Extra Credit 15
- Page ID
- 83245
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Q17.2.4
Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as they are never part of the half-reactions.
- \(\ce{Al}(s)+\ce{Zr^4+}(aq)⟶\ce{Al^3+}(aq)+\ce{Zr}(s)\)
- \(\ce{Ag+}(aq)+\ce{NO}(g)⟶\ce{Ag}(s)+\ce{NO3-}(aq) \hspace{20px} \textrm{(acidic solution)}\)
- \(\ce{SiO3^2-}(aq)+\ce{Mg}(s)⟶\ce{Si}(s)+\ce{Mg(OH)2}(s) \hspace{20px} \textrm{(basic solution)}\)
- \(\ce{ClO3-}(aq)+\ce{MnO2}(s)⟶\ce{Cl-}(aq)+\ce{MnO4-}(aq) \hspace{20px} \textrm{(basic solution)}\)
S17.2.4
a. \(\ce{Al}(s)+\ce{Zr^4+}(aq)⟶\ce{Al^3+}(aq)+\ce{Zr}(s)\)
Oxidation:
\(\ce{Al}(s)→\ce{Al^3+}(aq)\)
\(\ce{Al}(s)→\ce{Al^3+}(aq)+\ce{3e^-}\)
Reduction:
\(\ce{Zr^4+}(aq)→\ce{Zr}(s)\)
\(\ce{Zr^4+}(aq)+\ce{4e^-}→\ce{Zr}(s)\)
Overall Reaction:
\(\ce{3Zr^4+}(aq)+\ce{4Al}(s)→\ce{4Al^3+}(aq)+\ce{3Zr}(s)\)
Cell Notation:
Al(s)|Al3+(aq)||Zr4+(aq)|Zr(s)
b. \(\ce{Ag+}(aq)+\ce{NO}(g)⟶\ce{Ag}(s)+\ce{NO3-}(aq) \hspace{20px} \textrm{(acidic solution)}\)
Oxidation:
\(\ce{NO}(g)→\ce{NO_3-}(aq)+\ce{3e^-}\)
\(\ce{NO}(g)+\ce{2H_2O}→\ce{NO_3-}(aq)+\ce{3e^-}+\ce{4H^+}\)
Reduction:
\(\ce{Ag^+}(aq)→\ce{Ag}(s)\)
\(\ce{Ag^+}(aq)+\ce{e^-}→\ce{Ag}(s)\)
Overall Reaction:
\(\ce{3Ag+}(aq)+\ce{NO}(g)+\ce{2H_2O}(l)⟶\ce{3Ag}(s)+\ce{NO3-}(aq)+\ce{H^4+}(aq)\)
Cell Notation:
Pt(s)|NO(g)|NO3-(aq)||Ag+(aq)|Ag(s)
c. \(\ce{SiO3^2-}(aq)+\ce{Mg}(s)⟶\ce{Si}(s)+\ce{Mg(OH)2}(s) \hspace{20px} \textrm{(basic solution)}\)
Oxidation:
\(\ce{Mg}(s)⟶\ce{Mg(OH)_2}(s)\)
\(\ce{2H_2O}(l)+\ce{4OH^-}(aq)+\ce{Mg}(s)⟶\ce{Mg(OH)_2}(aq)+\ce{4H^+}(aq)+\ce{4OH^-}(aq)+\ce{4e^-}\)
Reduction:
\(\ce{SiO3^2-}(aq)⟶\ce{Si}(s)\)
\(\ce{6H^+}(aq)+\ce{6OH^-}(aq)+\ce{4e^-}+\ce{SiO3^2-}(aq)⟶\ce{Si}(s)+\ce{3H_2O}(l)+\ce{6OH^-}(aq)\)
Overall Reaction:
\(\ce{SiO3^2-}(aq)+\ce{Mg}(s)+\ce{H_2O}(l)⟶\ce{Si}(s)+\ce{Mg(OH)_2}(aq)+\ce{2OH^-}(aq)\)
Cell Notation:
Si(s)|SiO32-(aq)||Mg2+(aq)|Mg(s)
d. \(\ce{ClO3-}(aq)+\ce{MnO2}(s)⟶\ce{Cl-}(aq)+\ce{MnO4-}(aq) \hspace{20px} \textrm{(basic solution)}\)
Oxidation:
\(\ce{ClO3-}(aq)→\ce{Cl-}(aq)\)
\(\ce{ClO3-}(aq)+\ce{6e-}+\ce{6H+}→\ce{Cl-}(aq)+\ce{3H2O}\)
Reduction:
\(\ce{MnO2}(s)→\ce{MnO4-}(aq)\)
\(\ce{2MnO2}(s)+\ce{4H2O}→\ce{2MnO4-}(aq)+\ce{6e^-}+\ce{8H+}\)
Overall Reaction:
\(\ce{ClO3-}(aq)+\ce{2MnO2}(s)+\ce{2OH^-}(aq)→\ce{Cl-}(aq)+\ce{2MnO4-}(aq)+\ce{H_2O}(l)\)
Cell Notation:
Pt(s)|ClO3-(aq), Cl-(aq)||MnO4-(aq)|MnO2(s)
Q19.1.13
Find the potentials of the following electrochemical cell:
Cd | Cd2+ (M = 0.10) ‖ Ni2+ (M = 0.50) | Ni
S19.1.13
E= Eocell-(.0592/n)logQ
Q= [Reduction]/[Oxidation]
Cd2+ has a E°=-0.40
Ni2+ has a E°=-0.25
E°cell=E°(Cathode)-E°(Anode)
E°cell=(-0.25)-(-0.4)
E°cell=0.15V
\(\ce{Cd}\rightarrow\ce{Cd^2+}+\ce{2e^-}\)
\(\ce{Ni^2+}+\ce{2e^-}\rightarrow\ce{Ni}\)
We see that 2 electrons are transferred so n=2
logQ=(0.1)/(0.5)=0.20
logQ=0.69897
E= Eocell-(0.0592/n)logQ
E=0.15-(0.0592/2)*-0.69897
E=0.17V
Q19.3.5
Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain.
S19.3.5
No, it is not possible for a metal in the transition series to have six unpaired electrons. Metals in the transition series contain orbitals from the s and d orbitals. The s subshell contains 2 orbitals while the d subshell contains 3 orbitals for a total of 5. According to Hund's rule, every orbital in a subshell must be filled and paired before starting the next. However, if the ligand is high spin, the rule is ignored and every subshell in every orbital must have one electron before pairing. However, we can see in the diagram below that regardless of whether or not the molecule is high spin, the maximum number of orbitals is 5, and therefore, the maximum number of unpaired electrons can only be 5.
Q12.4.5
From the given data, use a graphical method to determine the order and rate constant of the following reaction:
2X→Y+Z
| Time (s) | 5.0 | 10.0 | 15.0 | 20.0 | 25.0 | 30.0 | 35.0 | 40.0 |
|---|---|---|---|---|---|---|---|---|
| [X] (M) | 0.0990 | 0.0497 | 0.0332 | 0.0249 | 0.0200 | 0.0166 | 0.0143 | 0.0125 |
S12.4.5
Since the coefficient of X is 2, the formula for the rate is Rate=k[X]2. If the rate is second order, we can graph the data using Time(s) vs. 1/[X] (M)
When we graph the data, we get:
The rate can now be calculated by finding the slope between two points.
If we use the points (5,10.1) and (40,80), we get...
k = (80-10.1)/(40-5)
k = 1.997 M-1 S-1
Q12.7.8
- Based on the diagrams in Question Q12.7.6, which of the reactions has the fastest rate? Which has the slowest rate?
- Based on the diagrams in Question Q12.7.7, which of the reactions has the fastest rate? Which has the slowest rate?
S12.7.8
1. Reaction in Graph A has the slowest rate because the activation energy is large and reaction in Graph B has the fastest rate because the activation energy is small.
2. Neither of the graphs in Question 12.7.7 is either the fastest or slowest because both of the reactions in the graphs have the same activation energy.
Q21.5.3
Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission?
S21.5.3
A very high temperature is required for fusion and not fission because in fusion you need to bring two or more protons close enough to each other for nuclear forces to overcome their electrostatic repulsion; and it doesn't take a lot of energy to split two atoms in fission.
Q20.4.1
Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?
S20.4.1
No, hydrogen is not chemically inert. It is a major disadvantage to use a hydrogen electrode because it is hard to set up due to the preparation of the platinized surface and control of the concentrations of the reactants.
Q20.5.30
Hydrogen gas reduces Ni2+ according to the following reaction: Ni2+(aq) + H2(g) → Ni(s) + 2H+(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.
- What is K for this redox reaction?
- Is this reaction likely to occur?
- What conditions can be changed to increase the likelihood that the reaction will occur as written?
- Is the reaction more likely to occur at higher or lower pH?
S20.5.30
a. The equation we can use in this problem is \(E^{o}_{cell}=\frac{0.025693V}{n}lnK\)
-0.25V=[(0.025693V)/2]*lnK
K=e-19.46
K=3.535x10-9
b. No the reaction is not likely to occur because ΔH = 54 kJ/mol. This means that this reaction is not energetically favorable. If the value was a negative number, then this reaction would be more likely to occur.
c. Heat is often a reactant in an endothermic reaction. Generally, the more reactants you have in a reaction, the more the products will be favored. Therefore, if temperature is increased, that will increase the likelihood of that reaction.
d. We can see from the equation that hydrogen ions (H+) are part of the products. According to le Chatlelier's principle, if we change the concentration of either the products/reactants, the reaction will adjust to reestablish chemical equilibrium. In terms of pH, the more H+ ions are present, the more acidic something is. In this example, we can decrease the concentration of H+ ions to make the equation move in the forward direction more favorable. This means the reaction is more likely to occur at a higher pH.