Extra Credit 42.

Q17.6.2- Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel. Mg and Zn

Solution: In order for a sacrificial anode to do its job it must be below the metal it is trying to protect in the Standard Reduction Potential Chart (which means its E⁰ must be less than E_steel). By below, we are looking at the standard reduction potential chart where Lithium is at the bottom. (Li⁺ + e^- ->Li). So we are trying to protect iron from corrosion here. Looking at the reduction potentials, only Mg and Zn are below Fe, according to the reduction potential table below, meaning they could act as sacrificial anodes. While Au, Ni, and Mg are all above the given standard reduction potential of iron which is -0.447 V.

Q17.6.2- Aluminum (E^∘_Al³⁺/Al=−2.07V) is more easily oxidized than iron (E^∘_Fe³⁺/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.

Solution: In this question we are looking at aluminum and iron. First off, Iron rusts when it is exposed to oxygen and water, as expected. Aluminum does not necessarily have the same reaction when exposed to oxygen or water (although aluminum still react with oxygen to form Al₂O₃). Aluminum has a certain corrosion resistance which Iron does not. Aluminum forms a thin layer on its surface that prevents oxygen from coming into contact with the surface. This layer that forms is called Aluminum oxide. When water comes into contact with this layer of aluminum oxide, the aluminum and oxygen atoms on the surface of the metal move apart. So when aluminum oxide gets hydrated or wet, the structure changes to become inert and because of this, it does not corrode to water or air, and it protected the inner layer of aluminum from corrosion.

Q12.3.5- How will each of the following affect the rate of the reaction: CO_(g)+NO_2(g)⟶CO_2(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.

Solution: For number 1, we will be looking at NO₂. Increasing the pressure of the reaction, will increase the rate of the reaction. Since the pressure goes from 0.1 to 0.3 atm, the rate of the reaction would triple because it is proportional. Also the rate of the reaction NO₂ consumption is first order, we can determine this by looking at the exponent value of the NO₂ which is just 1. For the second problem it is under the same circumstances as number 1. The order for CO consumption is 1, and since the concentration goes from 0.02 to 0.06, it is being tripled, meaning the rate will increase.

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.

Solution: In this reaction, our reactants would be Zinc and water. As the amalgam contains zinc, and the question says water is what is reacting with the Zinc. H₂ gas is what forms from the reaction with aqueous Zn²⁺. Since it is basic conditions though, when we split the reaction into its own half reactions, we have to account for the addition of the OH^- for the H⁺. The balanced reactions would be Zn_(s) -> Zn²⁺+2e^- 2H₂O_(l)+2e^- ->H_2(g)+2OH^-_(aq)

To get our second half reaction, we start with H₂O_(l) -> H_2(g),

We would then have to balance the oxygen's on each side. H₂O_(l) --> H_2(g) + H₂O_(l)

Now we must balance the H⁺ missing, 2H⁺_(aq) + H₂O_(l) --> H_2(g) + H₂O_(l)

Since our reaction is in basic conditions we must now add OH^- for every H⁺, 2OH^-_(aq) + 2H⁺_(aq) + H₂O_(l) --> H_2(g) + H₂O_(l) + 2OH^-_(aq)

We can now combine the H⁺ and the OH^-, 2H₂O_(l) + H₂O_(l) --> H2_(g) + H₂O_(l) + 2OH^-_(aq)

Adding the H₂O on the left give us 3, and we subtract the one on the right so we are left with 2 H₂O. It would also have 2e^- to balance the charges. 2H₂O_(l)+ 2e^- --> H2_(g) + 2OH^-_(aq)

With our second reaction being Zn_(s) -> Zn²⁺+2e^-, the 2e^- cancel out, and we can add the half reactions to get our full reaction.

Zn_(s) + 2H₂O_(l) --> Zn²⁺_(aq) + H_2(g) + 2OH^-_(aq)

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?

Solution: In octahedral complexes the t₂g orbital is on the bottom level and the eg orbital is on the top. In order for the configuration to be high-spin we must look at the ligand in the spectrochemical series. Weak field ligands would allow for the complex to be high spin. A strong field ligand means it has a large splitting, so electrons would favor the low spin. For example if the ligand was as strong as CN^-, the complex would be low spin. CN^-, is a strong field ligand which means it has a large splitting, so electrons would favor the low spin. While if we had the weak ligand Cl^-, the splitting is small, meaning electrons could go to the eg orbital meaning it would be high spin.

We can also compare the pairing energy P with splitting energy Δ_0. If P < Δ₀ , like the strong ligand, the d orbital will perform low spin; if P>Δ₀ , like the weak ligand, the d orbital will perform high spin.

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:

The reaction for this question is Co³⁺->Co²⁺+e^-. We are given K₁ and K₂ with T₁ and T₂. We must use an equation that involves these values as well as having the activation energy we are trying to determine.

K₁=0.054 s^-1 T₁=293K An equation we can use is as follows,

K₂=0.100 s^-1 T₂=298K

Our R is the gas constant, which is 8.314 J mol^-1K^-1, Plugging everything in, we get this.

Further simplifying, by taking simplifying fractions on both sides we get

The -6.88x10^-6 comes from the simplification of the fractions and than dividing that by 8.314.

We then divide -.616 by -6.88x10^-6.

Our E_a would then equal 89534 Jmol^-1, We can than simplify that by dividing by 1000, and getting 89.53 Kjmol^-1. E_a=89.53 Kj mol^-1

Q21.4.9- The following nuclei do not lie in the band of stability. How would they be expected to decay?

solution:

The way to proceed with these questions, we must look at the Neutron to proton ratio, which we will refer to as (N/Z).

If the N/Z ratio is high, alpha decay proceeds to decrease the N/Z ratio.

Positron emission increases the N/Z ratio.

Beta Decay decreases the N/Z ratio as well to obtain stability.

The graph on the right is a graph of nuclear stability. Where given protons and neutrons we can look at the graph

and determine what kind of decay would occur.

1. - Here we have 15 protons, and 13 neutrons. Graph it accordingly, we can tell ß decay will occur.
2. - Here we have 92 protons and 143 neutrons. We can't necessarily tell through the graph, but from what was said earlier, the N/Z ratio is high. It would need alpha decay to decrease it.
3. - With 20 protons and 17 neutrons, from the graph we can see, positron emission. It is very small, but it is the red on the graph.
4. - Here we have 3 protons, and 6 neutrons. This goes to the ß decay on the graph.
5. - We have 96 protons and 149 neutrons. A high ratio, but the graph also tells us that it is alpha decay.