 s−1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
- When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
- Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
A12.5.1
To find the activation energy:
\[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}}\]
k1 = 2.1 × 10−11 s−1 T1 = 27 °C = 300.15 K
k2 = 8.5 × 10−11 s−1 T2 = 37 °C = 310.15 K
\[E_a = \dfrac{\left( 8.314 J/mol*K\right) \ln \dfrac{8.5*10^{-11}s^{-1}}{2.1*10^{-11}s^{-1}}}{\dfrac{1}{300K}-\dfrac{1}{310K}}\]
Ea = 108,216.49 J = 108.216 kJ
To find the frequency factor:
\[k=Ae^{\frac{-E_a}{RT}}\]
\[2.1*10^{-11}s^{-1}=Ae^{\dfrac{-108216.48J}{\left(8.314J/mol*K\right)\left(300K\right)}}\]
2.1 × 10−11 s−1 = Ae-108,216.49 J/[(8.314 J/mol K)(300.15 K)]
A = 1.434 x 108 s-1
To find the rate constant at 47°C:
T = 47°C = 320.15 K
\[k=Ae^{\frac{-E_a}{RT}}\]
\[k=\left(1.434*10^8s^{-1}\right)e^{\dfrac{-108216.48J}{\left(8.314J/mol*K\right)\left(320K\right)}}\]
k = Ae-108,216.49 J/[(8.314 J/mol K)(320.15 K)]
k = 3.1675 x 10-10
To find time it takes to reach equilibrium at 27°C:
[A] = [Ao]e-kt
1.65 × 10−7 M = (0.150M)e(2.1 × 10^-11)t
1.1 × 10−6 = e(2.1 × 10^-11)t
t = 6.533 x 1011 seconds --> 7.56 x 106 days
By assuming that the reaction is irreversible in part (b), we are able to simplify our calculation because we do not have to consider any reactant that turns into product and then back into reactant, which in this case would be sucrose turning into glucose and fructose, and either or turning back into sucrose.
Q21.4.10
Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed
- \(^{6}_2He\)
- \(^{60}_{30}Zn\)
- \(^{235}_{91}Pa\)
- \(^{241}_{94}Np\)
- \(^{18}_{}F\)
- \(^{129}_{}Ba\)
- \(^{237}_{}Pu\)
A21.4.10
In order to determine what mode(s) of spontaneous radioactive decay each unstable isoptope might proceed, we must first look at the number of neutrons and protons in each isotope, and refer to the belt of stability to determine which types of decay will move the isotope to be more stable and lie on the belt.
1. 4 neutrons, 2 protons
You would want to increase the number of protons.
beta emission to \(^{6}_{3}Li\)
2. 30 neutrons, 30 protons
The ratio of neutrons to protons is lower on the belt of stability, so you would want to change the number of protons to move it up.
positron emission to \(^{60}_{29}Cu\) --> positron emission to \(^{60}_{28}Ni\).
3. 144 neutrons, 91 protons
You would want to increase the number of protons.
beta emission to \(^{235}_{92}U\)
4. 147 neutrons, 94 protons
You would want to increase the number of protons.
beta emission to \(^{241}_{95}Am\)
5. 9 neutrons, 9 protons
The ratio of neutrons to protons is lower on the belt of stability, so you would want to change the number of protons to move it up.
positron emission to \(^{18}_{8}O\)
6. 73 neutrons, 56 protons
The ratio of neutrons to protons is lower on the belt of stability, so you would want to change the number of protons to move it up.
positron emission to \(^{129}_{55}Cs\)
7. 143 neutrons, 94 protons
The ratio of neutrons to protons is lower on the belt of stability, so you would want to change the number of protons to move it up.
positron emission to \(^{237}_{93}Np\)
Q20.2.14
Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.
-
What has been oxidized? What has been reduced?
-
Balance the chemical equation.
A20.2.14
Unbalanced reaction: Cu (s) + HNO3 (aq) → NO (g) + Cu2+ (aq)
This reaction is an example of a redox reaction, in which electrons are transferred between reacting species. The oxidized species would be the one that is losing electrons, and the reduced species would be the one that is gaining the electrons.
The first step would be to split the overall reaction into two half-reactions.
Cu (s) → Cu2+ (aq)
HNO3 (aq) → NO (g)
Then, you would want to balance each half-reaction separately.
Cu (s) → Cu2+ (aq) + 2e-
HNO3 (aq) + 3H+ (aq) + 3e- → NO (g) + 2H2O (l)
You would then multiply each half-reaction by the appropriate coefficient so that the number of electrons in each reaction are the same,
3[Cu (s) → Cu2+ (aq) + 2e-]
2[HNO3(aq) + 3H+ (aq) + 3e- → NO (g) + 2H2O (l)]
which would yield the following two half-reactions.
3Cu (s) → 3Cu2+ (aq) + 6e-
2HNO3 (aq) + 6H+ (aq) + 6e- → 2NO (g) + 4H2O (l)
From these two half-reactions, we can see that the copper metal has been oxidized and the nitric acid, specifically the nitrogen, has been reduced. To find the balanced overall reaction, you would combine the two half-reactions together, and cancel out any common species that appear on both sides of the reaction arrow.
3Cu (s) + HNO3 (aq) + 6H+ (aq) → 3Cu2+ (aq) + 2NO (g) + 4H2O (l)
Q20.5.9
Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?
A20.5.9
A concentration cell is type of an electrolytic cell, which are driven by spontaneous chemical reactions and create an electric current through that reaction. Whereas a typical electrolytic cell will have two different species at each node, a concentration cell has the same species at each node, just in different concentrations. Thus, in a concentration cell, voltage is produced as particles begin to move in order to dilute the more concentrated side, while simultaneously making the dilute side more concentrated. This is achieved by moving electrons from the cell with the lower concentration to the cell with the higher concentration. V = 0 in a concentration cell when both sides or nodes of the cell have the same concentration and there is no longer any transfer of electrons between the two compartments. This happens when the concentration cell reaches equilibrium.
Q24.6.5
How can CFT explain the color of a transition-metal complex?
A24.6.5
The crystal field theory (CFT) looks at the degeneracy of the d-orbital in transition metal complexes in the presence of a ligand-field. This can be determined from the crystal field splitting, ∆o, which is caused by the presence of ligands, and is a form of energy. Strong-field ligands create low spin energy splitting diagrams, which is caused by a high crystal field splitting. Thus, strong-field ligands are able to absorb higher-energy lights, such as green, blue, and violet, which have shorter wavelengths and higher frequencies. On the other hand, weak-field ligands create high spin energy splitting diagrams, which is caused by a low crystal field splitting. Thus weak-field ligands absorb lower-energy lights, such as red, orange, and yellow, which have longer wavelengths and lower frequencies.
Transition-metal complexes display certain colors depending on the frequency and wavelength of light that they absorb. Compounds will absorb certain colored wavelengths and reflect back the complementary color of that wavelength. For example, a compound that absorbs red light will appear green, another compound that absorbs yellow light will appear violet, and so on. Depending on the ligand attached to the complex, the transition-metal complex will be able to absorb higher or lower energy light, and reflect back different colors depending on that light absorbed.