Extra Credit 38

Q17.5.6

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

S17.5.6

Batteries have the tendency to go dead because they only have a set amount of supply to use. They produce a current; converting chemical energy into electrical energy, a process that has to eventually stop because the battery is self-contained and has a set amount of material to use before the battery slows down and eventually dies. When a battery dies, it has reached the equilibrium since voltage drops to zero and the current drops. For example, this material can be thought about as the salt in a salt bridge. Fuel cells, on the other hand, have chemicals constantly flowing into the cell, and out of the cell. Fuel cells may use such things as hydrogen and oxygen, so essentially they never die because of the constant flow. And since there is a constant flow, the fuel cell will not reach equilibrium and not go dead.

Q12.3.1

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

S12.3.1

The rate of a reaction for a given chemical reaction is the change in concentration over the change in time measured in speed at which the chemical reaction proceeds. This can be split into two types; the rate of disappearance of reactants which is a negative rate at which the reactants disappear and the rate of formation of products which is the rate at which the products are formed. The rate constant, however, represented by the letter "k", expresses the relationship between the rate of a chemical reaction and the concentrations. It is a measure of the rate of the reaction, and the greater the rate constant, the faster the reaction. This does not depend upon the concentrations of the reactants and changes with temperature. The units for the rate constant dependent upon the order of reaction which can be determined experimentally.

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.

S12.2.10

We need to use an equation that incorporates two different temperatures, rate constants, and activation energy so ou would need to use the "two-point" form of the Arrhenius Equation:

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

Then plug in the information given in the questions to be able to solve for activation energy:

\(\mathrm{ln(\frac{1.1*10^{-2}\frac{L}{mol*s}}{4.95\frac{L}{mol*s}})}\)=\(\frac{-E_a}{8.3145\frac{J}{mol*K}}\)\(\mathrm{(\frac{1}{703K}-\frac{1}{865K})}\)

6.10924=\(\frac{-E_a}{8.3145\frac{J}{mol*K}}\)\((2.6640*10^{-4})\)

Isolate activation energy and solve:

-6.10924\((\frac{8.3145\frac{J}{mol*K}}{2.6640*10^{-4}K})\)=\(-E_{a}\)

-6.10924\((31209.912\frac{J}{mol})\)=\(-E_a\)

-190668.847=\(-E_a\)

Divide by negative one so your activation energy will be positive:

190668.847\(\frac{J}{mol}=E_a\)

Q21.4.5

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

S21.4.4

Electron capture is accompanied by the emission of an X-ray in the sense that as an electron is pulled inwards towards a nucleus, the emission is caused by the difference in energy, leading to X-ray emission, as the electron is captured. Although electron capture causes the emission of an x-ray, it is not the main cause behind the emission. The electron pulled into the nucleus was probably found in the first orbital. Therefore as the electrons falls from a high energy state to a low energy state, the difference in energy must be emitted as a high energy photon in the x-ray range, which is why the emission of an X-ray accompanies electron capture.

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

For these type of problems, you first want to identify the two half reactions, the oxidation and reduction parts. Oxidation is the loss of electrons that appear in the products and reduction is the gain of electrons in the reactants. An acronym for this is Oil Rig.

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

Break the overall reaction into two half reactions and balance the element and charge by adding electrons.

Oxidation: CuS —> Cu²⁺ + SO⁴_2– + 8e^–

Reduction: NO₃^– + 3 e^– —> NO

Next, balance the number of moles of Oxygens on both sides by adding H₂O to the side in need of Oxygen and then add Hydrogen atoms to the side needing Hydrogen.

Oxidation: CuS(s) + 4H2O(l)→ Cu2+(aq) + SO42−(aq) + 8e- + 8H+

Reduction: 3e−+ 4H+ + NO3−(aq) → 2H2O(l) + NO(g)

Now, balance the number of electrons in both the oxidation and reduction half so that they can cancel out.

oxidation: 3*(CuS(s) + 4H2O(l)→ Cu2+(aq) + SO42−(aq) + 8e- + 8H+)

reduction: 2*(3e−+ 4H+ + NO3−(aq) → 2H2O(l) + NO(g))

Multiply the oxidation half by 3 and reduction half by 2 so you will be able to cancel out electrons

oxidation: 3CuS(s) + 12H2O(l)→ 3Cu2+(aq) + 3SO42−(aq) + 24e- + 24H+

reduction: 24e−+ 32H+ +8 NO3−(aq) → 16H2O(l) + 8NO(g)

Cancel out electrons and other terms, such as 12H2O(l) and 24H+ on both sides.

balanced= 3CuS(s) + 8NO3−(aq) +8H+ → 3Cu2+(aq) + 3SO42−(aq) + 8NO(g) + 4H2O(l)

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

Break the overall reaction into two half reactions.

Oxidation: Ag (s) + HS⁻ (aq) → Ag₂S (s)

Reduction: CrO^2-₄ (aq) → Cr(OH)₃ (s)

Now, balance elements other than Oxygen and Hydrogen.

Oxidation: 2Ag (s) + HS⁻ (aq) → Ag₂S (s)

Reduction: CrO²⁻₄ (aq) → Cr(OH)₃ (s)

Proceed by balancing the number of Hydrogen and Oxygen atoms as well as the electrons so that they cancel in the oxidation and reduction reactions as done in example A.

oxidation: 3*(2Ag(s) + HS−(aq) → Ag2S(s) + H++ 2e-)

reduction: 2*(5H++ 3e-+ CrO42−(aq) → H2O(l)+ Cr(OH)3(s))

Multiply the oxidation half by 3 and reduction half by 2 so you will be able to cancel out electrons.

oxidation: 6Ag(s) + 3HS−(aq) → 3Ag2S(s) + 3H+ + 6e-

reduction: 10H+ + 6e- + 2CrO42−(aq) → 2H2O(l)+ 2Cr(OH)3(s)

Cancel out electrons and other terms, such as 3H+ on both sides.

Then add 7OH- to cancel out the 7H+ so you will be in basic solution

balanced in acid= 6Ag(s) + 3HS−(aq) + 2CrO42−(aq) +7H+→ 3Ag2S(s) + 2Cr(OH)3(s) + 2H2O(l)

+7OH- +7OH-

You get 7 waters on the reactant side and that can cancel with the 5 waters on product side leaving 2 waters on product side.

balanced in basic= 6Ag(s) + 3HS−(aq) + 2CrO42−(aq) + 5H2O(l)→ 3Ag2S(s) + 2Cr(OH)3(s) + 7OH-

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

Break the overall reaction into two half reactions and balance charge by adding electrons.

oxidation: Zn(s) → Zn2+(aq) + 2e-

reduction: 2H+ + 2e- + H2O(l) → H2(g) + H2O(l)

Cancel out the electrons

balanced= Zn(s) + 2H+ → Zn2+(aq) + H2(g)

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

Begin by identifying the oxidation and reduction half-reactions.

Reduction: O₂ → H₂O₂

Oxidation: Sb → SbO₂^-

Next, balance the number Oxygens and Hydrogens on both sides by adding H₂O and Hydrogen ions, followed by balancing the charge and making electrons cancel out.

reduction: 3*(2H+ + 2e- + O2(g) → H2O2(aq))

oxidation: 2*( 2H2O(l) + Sb(s) → SbO2−(aq) +4H+ + 3e-)

Multiply the oxidation half by 2 and reduction half by 3 so you will be able to cancel out electrons.

reduction: 6H+ + 6e- + 3O2(g) → 3H2O2(aq)

oxidation: 4H2O(l) + 2Sb(s) → 2SbO2−(aq) + 8H+ + 6e-

Cancel out electrons and other terms, such as 6H+ on both sides.

Then add 2OH- to cancel out the 2H+ so you will be in basic solution

3O2(g) + 4H2O(l) + 2Sb(s) → 3H2O2(aq) + 2SbO2−(aq) + 2H+

+2OH− +2OH−

You have 4 waters on the reactant side and that can cancel with the 2 waters on product side that you get after adding +2OH− leaving 2 waters on the reactant side.

balanced= 3O2(g) + 2Sb(s) +2H2O(l) +2OH− → 3H2O2(aq) + 2SbO2−(aq)

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

Begin by identifying the reduction and oxidation half-reactions. Then Balance the number of all elements but Hydrogen and Oxygen, then balance the number of Hydrogen, Oxygens, and number of electrons and multiply both reactions to cancel out the number of electrons.

reduction: 3*(4H+ + 2e- + UO22+(aq) → U4+(aq) + 2H2O(l))

oxidation: 4H2O(l) + Te(s) → U4+(aq) + TeO42−(aq) + 8H+ + 6e-

Multiply the reduction half by 3 so you will be able to cancel the electrons later.

reduction: 12H+ + 6e- + 3UO22+(aq) → 3U4+(aq) + 6H2O(l)

oxidation: 4H2O(l) + Te(s) → U4+(aq) + TeO42−(aq) + 8H+ + 6e-

Cancel out electrons and other terms, such as 4 H2O(l) and 8H+ on both sides.

balanced= 3UO22+(aq) + Te(s) + 4H+ → 3U4+(aq) + TeO42−(aq) + 2H2O(l)

Q20.5.4

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

S20.5.4

E is positive for any spontaneous redox reaction because for a reaction to be spontaneous it has to have a negative Gibbs free energy and it has that if E is positive.

When you plug in a positive E into G=-nFE you will then get a negative delta G because moles are always positive and F is Faraday's constant, a constant that is positive. So, when you multiple n (the number of electrons, a positive number), Faraday's constant (96,485c/mol), and when Gibbs free energy is negative, the result is a positive number. And when Gibbs free energy is negative it is spontaneous, making a positive E_cell spontaneous as well.

Q20.5.4
a-How many unpaired electrons are found in oxygen atoms?
b-How many unpaired electrons are found in bromine atoms?
c-Indicate whether boron atoms are paramagnetic or diamagnetic.
d-Indicate whether \(F^-\) ions are paramagnetic or diamagnetic.
e-Indicate whether \(Fe^{2+}\) ions are paramagnetic or diamagnetic.

S20.5.4

a-The oxygen atom has 2s22p4 as the electron configuration. Therefore, oxygen has 2

unpaired electrons in the p orbital.

b-A bromine atom has 1s22s22p63s23p64s23d104p5 as the electron configuration.

Therefore, bromine has 1 unpaired electron.

c-The boron atom has 1s22s22p1 as the electron configuration. Because it has one unpaired electron, it is paramagnetic.

d-The F- ion has 1s22s22p6 has the electron configuration. Because it has no unpaired electrons, it is diamagnetic.

e-The Fe2+ ion has 1s22s22p63s23p63d6 as the electron configuration. Because it

has 4 unpaired electrons, it is 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?

S14.7.8

Enzymes are biological so the only way to get them is to grow them, for example like yeast and bacteria, but for catalysts, you can make them in a factory (industrially) and don't need to wait for an organism to grow. However, enzymes are more efficient even though the process to obtain them is longer, and if we can better understand how enzymes work we can channel that into making catalyst more efficient since we can make them faster.