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.

Given: steel standard potential is -0.447 V

Question: what metals could be used as sacrificial anode

Answer:

A sacrificial anode is a metal that will oxidize faster then the metal that it is protecting thus a lower voltage--> oxodize faster--> sacraficial anode.

In our case we must find the metals with lower voltage

Q17.6.2

Aluminum E_Al³⁺/Al = -2.07V is more easily oxidized 1 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.

Given: Aluminum is more easily oxidized than iron

Answer: Aluminum has an oxide layer that prevents further oxidation, this layer is called "aluminum oxide", when water touches the surface of this layer the aluminum and oxygen atoms on the layer move apart. because the structure becomes inert it does not corrode when hydrated.

Phase II:

When iron is exposed to oxygen or water, it rusts. However, aluminum does not react in a similar fashion because of the fact that it has very good corrosion resistance. Specifically, aluminum undergoes the process of self-passivation, in which it reacts with oxygen and creates a thin layer of aluminum oxide on the surface of the metal. This, in essence, protects the actual aluminum metal from corroding as it prevents the oxygen from touching the metal.

Q12.3.5

How will each of the following affect the rate of the reaction:

\[\text{CO}(\mathit{g})+\text{NO}_2(\mathit{g})\rightarrow\text{CO}_2(\mathit{g})+\text{NO}(\mathit{g})\]

If the rate law for the reaction is

\[\text{Rate}{=K}{[CO][NO_2]}\]

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.

a) increasing the pressure of NO₂ from .1 atm to .03 atm means that the pressure on the system is increased 3x. this means that the particles have more kinetic energy and are moving faster and thus colliding at a higher rate. because this is a first order reaction the rate of reaction and increase in temperate or concentration will correlate directly. 3x increase in pressure means 3x increase in the rate of reaction.

\[\text{Rate}{=K}{[CO][0.1]}\]

\[\text{Rate}{=K}{[CO][0.3]}\]

it becomes apparent when solving that the rate will increase by 3x

-You must be careful here because this application only works for 1st order reactions, as the order changes the principles of proportionality changes as well. This same rule applies to part two of this problem.

b) Increase the concentration of the CO from .02 to .06 means that the concentration of the cobalt molecules increase 3x. This means that there is an increase in the amount of collisions because of the fact that there are simply more molecules to collide into and because this too is a first order reaction then the correlation between concentration increase and rate of reaction increase is direct. this means there is 3x increase in the rate of reaction as well.

\[\text{Rate}{=K}{[0.2][NO_2]}\]

\[\text{Rate}{=K}{[0.6][NO_2]}\]

Once you solve this out you can see that the rate will indeed increase 3x

Phase II:

Increasing the amount of reactants that are involved in the overall rate law directly affect the rate of the reaction. In this case, because each reactant in the rate law is in first order, how you manipulate the reactants is directly proportional to how the rate is affected. Therefore, for each question, multiplying the reactants by a factor of 3 will also multiply the rate by a factor of 3.

Q21.4.9

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

\(\ce{^{28}_{15}P}\)

\(\ce{^{235}_{92}U}\)

\(\ce{^{37}_{20}Ca}\)

\(\ce{^{9}_{3}Li}\)

\(\ce{^{245}_{96}Cm}\)

This is a plot of the number of neutrons versus the number of protons. In this plot you can simply plot the number of protons and number of neutrons on the x y axis. When this is plotted for isotopes of elements it can be seen that there is a correlation between the number of neutrons and protons and the respective stability of the molecule. This is shown here on the graph. The nuclide is most stable in 1:1 ratio and the respective colors represents the decay (can be seen in the key of the graph).

(......the top values is protons plus neutrons and bottom is only protons using subtraction you can find individual protons plus neutrons and can plot this on the x y axis to see the type of decay that would be plotted in our case we have.....)

(a) β decay; (b) α decay; (c) positron emission; (d) β decay; (e) α decay

As all of these follow the same general principle it will be useful to cover one of these in depth, i will be covering question (a)...

\(\ce{^{28}_{15}P}\)

Here we the proton + neutrons value:

\[protons + neutrons = 28\]

the protons value is found via the bottom number

\[protons= 15\]

using substitution:

\[15protons + neutrons = 28\]

thus using subtraction we can find neutrons:

neutrons= 13

Now we can plot on the graph for 13 neutrons and 15 protons, essentially the point (15,13), when plotted we can see that the decay falls under β (specifically positron emission) decay.

Phase II:

Using the same procedure as above for the rest of the problems, we see that:

b. \(\ce{^{235}_{92}U}\)

This isotope has 92 protons and 143 neutrons. The graph is slightly unclear, but because the N/Z is high, it undergoes alpha decay.

c. \(\ce{^{37}_{20}Ca}\)

Here, there are 20 protons and 17 neutrons, which means it undergoes β decay (positron emission).

d. \(\ce{^{9}_{3}Li}\)

There are 3 protons and 6 neutrons and this falls under β decay on the graph (electron emission).

e. \(\ce{^{245}_{96}Cm}\)

In this isotope, 96 protons and 149 neutrons are present. Again, this is a high N/Z ratio, which indicates alpha decay.

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.

Question: write the balanced chemical equation of the reaction between zinc and water under basic conditions, one product will be hydrogen gas.

Solution: In this reaction our reactants will be zinc, which is used in the amalgams fillings, and water which causes the reaction of Hydrogen gas release. One important factor to consider is this reaction is basic so we must look at OH- for H+ changes.

First and foremost we must write the half reaction equations:

oxidation and reduction respectively:

\[\text{Zn}(\mathit{s})\rightarrow\text{ZnO}\]

\[\text{OH}_2(\mathit{l})\rightarrow\text{H}_2\]

Now we must balance the oxygen out of both reactions by adding water molecules:

\[\text{Zn} (\mathit{s}) + \text{OH}_2\rightarrow\text{ZnO}\]

\[\text{OH}_2 (\mathit{l})\rightarrow\ce{H2}(\mathit{g})+\ce{H2O}(\mathit{l})\]

Now we can balance the hydrogen atoms:

\[\text{Zn}(\mathit{s}) + \ce{H2O}\rightarrow\text{ZnO} +\text{2H}^+\]

\[\text{OH}_2 (\mathit{l}) + \text{2H}^+\rightarrow\text{H}_2 +\ce{H2O}\]

Balance the charges of e- on each side

\[\text{Zn}(\mathit{s}) + \ce{H2O}(\mathit{l})\rightarrow\text{ZnO} +\text{2H}^+ + \text{2e}^-\]

\[\text{OH}_2 +\text{2H}^+ + \text{2e}^- \rightarrow\text{H}_2 +\ce{H2O}\]

Remember that we are solving for a basic soluition so we must had a hydroxidr for every hydrogen present.... these will combine to make water and will cancel out to reach the final solution

\[\text{Zn}(\mathit{s}) + \ce{H2O}(\mathit{l})\rightarrow\text{ZnO} +\text{2OH}_2 + \text{2e}^-\]

\[\text{2e}^- + \ce{2H2O}(\mathit{l}) + \ce{H2O}\rightarrow\text{H}_2 +\ce{H2O}\]

Now you combine the reaction and cancel our species from both sides.

\[\text{Zn}(\mathit{s}) + \ce{H2O}(\mathit{l})\rightarrow\text{Zn}^2 +\text{H}_2 + \text{OH}^-\]

Phase II: This is slightly incorrect! Here is how you do it:

As stated above, our reactants will be zinc and water in this reaction. The first step in attacking this problem is splitting it into two half-reactions:

\(\mathrm{Zn}(s)\rightarrow\mathrm{Zn^{+2}}(aq)+\mathrm{2e^-}\)

\(\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)\)

The first reaction is already balanced and all taken care of, so we move to the second reaction.

So, we start with what we have:

\(\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)\)

Next, we have to balance the oxygens on each side as by adding \(\mathrm{H_2O}(l)\) as follows:

\(\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)+\mathrm{H_2O}(l)\)

Now, we must balance the hydrogens by adding H⁺:

\(\mathrm{2H^+}(aq)+\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)+\mathrm{H_2O}(l)\)

Then, because our reaction is in basic conditions, we must add OH^- for every H⁺:

\(\mathrm{2OH^-}(aq)+\mathrm{2H^+}(aq)+\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)+\mathrm{H_2O}(l)+\mathrm{2OH^-}(aq)\)

We can now combine the hydrogen and hydroxide ions for create water molecules:

\(\mathrm{2H_2O}(aq)+\mathrm{H_2O}(l)\rightarrow\mathrm{H_2}(g)+\mathrm{H_2O}(l)+\mathrm{2OH^-}(aq)\)

Obviously we can combine the water molecules on the reactant side of the reaction to create three water molecules. However, because there is a water molecules on the product side of the reaction as well, we are left with only two water molecules on the left side of the equation. Therefore, we are left with:

\(\mathrm{2H_2O}(l)\rightarrow\mathrm{H_2}(g)+\mathrm{2OH^-}(aq)\)

Now we have to balance the charges by adding electrons. The left side of the equation has no charge, while the right side has a charge of -2, so we have to add two electrons to the left side of the equation. The overall reaction would be:

\(\mathrm{2H_2O}(l)+\mathrm{2e^-}\rightarrow\mathrm{H_2}(g)+\mathrm{2OH^-}(aq)\)

Finally, we have to add the half-reactions together (electrons cancel out) to receive an overall reaction of:

\(\mathrm{Zn}(s)+\mathrm{2H_2O}(l)\rightarrow\mathrm{Zn^{2+}}(aq)+\mathrm{H_2}(g)+\mathrm{2OH^-}(aq)\)

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).

Question: Explain how an electrode might work.

Answer: An electrode can either be an cathode or anode, this is determined by an area where current can enter and leave the system. When currents leaves electrode it is cathode and when it enters the electrode it is anode. When an electrode is close to to molecule then the electrode solution will donate electrons. This causes the atoms to become positive ions. When the electrode behaves as a cathode then the electrons are released from electrode and the solution around it is reduced.

The pH electrode can work by the following reaction:

\[\text{CO}_2(\mathit{g})+\text{OH}_2(\mathit{l})\rightarrow\text{H}^+_2(\mathit{aq})+\text{HCO}_3^-(\mathit{aq})\]

the carbon dioxide gas and water when close to the electrode the products most importantly the H+ ion concentration can be measured using the Nernst equation this can determined the voltage of the system and given voltage it is relatively easy to find pH from calculator or the electrode system itself.

Phase II:

This example would work similarly to a galvanic cell. Galvanic cells contain electrodes, which are electrical conductors used to make contact with a nonmetallic part of a circuit. An electrode in a galvanic cell is referred to as either an anode (where oxidation occurs) or a cathode (where reduction occurs). They can be inert (i.e. Pt or graphite), or they can be a metal that is part of the equation. In this case, there are two pH electrodes, one covered in a plastic membrane permeable to carbon dioxide and one covered in a plastic membrane permeable to oxygen. The carbon dioxide is at the anode (which is measured by the pH electrode), while the oxygen is at the cathode. As the \(\mathrm{CO_2}(g)\) is being oxidized, it is being converted to \(\mathrm{HCO_3^-}(aq)\), which changes the pH.

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:

To solve this type of problem you should use the arrhenius equation:

\[\text{K=A}^{-Ea/RT}\]

Given: two temperature readings and the rate readings at both of those specific temperatures

Solution: we must find the activation energy from the given data.

1) This equation can be derived from the arrhenius equations:

\[ln\frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})\]

2) Rearrange the equation to solve for the activation energy (E_a).

\[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}}\]

3) now we can just plug in our given values into the equation

\[ E_a = \dfrac{8.314 \ln \dfrac{0.100}{0.054}}{\dfrac{1}{298}-\dfrac{1}{293}}\]\[E_{a}= 89461.428\, J\, mol^{-1}\]

Phase II: This is slightly incorrect as the rate constants and the corresponding temperatures are mixed up.

Therefore, as state above, you rearrange the Arrhenius equation to solve for the activation energy:

\[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}}\]

And from here it is simply just plug and chug:

\[ E_a = \dfrac{8.314 \ln \dfrac{0.054}{0.100}}{\dfrac{1}{298}-\dfrac{1}{293}}\]

\[E_{a}= 89534\, J\, mol^{-1}\]

or

\[E_{a}= 89.53\, kJ\, mol^{-1}\]

k2k

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?

Given: the complex is octahedral, the configuration is d⁶.

There are 4 things that we should try to know before we make assumptions about whether the function is high spin or low spin. we must know the shape (octahedral in our case), we must know the configuration (d⁶ in our case), we must know the ligand ( unknown in our case) and lastly we must know the splitting energy (unknown in our case).

A high spin configuration would be favored if the ligand is a weak field ligand and the splitting energy must be small.

A low spin configuration would be favored if the ligand is a strong field ligand and the the splitting energy must be big.

To figure out if the ligand is weak/strong and splitting energy small/big we must use the spectrochemical series.

Given Here:

I⁻ < Br⁻ < S²⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < N₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < CH₃CN < py < NH₃

< en < bipy < phen < NO₂⁻ < PPh₃ < CN⁻ < CO

the stronger the ligand the more change that it can bring to the d orbital energy levels, stronger ligand means more energy. In our case the ligands after the H₂O molecules will become paired because they have more energy and it takes energy to pair two electrons together ( remember electrons are repulsive to each other) thus they produce a low spin complex. The molecules below H₂O will produce high spin because they do not have much energy and cannot pair until there is no more space inside the d orbital.

Phase II:

Crystal field theory describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. In octahedral complexes, the d-orbital splits into two different levels, three lower energy levels, collectively referred to as \(t_{2g}\), and the two upper energy levels, collectively referred to as \(e_g\). The difference in energy between the two sets of d orbitals is called the crystal field splitting energy, \(\Delta_o\) (or \(\Delta_{oct}\)). The magnitude of \(\Delta_o\) dictates whether a complex is high spin or low spin. Strong field ligands increase the distance between the t_2g orbitals and e_g orbitals (\(\Delta_o\)) and result in low-spin complexes, while weak field ligands result in high-spin complexes.

Q14.7.10

At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (ES) and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations:

enzyme (E) + substrate (S) ⇔ enzyme- substrate complex (ES) ⇔ enzyme (E) + Product (P)

This can also be shown as follows:

E + S ⇔ (forward k₁ reverse k_-1)ES ⇔ (forward k₂ reverse k_-2 ) E + P

Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?

For this problem first you must determine the rate constants for each step of the reaction.

\[\text{Rate}{=K_1}{[E][S]}\]

\[\text{Rate}{=K_{-1}}{[ES]}\]

\[\text{Rate}{=K_{-2}}{[E][P]}\]

\[\text{Rate}{=K_2}{[ES]}\]

Now we can focus on what we want to solve for, here the question is asking for us to write the expression for the rate of disappearance of the enzyme substrate complex.

Notice that this reaction is at equilibrium (double arrows) and that the forward reaction is positive ex.(k₁) and the reverse reactions are negative ex.(k_-1).

Notice that if the elementary step shows Es on the reactant side that means that the ES is disappearing.

If we combine the two disappearing elementary steps then we add the two rate laws.

The rate law would be:

\[\text{Rate}{=K_1}{[E][S]}+\text{K}_2\mathrm{[ES]}\]

once we factor [ES] out we can get the rate law for the reverse reaction of the formation of [ES]

\[\text{Rate}{=[ES]}{K_{-1}}+\text{K}_2\]

Notice that right now our value is negative, we want the rate law to be with respect to the appearance of [ES] so now we must add these rates together.

\[\text{Rate}{=}{K_{-1}}{[E][S]}+\text{K}_2{[E][P]}\]

this is forward reaction, now we add the reverse and forward together

\[\text{Rate}{=-[ES]}{K_{-1}}+\text{K}_2 +{K_1}{[E][S]}+\text{K_-2}{[E][P]}\]

b)

Now we must determine the reaction order. The reaction order would be zero. Considering the fact that the question states if you vary the substrate concentration and enzyme concentrations are small and substrate concentration high. Then it doesn't matter what the concentration is because the limiting factor is that there is simply not enough enzymes to handle the concentration of substrates regardless of how high or how low it is. Thus the answer is zero order the rate is independent of the concentrations.

Phase II: This is slightly incorrect!

As stated above, the rate laws for each individual step are as follows:

\(k_1[E][S]\)

\(k_{-1} [ES]\)

\(k_{-2} [E][P]\)

\(k_2[ES]\)

In this reaction, k₁ and k_-2 represent rates of appearances because those reactions form ES, while k_-1 and k₂ represent rates of disappearance because their reactions consume ES. If a reaction is showing the appearance of ES it will have a positive sign in the rate law and if the reaction is depicting disappearance, there will be a negative sign in the rate law. Therefore, once you combine these individual steps, the overall equation will be:

\(\frac {Δ[ES]}{Δt}=−(k_2+k_{-1})[ES]+k_1[E][S]+k_{-2}[E][P]\)

And according to this equation, the reaction order will be zero order, as the rate of appearance of ES equals rate of disappearance of ES.