Extra Credit 38

Q17.5.6

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

S17.5.6

A battery is a self-contained source of electrochemical power that utilizes redox reactions (galvanic cells). The forward reactions are thermodynamically favored; however, after a certain amount of uses, the electrodes are dissolved, and the cell can no longer be used. Typically these batteries die because the concentration of reactants reduces over time increasing the tendency for non-spontaneous reactions to occur. This is also the reason why \( E^{o}_{cell}\) drops in magnitude over time. It is important to note that secondary batteries can be recharged to allow the spontaneous reactions to run again. -Shail Trivedi For fuels cells, they are not self-contained. The electrodes are not eroded so long as a constant flow of gaseous reactants (hydrogen and oxygen) is supplied. Therefore, so long as there is this outside flow of reactants entering the system, the fuel cell will continue to run.

Q12.3.1

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

S12.3.1

The rate of the reaction varies with concentration (except zero-order reactions) and the value of the rate constant. The units of the rate of reaction will always be concentration over time (i.e. M/s) The rate constant value stays the same regardless of the concentration and is specific the a certain reaction. The units of the rate constant varies with the reaction order. (i.e. zero order - M/s; first order - 1/s; second order - 1/(Mxsec), etc.) A useful formula to determine the units of the rate constant is as follows:

\( \text{Let w = reaction order} \dots\)

\[ \text{units of k } \to M^{1-w} s^{-1}\]

-Shail Trivedi

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.

S12.5.10

The equation for relating the rate constant and activation energy of a reaction is the Arrhenius equation:

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

When given two rate constants at two different temperatures but for the same reaction, the Arrhenius equation can be rewritten as:

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

In this problem, all the variables are given except for the E_a (activation energy).

k₁ = 1.1 × 10⁻² L/mol/s

T₁ = 703 K

k₂ = 4.95 L/mol/s

T₂ = 865 K

R = 8.314 J/(mol K) (Ideal Gas Constant)

Now plug in all these values into the equation, and solve for E_a.

\[ln (\frac{4.95\frac{L}{mol×s}}{1.1 × 10^{-2}\frac{L}{mol×s}}) = \frac{E_a}{8.314 × 10^{-3}\frac{kJ}{mol×K}} (\frac{1}{703} - \frac{1}{865})\]

E_a = 190 kJ (2 sig figs)

I agree with the above explanation and feel that the work shown properly demonstrates how to do the problem. -Shail Trivedi

Q21.4.5

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

S21.4.5

During electron capture, one of the protons' charge is neutralized (the negative electron neutralizes the proton's positive charge). The electron reacts with a proton to produce a neutron.

When a nucleus captures an electron, the electron is pulled from the inner electron shell to the nucleus. This causes an empty space where the electron should be. To compensate for that space where the lower electron should have been, an electron from a higher energy level falls into the lower energy level inner shell. This fall in energy is emitted as a photon in the x-ray range. Also, note that the x-ray emitted has no atomic mass. -Shail Trivedi

Q20.2.9

Balance each redox reaction under the conditions indicated.

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

S20.2.9

To balance redox reactions in acidic conditions, follow these steps:

1. Write the 2 half reactions (oxidation and reduction) that make up the overall reaction
2. Balance all atoms that are not O or H
3. Balance O atoms by adding H₂O
4. Balance H atoms by adding H⁺
5. Balance charges by adding e^-
6. Multiply each half reaction by a coefficient to cancel out the e^-
7. Add the resulting half reactions

In basic conditions, follow the same procedure except add OH^- for every H⁺ added. When H⁺ and OH^- are on the same side of the reaction, they will make H₂O.

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

• CuS(s) + 4H₂O(l) → Cu²⁺(aq) + SO₄²⁻(aq)
  • CuS(s) + 4H₂O(l) → Cu²⁺(aq) + SO₄²⁻(aq) + 8H⁺(aq)
  • CuS(s) + 4H₂O(l) → Cu²⁺(aq) + SO₄²⁻(aq) + 8H⁺(aq) +8e^-
  • [CuS(s) + 4H₂O(l) → Cu²⁺(aq) + SO₄²⁻(aq) + 8H⁺(aq) +8e^-] x3
• NO₃⁻(aq) → NO(g) + 2H₂O(l)
  • NO₃⁻(aq) + 4H⁺(aq) → NO(g) + 2H₂O(l)
  • NO₃⁻(aq) + 4H⁺(aq) + 3e^-→ NO(g) + 2H₂O(l)
  • [NO₃⁻(aq) + 4H⁺(aq) + 3e^-→ NO(g) + 2H₂O(l)] x8
• Add the 2 equations
  • 3CuS(s) + 8NO₃⁻(aq) + 8H⁺(aq) → 3Cu²⁺(aq) +3 SO₄²⁻(aq) + 8NO(g) + 4H₂O(l)

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

• CrO₄²⁻(aq) → Cr(OH)₃(s)
  • CrO₄²⁻(aq) → Cr(OH)₃(s) + H₂O(l)
  • CrO₄²⁻(aq) + 5H⁺(aq) → Cr(OH)₃(s) + H₂O(l)
  • CrO₄²⁻(aq) + 5H⁺(aq) + 5OH^-(aq) → Cr(OH)₃(s) + H₂O(l) + 5OH^-(aq)
  • CrO₄²⁻(aq) + 5H₂O(l) → Cr(OH)₃(s) + H₂O(l) + 5OH^-(aq)
  • CrO₄²⁻(aq) + 5H₂O(l) + 3e^- → Cr(OH)₃(s) + H₂O(l) + 5OH^-(aq)
• Ag(s) + HS^- → Ag₂S(s)
  • 2Ag(s) + HS^-(aq) → Ag₂S(s)
  • 2Ag(s) + HS^-(aq) → Ag₂S(s) + H⁺(aq)
  • 2Ag(s) + HS^-(aq) + OH^-(aq)→ Ag₂S(s) + H⁺(aq) + OH^-(aq)
  • 2Ag(s) + HS^-(aq) + OH^-(aq)→ Ag₂S(s) + H₂O(l)
  • 2Ag(s) + HS^-(aq) + OH^-(aq)→ Ag₂S(s) + H₂O(l) + 2e^-
• Add the 2 equations
  • [CrO₄²⁻(aq) + 5H₂O(l) + 3e^- → Cr(OH)₃(s) + H₂O(l) + 5OH^-(aq)] x2
  • [2Ag(s) + HS^-(aq) + OH^-(aq)→ Ag₂S(s) + H₂O(l) + 2e^-] x3
  • 6Ag(s) + 3HS⁻(aq) + 2CrO₄²⁻(aq) + 5H₂O(l) → 3Ag₂S(s) + 2Cr(OH)₃(s) + 7OH^-(aq)

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

• Zn(s) → Zn²⁺(aq)
  • Zn(s) → Zn²⁺(aq) + 2e^-
• H₂O(l) → H₂(g) + OH^-(aq)
  • H₂O(l) → H₂(g) + H₂O(l)
  • H₂O(l) + 2H⁺(aq) → H₂(g) + H₂O(l)
  • 2H⁺(aq) + 2e^- → H₂(g)
• Add the two equations
  • Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

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

• O₂(g) → H₂O₂(aq)
  • O₂(g) + 2H⁺(aq) → H₂O₂(aq)
  • O₂(g) + 2H⁺(aq) + 2OH^-(aq) → H₂O₂(aq) + 2OH^-(aq)
  • O₂(g) + 2H₂O(l) + 2e^-→ H₂O₂(aq) + 2OH^-(aq)
• Sb(s) → SbO₂⁻(aq)
  • Sb(s) + 2H₂O(l) → SbO₂⁻(aq)
  • Sb(s) + 2H₂O(l) → SbO₂⁻(aq) + 4H⁺(aq)
  • Sb(s) + 2H₂O(l) + 4OH^-(aq) → SbO₂⁻(aq) + 4H⁺(aq) + 4OH^-(aq)
  • Sb(s) + 4OH^-(aq) → SbO₂⁻(aq) + 2H₂O(l) + 3e^-
• Add the 2 equations
  • [O₂(g) + 2H₂O(l) + 2e^-→ H₂O₂(aq) + 2OH^-(aq)] x3
  • [Sb(s) + 4OH^-(aq) → SbO₂⁻(aq) + 2H₂O(l) + 3e^-] x2
  • 3O₂(g) + 2Sb(s) + 2OH^-(aq) + 2H₂O(l) → 3H₂O₂(aq) + 2SbO₂⁻(aq)

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

• UO₂²⁺(aq) → U⁴⁺(aq)
  • UO₂²⁺(aq) → U⁴⁺(aq) + 2H₂O(l)
  • UO₂²⁺(aq) + 4H⁺(aq) → U⁴⁺(aq) + 2H₂O(l)
  • UO₂²⁺(aq) + 4H⁺(aq) + 2e^- → U⁴⁺(aq) + 2H₂O(l)
• Te(s) → TeO₄²⁻(aq)
  • Te(s) + 4 H₂O(l) → TeO₄²⁻(aq)
  • Te(s) + 4 H₂O(l) → TeO₄²⁻(aq) + 8H⁺(aq)
  • Te(s) + 4 H₂O(l) → TeO₄²⁻(aq) + 8H⁺(aq) + 6e^-
• Add the 2 equations
  • [UO₂²⁺(aq) + 4H⁺(aq) + 2e^- → U⁴⁺(aq) + 2H₂O(l)] x3
  • Te(s) + 4 H₂O(l) → TeO₄²⁻(aq) + 8H⁺(aq) + 6e^-
  • 3UO₂²⁺(aq) + Te(s) + 4H⁺(aq) → 3U⁴⁺(aq) + TeO₄²⁻(aq) + 2H₂O(l)

The steps above used to balance the redox reactions are detailed well in a formulaic manner. The steps used above are appropriate for balancing redox reactions in acidic or basic solutions. -Shail Trivedi

Q20.5.4

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

S20.5.4

The spontaneity of a reaction is determined by

To relate spontaneity and the cell potential, E, look towards the equations:

\[ΔG = ΔG^{o} + RTlnQ\]

\[E = E^{o} - \frac{RT}{nF} lnQ\]

ΔG=−nFE

(Cell potential is related to Gibbs energy by the above equation)

We know that a reaction is thermodynamically spontaneous when ΔG < 0. Therefore, we have to find the sign of E that will make ΔG negative. ΔG will be negative when E is positive because F (Faraday's constant) and n (number of moles) will always be positive. A positive cell potential The more negative the value of \( \Delta G\), the more spontaneous the reaction. The same concept holds true for \(E_{cell} \) where the more positive the value, the more spontaneous the reaction. -Shail Trivedi

Q24.5.1

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

S24.5.1

1. How many unpaired electrons are found in oxygen atoms ?
   • Oxygen has an electron configuration of [He] 2s²2p⁴. There are 4 electrons in the incompletely filled outer 2p shell. This leaves 2 e^- unpaired.

   • 

2. How many unpaired electrons are found in bromine atoms?
   • Bromine has an electron configuration [Ar] 4s²3d¹⁰4p⁵. The 4p shell is incompletely filled with 5 e^-. This leaves 1 unpaired e^-.

   • 

3. Indicate whether boron atoms are paramagnetic or diamagnetic.
   • Paramagnetic atoms have at least 1 unpaired electron. Diamagnetic atoms have no unpaired electrons. Boron has an electron configuration of [He] 2s²2p¹. There is one unpaired electron in the 2p orbital, so boron atoms are paramagnetic.
   • 
4. Indicate whether F^- ions are paramagnetic or diamagnetic.
   • The electron configuration of F^- is [Ne] because there is an extra electron making it an ion. All the orbitals are filled, and all electrons are paired. Therefore, F^- is diamagnetic.
   • 
5. Indicate whether Fe²⁺ ions are paramagnetic or diamagnetic.
   • The Fe normally has an electron configuration of [Ar] 4s²3d⁶. For the Fe²⁺ ion, 2 electrons are removed from the 4s orbital because it is more favorable energy-wise to do so. The electron configuration for Fe²⁺ is [Ar] 3d⁶. There are 6 e- in the 3d shell, which can normally hold 10 e-. Therefore, there are 4 unpaired electrons making Fe²⁺ paramagnetic.
   • (Some examples

Also note that diamagnetic substances have no unpaired electrons and are weakly repelled by magnets, while paramagnetic substances have one or more unpaired valence electrons that allow it to be weakly attracted by magnets. -Shail Trivedi

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?

S14.7.8

Enzymes are expensive to make, denature and fail at certain temperatures, are not that stable in a solution, and are very specific to the reaction it was made for. However, scientists can use the observations from enzymes to create catalysts that are more effective in aiding the reaction and cost less to produce. Overall, catalysts still play a large part in lowering the activation energy for reactions. Creating new catalysts can help in the improvement of areas such as medical, ecological, and even commercial products. Like the student mentioned, there is an interest to develop these enzymatic catalysts for use in industrial applications because these proteins are highly specific to a certain reaction, allowing a reaction to proceed at a higher rate as the activation energy of the reaction is dropped. -Shail Trivedi