Extra Credit 38

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Q17.5.6

Why do batteries go dead, but fuel cells do not?

Q17.5.6

Solution:

Batteries are self-contained and have a limited supply of reagents (to convert into electricity) before going dead. Once the limited supply of reagents run out, the battery dies. As the reagents are used up during the discharge, there is a loss voltage. This loss of voltage forms a brown sediment that builds up at the bottom of the case which interferes with the reaction and can cause a shortage. This shortage will drain the battery. A fuel cell, on the other hand, receives the chemicals it uses from the outside (such as hydrogen and oxygen); therefore, it will not run out. Fuel cells can generate power almost indefinitely, as long as they have fuel to use.

Q12.3.1

How do the rate of a reaction and its rate constant differ?

Q12.3.1

Solution:

The rate of a reaction is the change in concentration of a reactant or product per unit time, while the rate constant is a constant of proportionality in the rate law equation. The instantaneous rate depends on the concentrations of the reactants at that time. In contrast, the rate constant of a particular reaction at a particular temperature remains constant and does not depend upon the concentrations of the reactants. The rate of a reaction is always expressed as M/s, while the rate constant units depend on the order of the reaction.

Q12.5.10

The rate constant for the decomposition of acetaldehyde, CH₃CHO, to methane, CH₄, and carbon monoxide, CO, in the gas phase is 1.1 × 10⁻² L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.

Q12.5.10

Solution:

Known:

k₁= 1.1x10^-2 1/Ms T₁= 703K

k₂= 4.95 1/Ms T₂= 865K

R= 8.314 J/Kmol (Gas constant)

Unknown:

Ea=?

Equation:

\[ln(\frac{k_2}{k_1})=-\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})\]

This equation is derived from Arrhenius' equation:

\[k=Ae^\frac{-Ea}{RT}\]

Looking at this equation, the only unknown is Ea. So we can just plug in our known values, and solve for Ea.

\[ln(\frac{4.95M^{-1}s^{-1}}{1.1\times10^{-2} M^{-1}s^{-1}})=\frac{-Ea}{8.314JK^{-1}mol^{-1}}(\frac{1}{865K}-\frac{1}{703K})\]

Ea ≈ 190669.0845 J

Ea ≈ 1.91×10⁵ J = 191 kJ (3 sig figs)

Q21.4.5

Why is electron capture accompanied by the emission of an X-ray?

Q21.4.5

Solution:

During electron capture, one of the inner electrons of an atom is absorbed into the nucleus. The absorbed electron combines with a proton to form a neutron. This results in the emission of a neutrino and an emission of an X-ray.

\[{^0_{-1}}e^{-}+{^1_{1}}p^{+}⟶ {^1_{0}}n+{v}_e + xray\]

V_e stands for neutrino.

Because one of the electrons was pulled from an inner shell to the nucleus, an electron is "missing" from an inner shell. Thus, an electron from a higher level falls to replace that "missing" electron. Because an electron is falling from a higher energy state to a lower energy state, the energy difference must be emitted. It is emitted as a high energy photon in the x-ray range.

Q20.2.9

Balance each redox reaction under the conditions indicated.

CuS(s) + NO₃⁻(aq) → Cu²⁺(aq) + SO₄²⁻(aq) + NO(g); acidic solution
Ag(s) + HS⁻(aq) + CrO₄²⁻(aq) → Ag₂S(s) + Cr(OH)₃(s); basic solution
Zn(s) + H₂O(l) → Zn²⁺(aq) + H₂(g); acidic solution
O₂(g) + Sb(s) → H₂O₂(aq) + SbO₂⁻(aq); basic solution
UO₂²⁺(aq) + Te(s) → U⁴⁺(aq) + TeO₄²⁻(aq); acidic solution

Q20.2.9

Solution:

1. CuS(s) + NO₃⁻(aq) → Cu²⁺(aq) + SO₄²⁻(aq) + NO(g); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: \[{NO_3^-}_{(aq)}⟶ NO_(g)\]

Oxidation: \[{CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}\]

2. In each half reaction, balance all elements. Balance O with H₂O and H with H⁺. Balance the charge on each side using e^-.

Reduction: \[{3e^-}+{4H^+}_{(aq)}+{NO_3^-}_{(aq)}⟶ {NO}_{(g)}+{2H_2O}_{(l)}\]

Oxidation: \[{4H_2O}_{(l)}+{CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}+{8H^+}_{(aq)}+8e^-\]

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: \[8({3e^-}+{4H^+}_{(aq)}+{NO_3^-}_{(aq)}⟶ {NO}_{(g)}+{2H_2O}_{(l)})\]

\[{24e^-}+{32H^+}_{(aq)}+{8NO_3^-}_{(aq)}⟶ {8NO}_{(g)}+{16H_2O}_{(l)}\]

Oxidation: \[3({4H_2O}_{(l)}+{CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}+{8H^+}_{(aq)}+8e^-)\]

\[{12H_2O}_{(l)}+{3CuS}_{(s)}⟶ {3Cu^{2+}}_{(aq)}+{3SO_4^{2-}}_{(aq)}+{24H^+}_{(aq)}+24e^-\]

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: \[{8H^+}_{(aq)}+{8NO_3^-}_{(aq)}+{3CuS}_{(s)}⟶ {8NO}_{(g)}+{4H_2O}_{(l)}+{3Cu^{2+}}_{(aq)}+{3SO_4^2-}_{(aq)}\]

5. Check if number of atoms and charges balance.

The charges on both sides equal to 0, and the number of atoms are equal on both sides. ✓

2. Ag(s) + HS⁻(aq) + CrO₄²⁻(aq) → Ag₂S(s) + Cr(OH)₃(s); basic solution

1. Write the oxidation and reduction half reactions.

Reduction: \[{CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}\]

Oxidation: \[{Ag}_{(s)}+{HS^-}_{(aq)}⟶{Ag_2S}_{(s)}\]

2. In each half reaction, balance all elements. Balance O with H₂O and H with H⁺. Balance the charge on each side using e^-.

Reduction: \[{3e^-}+{5H^+}_{(aq)}+{CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}+{H_2O}_{(l)}\]

Oxidation: \[{2Ag}_{(s)}+{HS^-}_{(aq)}⟶ {Ag_2S}_{(s)}+{H^+}_{(aq)}+2e^-\]

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: \[2({3e^-}+{5H^+}_{(aq)}+{CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}+{H_2O}_{(l)})\]

\[{6e^-}+{10H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}\]

Oxidation: \[3({2Ag}_{(s)}+{HS^-}_{(aq)}⟶ {Ag_2S}_{(s)}+{H^+}_{(aq)}+2e^-)\]

\[{6Ag}_{(s)}+{3HS^-}_{(aq)}⟶ {3Ag_2S}_{(s)}+{3H^+}_{(aq)}+6e^-\]

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: \[{7H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}+3Ag_2S_{(s)}\]

5. Basic solution: add a number of OH^- ions equal to the number of H⁺ ions.

\[{7OH^-}_{(aq)}+{7H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}+3Ag_2S_{(s)}+{7OH^-}_{(aq)}\]

6. Basic solution: Combine H⁺ and OH^- ions to form water molecules. Cancel common water molecules, and leave a remainder H₂O molecules on just one side.

\[{5H_2O}_{(l)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+3Ag_2S_{(s)}+{7OH^-}_{(aq)}\]

7. Check if number of atoms and charges balance.

The charges on both sides equal to -7, and the number of atoms are equal on both sides. ✓

3. Zn(s) + H₂O(l) → Zn²⁺(aq) + H₂(g); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: \[{H_2O}_{(l)}⟶ {H_2}_{(g)}\]

Oxidation: \[{Zn}_{(s)}⟶ {Zn^{2+}}_{(aq)}\]

2. In each half reaction, balance all elements. Balance O with H₂O and H with H⁺. Balance the charge on each side using e^-.

Reduction: \[{2e^-}+{2H^+}_{(aq)}+{H_2O}_{(l)}⟶{H_2}_{(g)}+{H_2O}_{(l)}\]

Oxidation: \[{Zn}_{(s)}⟶{Zn^{2+}}_{(aq)}+2e^-\]

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: \[{2e^-}+{2H^+}_{(aq)}+{H_2O}_{(l)}⟶{H_2}_{(g)}+{H_2O}_{(l)}\]

Oxidation: \[{Zn}_{(s)}⟶{Zn^{2+}}_{(aq)}+2e^-\]

They have the same number of electrons. No changes needed.

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: \[{2H^+}_{(aq)}+{Zn}_{(s)}⟶ {H_2}_{(g)}+{Zn^{2+}}_{(aq)}\]

5. Check if number of atoms and charges balance.

The charges on both sides equal to +2, and the number of atoms are equal on both sides. ✓

4. O₂(g) + Sb(s) → H₂O₂(aq) + SbO₂⁻(aq); basic solution

1. Write the oxidation and reduction half reactions.

Reduction: \[{O_2}_{(g)}⟶ {H_2O_2}_{(aq)}\]

Oxidation: \[{Sb}_{(s)}⟶{SbO_2^-}_{(aq)}\]

2. In each half reaction, balance all elements. Balance O with H₂O and H with H⁺. Balance the charge on each side using e^-.

Reduction: \[{2e^-}+{2H^+}_{(aq)}+{O_2}_{(g)}⟶ {H_2O_2}_{(aq)}\]

Oxidation: \[{2H_2O}_{(l)}+{Sb}_{(s)}⟶ {SbO_2^-}_{(aq)}+{4H^+}_{(aq)}+3e^-\]

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: \[3({2e^-}+{2H^+}_{(aq)}+{O_2}_{(g)}⟶ {H_2O_2}_{(aq)})\]

\[{6e^-}+{6H^+}_{(aq)}+{3O_2}_{(g)}⟶ {3H_2O_2}_{(aq)}\]

Oxidation: \[2({2H_2O}_{(l)}+{Sb}_{(s)}⟶ {SbO_2^-}_{(aq)}+{4H^+}_{(aq)}+3e^-)\]

\[{4H_2O}_{(l)}+{2Sb}_{(s)}⟶ {2SbO_2^-}_{(aq)}+{8H^+}_{(aq)}+6e^-\]

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: \[{3O_2}_{(g)}+{4H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}+2H^{2+}_{(aq)}\]

5. Basic solution: add a number of OH^- ions equal to the number of H⁺ ions.

\[{2OH^-}_{(aq)}+{3O_2}_{(g)}+{4H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}+2H^{2+}_{(aq)}+{2OH^-}_{(aq)}\]

6. Basic solution: Combine H⁺ and OH^- ions to form water molecules. Cancel common water molecules, and leave a remainder H₂O molecules on just one side.

\[{2OH^-}_{(aq)}+{3O_2}_{(g)}+{2H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}\]

7. Check if number of atoms and charges balance.

The charges on both sides equal to -2, and the number of atoms are equal on both sides. ✓

5. UO₂²⁺(aq) + Te(s) → U⁴⁺(aq) + TeO₄²⁻(aq); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: \[{UO_2^{2+}}_{(aq)}⟶ {U^{4+}}_{(aq)}\]

Oxidation: \[{Te}_{(s)}⟶ {TeO_4^{2-}}_{(aq)}\]

2. In each half reaction, balance all elements. Balance O with H₂O and H with H⁺. Balance the charge on each side using e^-.

Reduction: \[{2e^-}+{4H^+}_{(aq)}+{UO_2^{2+}}_{(aq)}⟶{U^{4+}}_{(aq)}+{2H_2O}_{(l)}\]

Oxidation: \[{4H_2O}_{(l)}+Te_{(s)}⟶{TeO_4^{2-}}_{(aq)}+8H^+_{(aq)}+6e^-\]

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: \[3({2e^-}+{4H^+}_{(aq)}+{UO_2^{2+}}_{(aq)}⟶{U^{4+}}_{(aq)}+{2H_2O}_{(l)})\]

\[{6e^-}+{12H^+}_{(aq)}+{3UO_2^{2+}}_{(aq)}⟶{3U^{4+}}_{(aq)}+{6H_2O}_{(l)}\]

Oxidation: \[{4H_2O}_{(l)}+Te_{(s)}⟶{TeO_4^{2-}}_{(aq)}+8H^+_{(aq)}+6e^-\]

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: \[{4H^+}_{(aq)}+{3UO_2^{2+}}_{(aq)}+{Te}_{(s)}⟶ {3U^{4+}}_{(aq)}+{2H_2O}_{(l)}+{TeO_4^{2-}}_{(aq)}\]

5. Check if number of atoms and charges balance.

The charges on both sides equal to +10, and the number of atoms are equal on both sides. ✓

Q20.5.4

For any spontaneous redox reaction, E is positive. Use thermodynamic arguments to explain why this is true.

Q20.5.4

Solution:

For a redox reaction to be spontaneous at a constant temperature and pressure, ΔG must be negative. Based on the thermodynamic equation \[\Delta G=-nFE_{cell}\] n is the number of electrons transferred in the reaction, which is always a nonnegative number. F is Faraday’s constant (96,485 C/mol e-), which is also nonnegative. Therefore, the only contribution to the sign of ΔG is E. If E is positive, ΔG will be negative, resulting in a spontaneous reaction. If E is negative, ΔG will be positive, resulting in a nonspontaneous reaction. Therefore, E must be positive to give a negative ΔG.

Q24.5.1

How many unpaired electrons are found in oxygen atoms ?
How many unpaired electrons are found in bromine atoms?
Indicate whether boron atoms are paramagnetic or diamagnetic.
Indicate whether F^- ions are paramagnetic or diamagnetic.
Indicate whether Fe²⁺ ions are paramagnetic or diamagnetic.

Q24.5.1

Solution:

Oxygen atoms have 6 valence electrons. Its electron configuration is 1s²2s²2p⁴.

Therefore, Oxygen has 2 unpaired electrons.

Bromine atoms have 7 valence electrons. Its electron configuration is [Ar]4s²3d¹⁰4p⁵.

Therefore, bromine has 1 unpaired electron.

3. Boron atoms have 3 valence electrons. Its electron configuration is 1s²2s²2p¹.

Boron has 1 unpaired e-. Therefore, it is paramagnetic.

4. F^-electron configuration is 1s²2s²2p⁶.

F- has zero unpaired electrons, which means every shell is filled. Therefore, diamagnetic.

5. Fe²⁺ electron configuration is [Ar]3d⁶.

Fe²⁺ has 4 unpaired electrons. Therefore, paramagnetic.

Q14.7.8

The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why?

Q14.7.8

Solution:

There is a keen interest in understanding how enzymes work for designing catalysts for industrial applications because if we study the principles of catalysts and how they operate, we can develop new man made catalysts.

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