Extra Credit 44

Question 17.6.3

If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.

Solution 17.6.3

The reactions mentioned above are examples of cathodic protection. Cathodic protection is a way to protect metals from corrosion. Oxidation (corrosion) occurs at the anode, which is the metal that has the more negative standard reduction potential. Therefore, to protect a metal from corrosion you use a metal that is more reactive( to protect a metal from corrosion you use a metal that is preferentially oxidized although more reactive is still good). In general, the sacrificial anode which protects the parent metal should be lower(more negative) in the activity series. (you use an anode that is lower than the parent metal because when it's lower in the activity series, it will supply the electron and allow the parent metal to act as the inert electrode and avoid oxidation) (See Standard Reduction Potential for more on using the activity series.)

In the first reaction between iron and zinc, zinc (−0.7618 V) has a more negative standard reduction potential (SRP) than iron (−0.447 V) which indicates that it's the anode. Zinc acts as the sacrificial anode and is the metal that corrodes (gets oxidized), iron is an inert (chemically inactive) metal.

In the second case between iron (−0.447 V) and copper (0.34 V), iron has the more negative SRP and serves as the (sacrificial) anode which means it undergoes corrosion (gets oxidized). The copper is protected from corrosion and remains inert.

Red: Brandon Chan corrections although most were just clarifications.

Question 12.3.7

Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:

\[ ^{32}_{15}\text{P} \rightarrow ^{32}_{16}\text{S} +e^−\]

Rate = 4.85 × 10⁻² \(\text{day}^{−1} [^{32}\text{P}]\)

What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?

Solution 12.3.7

This question is asking to find the rate of production of electrons with a given concentration and rate constant of phosphorus. Because radioactive decay is a first order process, the rate law for this reaction is \[\text{rate} = k[^{32}_{15}\text{P}]\] where k is the rate constant given: 4.85 × 10⁻² \(\text{day}^{−1} [^{32}\text{P}]\), and the concentration of phosphorus is 0.0033M. Plug in the numbers into the rate law equation gives you

\(\text{rate}\) = (4.85 × 10⁻² \(\text{day}^{−1}\)) x (.0033M) = 1.6005 x 10^-13 \(M \text{day}^{−1}\)

I would expand the solution to this:

I'm not sure if this is a typo because the question gives us the rate, but it also gives us the concentration of P, so I'm not sure if the 4.85 × 10⁻² should be treated as a rate constant, k, or as the actual rate because later on it gives us the concentration of P which would be redundant if we are given the rate. But if we are to interpret this question the person above interpreted it I would add this:

We can compare the rate of formation of reactants and products for any equation like:

aA ⟶ bB + cC

Where rate = -(1/a)*(Δ[A]/Δt) = (1/b)*(Δ[B]/Δt) = (1/c)*(Δ[C]/Δt)

rate=k[P] =(1/n)(Δ[e^-]/Δt) where n is the stoichiometric coefficient of electrons in the equation, which is equal to 1. Δ[e^-]/Δt is equal to the instantaneous rate of production of electrons.

We are given Given k and [P] we can solve for the equation

rate= (4.85 × 10⁻² 1/(day))(0.0033 M)==(1/1)(Δ[e^-]/Δt)= 1.60x10^-4 M/day

Also I'm not sure if the rate constant is correct because if this is a first order reaction, wouldn't the rate constant be 1/time rather than

The rate I got was different from the person above.

The other way this question could be interpreted is as follows: the rate they give us is the actual rate of the reaction which would coincide with the rate they give us because the rate's units are M/day. If we follow this interpretation, we know that the rate of the rxn is

rate=(Δ[e^-]/Δt)

so in this case, we would consider the instantaneous rate of electron production equal to the rate they give us.

so (Δ[e^-]/Δt)= 4.85 × 10⁻² M/(day

Question 12.5.16

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first \(A + BC\rightarrow \ AB + C\) reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?

Solution 12.5.16

This question asks to figure out what happens in two different scenarios: when (1) total energy is above the transition state and (2) when total energy is below the transition state. The two postulates of the Collision Theory that are relevant to this question is

1. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
2. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).

Below is a standard energy diagram

The transition state is the peak of the graph. For a reaction to occur, it (the reactants A and B) has to have adequate energy to overcome the activation energy (E_a) for the reactants A and B to turn into C and D. (so the reactants must have an energy level above the transition state to proceed)

The first scenario (total energy is above the transition state)

Here you can see that a reaction occurred between molecule A and molecule BC. The molecules AB(yellow and purple) and C(navy blue) is created. There is definitely enough energy for a reaction to occur in this case because total energy is well above the transition state. However, not every collision results in a reaction regardless of enough energy or not. Sometimes a reaction doesn't occur because the molecules didn't collide at the proper orientation.

The second scenario (total energy below transition state)

Nothing changes because there is not enough energy to overcome the activation energy to get past the transition state. This results in no reaction occurring regardless of the orientation that it collides.

I would swap these two solutions so that they would coincide with the order of the question. But the solutions are correct.

Question 21.4.11

Write a nuclear reaction for each step in the formation of \( ^{218}_{84}\text{Po}\) from \( ^{238}_{92}\text{U}\), which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order.

Solution 21.4.11

This question is asking to write a series of nuclear reactions for the radioactive decay of \( ^{238}_{92}\text{U}\) into \( ^{218}_{84}\text{Po}\). For this question you need to know about two radioactive decay reactions: alpha and beta decay.

Alpha decay results in an alpha particle (helium) being emitted from the nucleus. The mass number of the atom is reduced by 4 and the atomic number reduced by 2. The general formula for writing an alpha decay reaction is: \[^{A}_{Z}\text{X} \rightarrow ^{A–4}_{Z–2}\text{Y}+ ^{4}_{2}\text{He}\]

Beta decay is the emission of an electron from the nucleus. (I would also include that full reaction of decay i.e. n⁰-->p⁺ +e^- + antineutrino) This results in a change in atomic number, but no change in mass number. The general formula for writing a beta decay reaction is: \[^{A}_{Z}\text{X} \rightarrow ^{A}_{Z+1}\text{Y}+ ^{0}_{–1}\text{e}\]

Note that when writing nuclear nomenclature, the mass number and atomic number must be the same on both sides of the equation. When writing a series of decay reactions involving the step-wise emission of particles, use the resulting atom (the product) from the previous step as the reactant in the next step.

The nuclear nomenclature of the step-wise emission of α, β, β, α, α, α, α particles:

α: \( ^{238}_{92}\text{U}\rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\text{He}\)

β: \(^{234}_{90}\text{Th}\rightarrow^{234}_{91}\text{Pa} + ^{ 0}_{–1}\text{e}\)

β: \(^{234}_{91}\text{Pa}\rightarrow^{234}_{92}\text{U} + ^{ 0}_{–1}\text{e}\)

α: \( ^{234}_{92}\text{U}\rightarrow ^{230}_{90}\text{Th} + ^{4}_{2}\text{He}\)

α: \( ^{230}_{90}\text{Th}\rightarrow ^{226}_{88}\text{Ra} + ^{4}_{2}\text{He}\)

α: \( ^{226}_{88}\text{Ra}\rightarrow ^{222}_{86}\text{Rn} + ^{4}_{2}\text{He}\)

α: \( ^{222}_{86}\text{Rn}\rightarrow ^{218}_{84}\text{Po} + ^{4}_{2}\text{He}\)

Correct

Question 20.2.15

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. Pt²⁺(aq) + Ag(s) →
2. HCN(aq) + NaOH(aq) →
3. Fe(NO₃)₃(aq) + NaOH(aq) →
4. CH₄(g) + O₂(g) →

Solution 20.2.15

Common features of chemical reactions:

Acid-base reaction: the neutralization of an acid and a base results in formation of water.

Precipitation reaction: reactions that involve solid formation

Redox reaction: reactions that involve electron transfer ie. Metal and Nonmetal, when O₂ is a product or reactant. A gain of electron is called reduction, and a loss of electron is oxidation.

Another important point to consider is if the reaction will even occur or not. To do this, consult the Activity Series of Metals. Generally, when you see a metal in its elemental form and a metal in ion solution check the activity series to see if the metal can displace the metal ion and swap places. A metal can displace metal ions listed below it.

a. Pt²⁺(aq) + Ag(s) → Pt(s) + Ag²⁺(aq) redox reaction

Since Ag is above Pt in the series this reaction will occur and there will be a transfer of electrons which means it's a redox reaction. (can also be classified by a change in oxidation state of the molecules involved in the reaction. Pt changes its oxidation state from 2+ to 0)

b. HCN(aq) + NaOH(aq) → NaCN(aq) + H₂O(l) acid-base reaction

The first thing to note is that the reactants are an acid and a base. This should tell you that an acid-base reaction will occur and they will neutralize each other and form water. In this reaction, Na⁺ actually acts as a spectator ion because NaOH dissociates completely in solution because it is a strong base. NaCN is in aqueous solution so it also can be written as Na⁺ and CN^-. HCN is a weak acid so it won't dissociate into completely two separate ions.From the chemical equation written out completely \[\text{HCN}_ {(aq)} + \text{Na}^+_{(aq)} +\text{OH}^-_{(aq)}\rightarrow \text{Na}^+_{(aq)} +\text{CN}^-_{(aq)}+ \text{water}_{(l)}\] The net ionic equation would be the same as the one above without Na⁺ because it is a spectator ion, meaning it doesn't participate in the reaction.

c. Fe(NO₃)₃(aq) + 3NaOH(aq) →3Na(NO₃)(aq)+Fe(OH)₃(s) precipitation reaction

This is an example of a double displacement reaction because both of the reactants are soluble ionic compounds. To predict the products of a double displacement reaction the cations switch their anions switch places. The general formula can be written as AX + BY → AY +BX. In order for the reaction to take place, one of the products has to be soluble and the other must be insoluble. The insoluble product is called the precipitate, hence the name of the reaction. Use the Solubility Rules to predict whether a compound is soluble or not. (rule 6 states that "Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al³⁺ are insoluble. Thus, Fe(OH)₃, Al(OH)₃, Co(OH)₂ are not soluble." so we know that the product Fe(OH)₃ is insoluble and this is a precipitation reaction)

d. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) redox reaction

Combustion reactions always involve molecular O₂. Oxygen is used to burn the compound and always creates CO₂(g) and H₂O(g). Combustion reactions are common redox reactions. Here it is harder to track the transfer of electrons, but usually find the pure element (which has oxidation state of 0) and track the electrons it has once the reaction occurs. Here, O₂ on the reactant side has oxidation state of 0, but on the product side is has oxidation state -2 in CO₂ and H₂O. See Oxidation States (Oxidation Numbers) for rules on assigning oxidation numbers.

Correct

Question 20.5.10

Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.

Solution 20.5.10

First, we have to define a few terms. An electrochemical cell is a device used to measure the electrical energy produced from a spontaneous chemical reaction or introducing energy into the cell to make a non-spontaneous reaction occur. Many salts are sparingly soluble in water, meaning is remains mostly as an insoluble salt with only a few ions released into the solution. The solubility of a sparingly soluble salt corresponds to its K_sp value. Because salts are sparingly soluble, it is hard to accurately determine their concentrations which means hard to measure their solubility. A common procedure to determine solubility is using an electrochemical cell. One reaction needs to involve the insoluble salt, and the other reaction has to make the net cell reaction be the dissociation of the salt. For example to find the solubility of AgCl(s):\[\text{Ag}_{(s)}+\text{Cl}^-_{(aq)}\rightarrow\text{AgCl}_{(s)}+\text{e}^-\] this reaction at the anode involves the insoluble salt AgCl(s) \[\text{Ag}^+_{(aq)}+\text{e}^-\rightarrow\text{Ag}_{(s)}\] and this reaction at the cathode creates the combined net cell reaction \[\text{Ag}^+_{(aq)}+\text{Cl}^-_{(aq)}\rightarrow\text{AgCl}_{(s)}\]

From the equilibirum expression \(\text{AgCl}_{(s)}\rightleftharpoons\text{Ag}^+_{(aq)}+\text{Cl}^-_{(aq)}\) we can get the solubility of AgCl to be K_sp=[Ag⁺][Cl^-]. (which can then be used to determine the solubility depending on the concentration of Ag⁺ and Cl^-)

In case this person isnt correct, here's another example.

If we want to determine the solubility of AgI at standard conditions, we can create the following half reactions

AgI(s) + e¯ ---> Ag(s) + I¯ E°=-0.15

Ag⁺ + e¯ ---> Ag(s) E°=0.80

We calculate E°cell=-0.95

we also know that the chemical equation for AgI disolving is

AgI(s) ⇌ Ag⁺(aq) + I¯(aq)

So when we solve for Ksp, we get

K_sp = [Ag⁺] [I¯]

So using the nerst equation at standard conditions we get the following

E_cell = E° - (0.0591/ n) log K_sp

1 mole of electron was transferred so n=1

0 =-0.95 - (0.0591/1)log K_sp

(0.95 /-0.0591)=log K_sp

log K_sp = -16.07

K_sp = 8.51 x 10¯¹⁷

Correct.

Q24.6.6

Do strong-field ligands favor a tetrahedral or a square planar structure? Why?

S24.6.6

Strong-field ligands have a large crystal field splitting energy which is the difference in energy between the two sets of d orbitals. (Δ_o octahedral complexes, Δ_t for tetrahedral). This means the strong-field ligands tend to be low spin. Electrons tend to fall in lowest energy energy state, P<Δ_o (pairing energy is less than crystal field splitting energy) meaning it is more energetically favorable to completely fill up low energy orbitals first.

Looking at the crystal field splitting of both tetrahedral and square planar structures:

The Δ_sp is typically larger than the pairing energy (P<Δ_sp) meaning square planar complexes are usually low spin complexes (strong-field ligands). The Δ_t is typically smaller than the pairing energy (P>Δ_t) meaning tetrahedral complexes favor high spin complexes (since strong field ligands tend to be low spin, strong field ligands favor square planar structure because it allows them to be a low spin).

In a tetrahedral complex, Δ_t is relatively small even with strong-field ligands because fewer ligands to bond with. It is rare for the Δ_t of tetrahedral complexes to exceed the pairing energy. Usually, electrons will move up to the higher energy orbitals rather than pair. Because of this, most tetrahedral complexes are high spin.

In square planar complexes Δ will almost always be large, even with a weak-field ligand. Electrons tend to be paired rather than unpaired because paring energy is usually much less than Δ. Therefore, square planar complexes are usually low spin.

Q14.7.11

A particular reaction was found to proceed via the following mechanism:

• \(\text{A}+\text{B}\)→\(\text{C}+\text{D}\) (slow)
• \(2\text{C}\)→\(\text{E}\) (fast)
• \(\text{E}+\text{A}\)→\(\text{B}+\text{F}\) (fast)

What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates

S14.7.11

The overall reaction (the mechanism) is the sum of the individual elementary reactions. The mechanism must give the balanced chemical reaction for the overall reaction. Adding up all the reactants and the products gives you \[2\text{A}+\text{B}+2\text{C}+\text{E}\rightarrow\text{B}+\text{C}+\text{D}+\text{E}+\text{F}\] Simplifying the reaction gives you the overall reaction \[2\text{A}+\text{C}\rightarrow\text{D}+\text{F}\]

Both catalysts and intermediates do not appear in the overall reaction. B and E do not appear in the overall reaction, meaning one is a catalyst and one is an intermediate. Intermediates are produced in one step and used up in another. Therefore, E is an intermediate because it is a product in the second step and a reactant in another. It is used up. However, catalysts cannot be "used up" they are constantly recycled. This indicates that catalysts are reactants in one step and products in another. Examining the elementary steps given, it is evident that B is the catalyst because it is a reactant in the first step and a product in the last step. It is not used up.

Correct