Extra Credit 46

Question 17.6.5

Why would a sacrificial anode made of lithium metal be a bad choice despite its \(E^∘_{Li^{+}/Li}=-3.04\space V\), which appears to be able to protect all the other metals listed in the standard reduction potential table?

Solution 17.6.5

First, it's essential to understand the concept of a standard reduction potential.

A sacrificial anode is a metal that protects a less active metal from corroding due to external sources. Lithium has the lowest standard reduction potential value, so technically it should be the best sacrificial anode and should be able to protect the other metals. However, we need to keep in mind how reactive lithium is. Lithium is an Alkali Metal, which are known to be highly reactive due to their low ionization energy. This can be attributed to the fact that they only have one valence electron; they would prefer to lose this electron because then they'd end up with a full valence shell. Consequently, the alkali metals are extremely reactive with water and are likely to explode as a result. Due to the high reactivity of lithium, it corrodes very quickly, meaning that it will not properly protect the metal it was supposed to protect. Therefore, lithium would not be an effective sacrificial anode.

In regards to the \(E^∘_{Li^{+}/Li}=-3.04\space V\) value, this is a reduction potential. The negative value indicates that it is very bad at being reduced and that the process of the reduction of Lithium is non-spontaneous, further proving that a sacrificial anode made of lithium metal would be unreasonable.

Question 12.3.9

What is the instantaneous rate of production of \(N\) atoms in a sample with a carbon-14 content of \(1.5\times10^{-9} M\)?

Editor's Note: You answered the question for \(6.5\times10^{-9} M\), but the question asked for \(1.5\times10^{-9} M\).

Solution 12.3.9

The question asks for the instantaneous rate of production of \(N\) atoms from a sample of carbon-14.

We know that nitrogen is produced from carbon-14, if carbon-14 goes through a beta decay by emitting an electron. This process is described in the equation below.

\(^{14}_{6}C \rightarrow ^{14}_{7}N + ^0_{-1}\beta\)

This is a unimolecular reaction. Otherwise known as a elementary reaction with only one reactant. Therefore, we can determine the rate law from the equation.

We know that the rate constant of the radioactive decay of carbon 14 is \(k=1.21\times10^{-4}\space year^{-}\)

So, the rate law would be \(rate=k[^{14}_{6}C]\)

Now, let's add in the rate constant:

\(rate=1.21\times10^{-4}[^{14}_{6}C] \)

The carbon-14 sample has a concentration of \(1.5\times10^{-9} M\), so we can substitute that in as well.

\(rate=1.21\times10^{-4}[1.5\times10^{-9} M] \)

\(rate=1.815\times10^{-13}\space M/year\)

Because our rate had three significant figures, we must stay consistent with the significant figures: \(rate=1.82\times10^{-13}\space M/year\)

Question 12.6.1

Why are elementary reactions involving three or more reactants very uncommon?

Solution 12.6.1

An elementary reaction involving three reactions is referred to as a termolecular reaction. Elementary reactions with three or more reactants are very uncommon. This is because in order for the reaction to take place, all of the reactants (3 or more) must collide with each other at the same time in a specific orientation and with sufficient energy (Collision Theory). A bimolecular reaction, which is an elementary reaction with two reactants, is is more likely to happen because the lower amount of reactants will have a greater chance colliding with each other at the same time in a specific orientation and with sufficient energy. In essence, it's easier to combine two things with proper energy and orientation than three.

Question 21.4.13

Define the term half-life and illustrate it with an example.

Solution 21.4.13

The half life (denoted as \(t_{1/2}\)) of a reaction is the amount of time it takes for the concentration of the reactant to decrease by half of its original concentration.

For example, Uranium-238 has a half life of 4.468 billion years. If we start with an initial sample that has 50 grams of Uranium-238, then after 4.468 billion years, there will be 25 grams of the sample left. This example is illustrated below.

\([A]\ =\ [A_{0}](\frac{1}{2})^{\frac{t}{h}}\)

\([A]\ =\ [50](\frac{1}{2})^{\frac{4.468}{4.468}}\) (note that \(t\) and \(h\) are in units of billion years)

\([A]\ =\ [50](\frac{1}{2})^{1}\)

\([A]\ =\ 25\)

Question 20.3.1

Is \(2NaOH(aq) + H_2SO_4(aq) \rightarrow Na_2SO_4(aq) +2H_2O(l)\) a oxidation-reduction reaction? Why or why not?

Solution 20.3.1

In order to determine if this reaction is a redox (oxidation-reduction) reaction, we must first determine the oxidation numbers of each individual element. It is recommended that you become familiar with the rules for assigning oxidation numbers.

Let's start with the \(2NaOH(aq)\).

According to the rules for assigning oxidation numbers, Na: +1 since it is a group 1 metal, O: -2 since the oxygen here does not act as a peroxide, and H:+1.

Let's verify that these oxidation numbers are correct. We know that the \(NaOH\) is neutral. So if we add the oxidation numbers of each element of the compund, we should get zero. 1-2+1=0

For \(H_2SO_4(aq)\), H:+1, O: -2, and S: +6. We know the oxidation number of H and O from the oxidation number rules, but in order to determine the oxidation number of sulfur, we need to use the oxidation numbers of the other elements. As previously mentioned, O: -2 and H: +1. We also know that \(H_2SO_4\) is neutral. Therefore, the sum of the oxidation numbers should equal to zero. However, we have 2 hydrogen atoms and 4 oxygen atoms. So we need to multiply each oxidation number of an element with the amount of atoms present in the compound.

2(+1) + (oxidation # for Sulfur) + 4(-2) = 0

We now know that the oxidation number for sulfur is +6.

For \(Na_2SO_4(aq)\), Na: +1, O: -2, and S: +6

Let's verify these oxidation numbers. 2(+1) + (+6) + 4(-2) = 0

For \(2H_2O(l)\), H: +1 and O: -2

Let's verify these oxidation numbers. (+1) + 2(-2) = 0

As you can see, none of the oxidation numbers have been changed. This means that this reaction is not a redox reaction. If this reaction was a redox a reaction, then one of the reactants would have been reduced (a decrease in oxidation number) and another reactant would be oxidized (an increase in oxidation number). In the reactant side, the oxidation number of hydrogen is +1, oxygen is -2, sodium is +1, and sulfur is +6. In the product side of the reaction, the oxidation number for hydrogen is +1, oxygen is -2, sodium is +1, and sulfur is +6 as well. None of the species lost or gained any electrons. Therefore, this reaction is not a redox reaction.

Question 20.5.12

How many electrons are transferred during the reaction \(Pb(s) + Hg_2Cl_2(s) \rightarrow PbCl_2(aq) + 2Hg(l)\)? What is the standard cell potential? Is the oxidation of \(Pb(s)\) by \(Hg_2Cl_2\) spontaneous? What is \(\Delta G^{°}\) for this reaction?

Solution 20.5.12

To find out how many electrons are transferred during this redox reaction, we must first balance the redox reaction. It is recommended to become familiar with the rules for balancing redox reactions.

We will use the half reaction method for this reaction. To do this, we must split the original equation into two equations. One equation is for oxidation and the other is for reduction.

Oxidation: \(Pb(s) \rightarrow PbCl_2(aq)\) (Pb goes from an oxidation number of 0 to +2 which means it has lost electrons)

Reduction: \(Hg_2Cl_2(s) \rightarrow 2Hg(l)\) (Hg goes from an oxidation number of +1 to 0 which it means it gained electrons)

One way to differentiate between oxidation and reduction is to remember OIL RIG: Oxidation is Loss (of electrons) and Reduction is Gain (of electrons).

Now that we know the half reactions, we must balance them. Let's start with the oxidation reaction.

\(2Cl^{-}(aq) + Pb(s) \rightarrow PbCl_2(aq)\) We add two chlorine ions on the reactant side to balance the 2 atoms of chlorine on the product side. Now the reactant side has a charge of -2 due to the 2 chlorine ions.

\(2Cl^{-}(aq) + Pb(s) \rightarrow PbCl_2(aq) + 2e^{-} \) We add two electrons on the product side to balance the -2 charge on the reactant side. Now both sides of the equation have a charge of -2.

Now let's balance the reduction reaction.

\(Hg_2Cl_2(s) \rightarrow 2Hg(l) + 2Cl^{-}(aq)\) We add two chlorine ions on the product side to balance the 2 atoms of chlorine on the reactant side. Now the product side has a charge of -2 due to the 2 chlorine ions.

\(2e^{-} + Hg_2Cl_2(s) \rightarrow 2Hg(l) + 2Cl^{-}(aq)\) We add two electrons on the reactant side to balance the -2 charge on the product side. Now both sides of the equation have a charge of -2.

We can see that both the reduction and oxidation half reactions have two electrons in them. This means that two electrons were transferred in this reaction.

To find the standard cell potential, we must use the equation \(E^{°}_{cell} = E^{°}_{reduction}+E^{°}_{oxidation}\)

Keep in mind that \(E^{°}_{oxidation} = -E^{°}_{reduction}\)

To use this equation, we must find out the half reactions of the redox reaction which we have already done earlier.

Once we determine the half reactions, we must look at the list of Standard Reduction Potentials.

Since Pb is oxidized, \(E^{°}_{oxidation}\) = +0.13 V

Since Hg is reduced, \(E^{°}_{reduction}\) = +0.85 V

Now we add the two values \(E^{°}_{cell}\)= 0.13V + 0.85V = 0.98V

Now, let's answer another question: Is the oxidation of \(Pb\) by \(Hg_2Cl_2\) spontaneous?

Here is the oxidation reaction: \(Pb(s) \rightarrow PbCl_2(aq)\)

We can discuss the change in entropy to decide if this reaction is spontaneous. The reaction shows the reactant is in a solid phase and the product is in an aqueous phase. Going from the solid phase to the aqueous phase creates more disorder which means this reaction is spontaneous. Keep in mind that "disorder" is not an accurate way to define entropy, however, it is used here to simplify the solution so it is more clearly understood. Additionally, we can say that because \(E^{°}_{cell}\) is positive, it is a spontaneous reaction.

To find \(\Delta G^{°}\), we need to use this equation: \(\Delta G^{°} = -nFE^{°}_{cell}\)

We already found all the values we need for this equation:

• n = the moles of electrons that were transferred in the redox reaction
• F= Faraday's constant = 96486 C/mole e-
• \(E^{°}_{cell}\) = standard cell potential = 0.98 V

\(\Delta G^{°}\) = -(2)(0.98)(96486) = -189112.56 J/mole = -189.11 kJ/mole

As you can see, \(\Delta G^{°}\) is negative which further supports that this reaction is spontaneous.

Question 24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

a. \([Cu(NH_3)_4]^{2+}\)

b. \([Ni(CN_4)]^{2-}\)

Solution 24.6.8

For Part A

The coordination number for this complex is 4 since there are 4 ammine groups (which are monodentate, meaning they only "bite" once) attached to the copper (II) ion. If the coordination number was 6, then it would have an octahedral structure. Thus, the structure of this complex can either be square planar or tetrahedral. To determine this, we need to take a look at the ligands attached to the copper (II) ion. We know from the spectrochemical series that ammine or \(NH_3\) is a strong field ligand. Strong field ligands also produce a low spin configuration. We also know that the square planar structure favors a low spin configuration while the tetrahedral structure favors a high spin configuration. Since \(NH_3\) is a strong field ligand and produces a low spin configuration, we can conclude that this complex is square planar. The number of unpaired electrons can be determined from drawing a diagram of the square planar field as shown below.

In addition to looking at the type of ligand to determine complex structure, we can also use the number of d-electrons in the complex to determine the structure. For instance, coordination complexes with \(d^{0}\) or \(d^{10}\) electron configuration tend to be tetrahedral, whereas \(d^{8}\) leans more toward the square planar geometry. In this case, where we have \(d^{9}\), it may not be as easy to determine the shape, so further steps must be taken (those in the previous paragraph).

Since this complex has a low spin configuration, the lower energy levels must be filled before the higher energy levels. The copper (II) ion also has 9 d electrons, so 9 electrons must be filled in the diagram. From the diagram, we can see that there is one unpaired electron in this complex.

For Part B

This complex also has a coordination number of 4. There are 4 \(CN^{-}\) (monodentate) ligands. We know that this complex cannot have a octahedral complex since the complex would require a coordination number of 8. Therefore, this complex can either have a tetrahedral or a square planar structure. The \(CN^{-}\) ligand is a strong-field ligand which means it produces a low spin electron configuration. As mentioned in Part A, the square planar structure favors a low spin configuration. Therefore, we can predict that this complex will have a square planar structure as well. To determine the number of unpaired electrons in this complex, we must draw the diagram of the square planar field.

Since this complex has a low spin configuration, the lower energy levels must be filled before the higher energy levels. Its low spin configuration can also be attributed to the fact that it has a lower pairing than splitting energy, meaning that it requires less energy to pair the electrons than it does to split them into separate orbitals. Low spin configuration also follows the Aufbau Principle. The nickel (II) ion has 8 d electrons, so 8 electrons must be filled in the diagram. From the diagram, we can see that there are no unpaired electrons in this complex.

Question 14.7.12

A particular reaction has two accessible pathways (A and B), each of which favors conversion of \(X\) to a different product (Y and Z, respectively). Under uncatalyzed conditions, pathway A is favored, but in the presence of a catalyst, pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.

Solution 14.7.12

The product "Y" from the pathway A is favored both in the presence of a catalyst and without a catalyst.

If a catalyst is not present, then pathway A will be favored since the activation energy for pathway A will be lower. This means that the reactant "X" will be irreversibly converted to the product "Y."

If a catalyst is present, then pathway B will be favored since the activation energy for pathway B will be lower. This means that the reactant "X" will be converted to the product "Z". However, the reaction for pathway B is reversible, meaning that the product Z will be converted back to X. Once this conversion takes place, the reaction can take place through pathway A, resulting in the reactant X being converted to the product Y.