Extra Credit 42.

Q17.6.2

Aluminum (E∘Al3+/Al=−2.07V)(EAl3+/Al∘=−2.07V) is more easily oxidized than iron (E∘Fe3+/Fe=−0.477V)(EFe3+/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.

S17.6.2

Untreated aluminum has very good corrosion resistance over iron, because aluminum undergoes self passivation. It reacts with the O₂ in the air to create Al₂O₃ on the surface of the aluminum metal. This creates a barrier between the O₂ in the air and the pure aluminum metal, which is a great way to avoid corrosion. Iron, on the other hand, can't undergo self passivation, so the only way to protect the iron is to coat it in another metal or paint, or use cathodic protection. With neither of those things available, iron will continue to react with the O₂ in the air and corrode.

Q12.3.5

How will each of the following affect the rate of the reaction: CO(g)+NO2(g)⟶CO2(g)+NO(g)CO(g)+NO2(g)⟶CO2(g)+NO(g) if the rate law for the reaction is rate=k[NO₂][CO]?

1. Increasing the pressure of NO₂ from 0.1 atm to 0.3 atm

2. Increasing the concentration of CO from 0.02 M to 0.06 M.

S12.3.5

1. \[\text{rate} = k[NO_2][CO] \]

The rate with NO₂=0.1 is:

• \[\text{rate} = k[.1][CO] \]

You can use the partial pressure of the gas in this equation for concentration

The rate with NO₂=0.3 is:

• \[\text{rate} = k[.3][CO] \]

Here, it is clear that 0.3k[CO]>0.1k[CO], therefore, when you increase the pressure of NO₂ from 0.1 atm to 0.3 atm, the rate of the reaction increases. Another way to think about it is that increasing the pressure of the gas increases the chances of collisions of particles. With a higher chance of collision, the reaction will proceed at a faster rate.

2. \[\text{rate} = k[NO_2][CO] \]

The rate with CO=.02:

• \[\text{rate} = k[NO_2][.02] \]

The rate with CO=.06:

• \[\text{rate} = k[NO_2][.06] \]

You substitute the concentration of CO in each equation for the concentration given in the question, because the rate is determined by concentration of each molecule, and the order of reaction, which is 1 for each individual species.

It is clear that .06k[NO₂]>.02k[NO₂], therefore, increasing the concentration of CO from .02M .06M will increase the rate of reaction. Another way to think about it is that increasing the concentration of CO increases the number of CO molecules, which will in turn increase the changes of collisions happening. The higher chance of collision, the faster the reaction will proceed.

Q12.5.14

The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:

T (K) k (s−1)

293 0.054

298 0.100

S12.5.4

In order to determine the activation energy of a reaction when given the two temperatures and corresponding k values, you use the Arrhenius equation. We can assume: T₁=298, k₁=.1, T₂=293, and k₂=.054. It doesn't matter how you assign these, as long as your corresponding values are both labeled as 1 or 2.

Arrhenius Equation:

\[ \ln \dfrac{k_{1}}{k_{2}} = -\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )\]

• where:
  • E_a=activation energy
  • k=rate constant
  • T=temperature
  • R=.0083145 kJ*mol^-1*K^-1

Next, you plug in your given information to the equation

\[ \ln \dfrac{.1}{.054} = -\dfrac{E_{a}}{.0083145} \left (\dfrac{1}{298}-\dfrac{1}{293} \right )\]

• Then, you do the necessary algebra, and find that:
• E_a=89.47 kJ

Q 21.4.9

The following nuclei do not lie in the band of stability. How would they be expected to decay?

1. 28-P

2. 235-U

3. 37-Ca

4. 9-Li

5. 245-Cm

S21.4.9

To answer these questions you need to look at the belt of stability. The belt of stability is a graph, with the proton number on the x-axis, and the neutron number on the y-axis. In order to check if your element is on the belt of stability, you just go to the graph, locate the number of protons on the graph, locate the number of neutrons, and where those two lines intersect is where the point you are looking at is. If the element lies above the belt of stability, it favors beta emission. If the element lies below the belt of stability, it favors positron emission, and if the atomic number is greater than 83, it favors alpha decay

1. This would undergo positron emission, because it lies below the belt of stability.
2. This would undergo alpha decay, because its atomic number is 82, which is greater than 83.
3. This would favor positron emission, because it lies below the belt of stability.
4. This would favor beta decay, because it lies above the belt of stability.
5. This would favor alpha decay, because its atomic number is 96, which is greater than 83

Q20.2.13

Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.

S20.2.13

This is a redox reaction that we must balance under basic conditions

Overall Equation: Zn(s)+H₂O(l)-->Zn²⁺(s)+H₂(g)

1. Split the reaction into the oxidation and reduction reactions. You can remember this using OIL RIG (oxidation is losing electrons, reduction is gaining electrons). You look at the overall reaction, assign oxidation numbers to the components of the reaction, and decide which element is undergoing reduction and oxidation

• You can see that because Zn starts in the elemental form, it has an oxidation number of 0. It then goes to +2 because the O is -2, and the oxidation state must add to 0.
• H has an oxidation number of +1 as a rule
• O has an oxidation number of -2 as a rule
• So, from this, you can see that because Zn goes from 0 to 2, it is being oxidized, because it is losing electrons
• The Hydrogen goes from +1 to 0, so it is being reduced, and gaining electrons

Half reactions:

• Oxidation: \[ Zn(s)\rightarrow Zn^{2+}\]
• Reduction: \[ 2H^+(l)\rightarrow H_2(g)\]

2. Balance oxygen by adding water to the side of the equation without enough oxygen

• Oxidation: \[ Zn(s)+H_2O\rightarrow ZnO\]
• Reduction: \[ H_2O\rightarrow H_2+H_2O\]

3. Balance the hydrogen atoms

• Oxidation: \[ Zn(s)+H_2O\rightarrow ZnO+2H^+\]
• Reduction: \[ H_2O\rightarrow H_2+H_2O\]

4.Balance the charges of each side of each of the half reactions

• Oxidation: \[ Zn(s)+H_2O\rightarrow ZnO+2H^++2e^-\]
• Reduction: \[ H_2O+2H^++2e^-\rightarrow H_2+H_2O\]
• If the electrons were not equal in both equations, you would have to multiply everything in the equation in order to make the electrons in both half reactions equal

5. Because it is a basic solution, we are going to add a hydroxide for every hydrogen present in the half reaction

• Oxidation: \[ Zn(s)+H_2O\rightarrow ZnO+2H^++2e^-+2OH^-\]
• Reduction: \[ H_2O+2H^++2OH^-+2e^-\rightarrow H_2+H_2O\]

6. You can combine the hydroxide and hydrogen to create water

• Oxidation: \[ Zn(s)+H_2O\rightarrow ZnO+2H_2O+2e^-\]
• Reduction: \[ H_2O+2H_2O+2e^-\rightarrow H_2+H_2O\]

7. Now you can combine the two reactions together

• \[ Zn(s)+H_2O+H_2O+2H_2O+2e^-\rightarrow ZnO++2H_2O+H_2+H_2O+2e^-\]

8.Cancel out like species from each side

• \[ Zn(s)+H_2O\rightarrow Zn^{2+}+H_2+OH^-\]

Q20.5.8

Blood analyzers, which measure pH, P_CO2 , and P_O2 , are frequently used in clinical emergencies. For example, blood P_CO2 is measured with a pH electrode covered with a plastic membrane that is permeable to CO₂. Based on your knowledge of how electrodes function, explain how such an electrode might work. Hint: CO₂(g) + H₂O(l) → HCO₃^-(aq) + H⁺(aq).

S20.5.8

This blood analyzer would work similarly to a galvanic cell

• The anode is where oxidation occurs, and is on the left side of the cell
• The cathode is where reduction occurs and is on the right side of the cell
• Electrons flow from the anode to the cathode
• Ions flow from the salt bridge
• the electrode can be something inert such as Pt or C(graphite), or part of the equation
  • In this case, it is a pH electrode covered in a plastic membrane permeable to carbon dioxide
  • the other pH electrode is covered in a plastic membrane permeable to oxygen gas

In this specific example, we are given a chemical reaction equation as a hint

• The carbon dioxide is at the anode, because it is oxidized as it changes from CO₂ to HCO₃^-
  • the Carbon goes from 2 to 4, so it is losing electrons
• So, the pH electrode at the anode will measure the concentration of the CO₂ and the pH will change as the concentration of carbon dioxide changes
• The oxygen gas is at the cathode, because the CO₂ is at the anode
• The pH electrode at the cathode will measure the concentration of O₂ and the pH will change as the concentration of O₂ changes

So, you add blood to the galvanic cell equipment, and turn it on. As the CO₂ is being oxidized, it is converting to HCO₃^-, and this will change the pH. The magnitude of difference in the starting pH, to the ending pH, will show how much CO₂ was converted to the HCO₃^-, so it can measure how much CO₂ you have in your blood.

Q24.6.4

For an octahedral complex of a metal ion with a d6 configuration, what factors favor a high-spin configuration versus a low-spin configuration?

A24.6.4
For an octahedral complex of a metal ion with a d6 configuration, there are many factors that favor a high spin versus low spin

• The ligands attached to it play a huge role. Strong field ligands favor low spin configuration, and weak field ligands favor high spin configuration.
  • This is because The ligands that are bound very tightly make the crystal field splitting energy very high, so it requires less energy to start double filling the bottom three orbitals
  • Ligands that are bound very weakly
• As a general rule for high spin versus low spin, a configuration is low spin when the pairing energy is lower than the crystal field splitting energy, and is high spin when the pairing energy is higher than the crystal field splitting energy.
  

Q14.7.1

What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor (A)? What effect does it have on the change in potential energy for the reaction?

A catalyst lowers the activation energy of the reaction. It actually provides a different mechanism for the reaction to undergo, which happens to have a lower activation energy. The change in potential energy for the reaction stays the same, because the catalyst does no alter the reactant or product position or energy, it only affects the activation energy and mechanism. It is highly likely that the frequency factor will change in the presence of a catalyst, because if you lower the activation energy, and plug it into the Arrhenius equation, it means that another number must change to keep the equation equal. This means that it is likely A will change.

Arrhenius Equation: \[k = A e^{-E_a/RT}\]

So, because E_a is changing, while T and k stay the same, and R is a constant, it means that A must change in order to keep the other values in the reaction the same.