Extra Credit 18

S17.2.7

In galvanic cells like the one pictured above, the salt bridge is necessary to maintain a balance of charges within the internal circuit, as electrons are continuously moving from one half cell to the other. Specifically, in the cell, electrons flow from the anode to the cathode. The oxidation reaction that occurs at the anode generates both electrons and positively charged ions. The electrons move through the wire toward the cathode, leaving an unbalanced positive charge in this vessel (anode). Therefore, in order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell. A similar (but reversed) situation is found in the cathodic cell, where the migration of positively charged ions from the salt bridge are moving into this half cell in order to counteract the influx of electrons. In the absence of a salt bridge, opposing charges will continue to build up in each half cell and the reaction will be brought to a halt due to the charge imbalance. In essence, with a salt bridge, which is usually composed of an inert electrolyte such as sodium nitrate (which is pictured above), each half-cell remains electrically neutral and a current can continue to flow through the circuit.

S19.1.16

When predicting the products of chemical reactions, you can usually start off by identifying the type of reaction involved. The primary reaction types students encounter are displacement (single and double), acid-base, synthesis, and combustion. From there, each type of reaction has certain ways to go about solving it. We will guide you on how to solve each of the problems below, in a step-by-step fashion.

1. \( \mathrm{MnCO_3}(s)+\mathrm{HI}(aq)(aq)\rightarrow\)

• As stated above, the first step is identifying the type of reaction we are dealing with. In this case, it is an acid-base as there is clearly an acid involved.
• However, since there are two species on the reactant side of the equation, each composed of two different ions, it would be easier to treat this problem as a double replacement reaction.
• In double replacement reactions, the cation and anions switch places. It is helpful to separate each compound into their cation and anionic parts with their charges as such:

\( \mathrm{HI}(aq)\) is composed of \( \mathrm{H^+}\) and \(\mathrm{I^-}\), while \(\mathrm{MnCO_3}(s)\) is composed of \( \mathrm{Mn^{2+}}\) and \(\mathrm{CO_3^{-2}}\).

• From here, you simply switch the places of the ions to form new compounds. Therefore, you would get the overall reaction of:

\( \mathrm{MnCO_3}(s)+\mathrm{HI}(aq)\rightarrow\mathrm{H_2CO_3}(aq)+\mathrm{MnI_2}(s) \).

• The last step in any chemical reaction is balancing! However many moles of each element there are on the reactant side, there should be the same amount of moles on the product side. Everything seems to be balanced except for iodine! If there are now 2 moles of iodine in the products, there needs to be two moles of iodine in the reactants. So, the new reaction should be:

\( \mathrm{2HI}(aq)+\mathrm{MnCO_3}(s)\rightarrow\mathrm{H_2CO_3}(aq)+\mathrm{MnI_2}(s) \)

• Usually, you would be done with the problem here! However, in this special case, \(\mathrm{H_2CO_3^{-2}}\) is unstable at STP, which means that it will not stay in this state. It will decompose as follows:

\( \mathrm{H_2CO_3}(s)\rightarrow\mathrm{H_2O}(l)+\mathrm{CO_2}(g) \).

• Therefore, the final answer is:

\( \mathrm{MnCO_3}(s)+\mathrm{2HI}(aq)\rightarrow\mathrm{CO_2}(g)+\mathrm{MnI_2}(s)+\mathrm{H_2O}(l) \)

2. \( \mathrm{CoO}(s)+\mathrm{O_2}(g)\rightarrow\)

• Identify the type of reaction: In this case, it is a synthesis reaction.
• In synthesis reactions, two or more chemical species combine to form a more complex product. One way to think of synthesis reactions is that they are the reverse of a decomposition reaction.
• In this specific example, a metal oxide is being reacted with oxide. Therefore, the product will be a more complex metal oxide (these usually follow the generic chemical formula of \( \mathrm{M_xO_y}\)). When you react a metal with an oxide, the metal tends to become more oxidized. So, in terms of this reaction, the oxidation state for the metal (\( \mathrm{Co}(s)\)) increases from 2⁺ to 3⁺, resulting in \( \mathrm{Co^{3+}}\).
• Since that is the resulting metallic part of the entire metal oxide, you then combine this with oxide to receive the full product of \(\mathrm{Co_2O_3(s)}\), which results in the chemical reaction of:

\( \mathrm{CoO}(s)+\mathrm{O_2}(g)\rightarrow\mathrm{Co_2O_3}(s)\).

• Balance! The final product will thus be:

\( \mathrm{4CoO}(s)+\mathrm{O_2}(g)\rightarrow\mathrm{2Co_2O_3}(s)\)

3. \(\ce{La}(s)+\ce{O2}(g)⟶\)

• Identify the type of reaction: Again, this is a synthesis reaction.
• However, this is a simpler example of a synthesis reaction. In the simplest synthesis reactions, two elements simply combine to form a binary compound (a compound made of two elements).
• Again, a metal is reacting with oxide and will therefore form a metal oxide. And, also again, in reactions where a metal oxide is being formed, the product has the formula of \( \mathrm{M_xO_y}\). Therefore, because the reactants have oxidation states of 3⁺ and 2^- respectively, the result will be:

\( \mathrm{La}(s)+\mathrm{O_2}(g)\rightarrow\mathrm{La_2O_3}(s)\)

• Don't forget to balance! In the end, the final product is:

\( \mathrm{4La}(s)+\mathrm{3O_2}(g)\rightarrow\mathrm{2La_2O_3}(s)\)

4. \( \mathrm{V}(s)+\mathrm{VCl_4}(s)\rightarrow\)

• Identify the type of reaction: This is, once again, a synthesis reaction as it involves the joining together of two reactants, or compounds, to produce a complex product.
• Following the procedure for synthesis reactions, you end up with the reaction:

\( \mathrm{V}(s)+\mathrm{VCl_4}(s)\rightarrow\mathrm{VCl_3}(s)\)

• Balance! The overall reaction is:

\( \mathrm{V}(s)+\mathrm{3VCl_4}(s)\rightarrow\mathrm{4VCl_3}(s)\)

5. \( \mathrm{Co}(s)+xs\ce{F_2}(aq)\rightarrow\)

• Identify the type of reaction: A synthesis reaction, yet again.
• However, this certain reaction employs something that makes it different from the rest, the symbol "xs" which is shorthand for "excess." Since it is in front of a reactant, it signifies that you need to use more than the reaction stoichiometry would normally call for. Usually, it is done to drive a reaction to completion, or to prevent side reactions. Therefore, it indicates that F₂ is added in excess, so it is favored more as the reaction is driven to completion.
• Again, following a similar procedure as other simple synthesis reactions, the two reactants simply combine and you receive the reaction of:

\( \mathrm{Co}(s)+\mathrm{F_2}(aq)\rightarrow\mathrm{CoF_2}(s)\)

• Everything is balanced! So, the reaction above is, indeed, the correct answer:

\( \mathrm{Co}(s)+\mathrm{F_2}(aq)\rightarrow\mathrm{CoF_2}(s)\)

6. \( \mathrm{CrO_3}(s)+\mathrm{CsOH}(aq)\rightarrow\)

• Identify the type of reaction: Again, this is an acid-base reaction as it clearly involves a base.
• In acid-base reactions, the products are usually water and a salt. Using this knowledge, the reaction is:

\(\mathrm{CrO_3}(s)+\mathrm{CsOH}(aq)\rightarrow\mathrm{Cs_2CrO_4}(s)+\mathrm{H_2O}(l)\)

• Balance! The overall reaction is:

\(\mathrm{CrO_3}(s)+\mathrm{2CsOH}(aq)\rightarrow\mathrm{Cs_2CrO_4}(s)+\mathrm{H_2O}(l)\)

S19.3.8

Crystal field theory describes the breaking of orbital degeneracy in transition metal complexes due to the presence of ligands. In the questions above, we are dealing with octahedral complexes and in an octahedral complex, there are six ligands attached to the central transition metal. 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 \(\Delta_{oct}\)). Because the \(t_{2g}\) orbitals decrease with respect to this normal energy level, they become more stable. As stated in the question, stability increases as the number of electrons in the \(t_{2g}\) orbitals increases. But why? It is because the magnitude of \(\Delta_o\) increases as well. The magnitude of \(\Delta_o\) dictates whether a complex is high spin or low spin, which directly affects its stability. Larger values of \(\Delta_o\) yield a low-spin complex (more stable), whereas smaller values of \(\Delta_o\) produce a high-spin complex (less stable). The magnitude of \(\Delta_o\) depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand.

In the question, it is stated that for each problem given, you are comparing two complexes with the same central transition metal with no change in oxidation state. Therefore, the only factor you have to take into account when determining which complex is more stable is the nature of the ligand, which is actually quite easily done. All you have to do is look at a spectrochemical series! A spectrochemical series is a list of ligands ordered on ligand strength, ranging from weak field ligands to strong field ligands. Strong field ligands increase the distance between the t_2g orbitals and e_g orbitals (\(\Delta_o\)) and result in low-spin complexes, which, as stated previously, are more stable. So, for each problem, when looking at the two complexes given, whichever has the stronger ligand attached will correspond to the more stable complex.

a. \( \mathrm{[Fe(H_2O)_6]^{2+}}\) or \( \mathrm{[Fe(CN)_6]^{4-}}\)

Again, for both these octahedral complexes, because the metal and oxidation state of the metal are the same, we must compare the ligands in order to determine which complex has a larger \(\Delta_o\) and is thus more stable. In this case, the two complexes have the ligands \(\mathrm{H_2O}\) and \( \mathrm{CN^-}\) respectively. By directly looking at the spectrochemical series, we see that \( \mathrm{CN^-}\) is a stronger field ligand than \(\mathrm{H_2O}\). Therefore, \(\mathrm{[Fe(CN)_6]^{4-}}\) is more stable.

b. \( \mathrm{[Co(NH_3)_6]^{3+}}\) or \( \mathrm{[Co(F)_6]^{3-}}\)

Following the same procedure, the two complexes have the ligands \( \mathrm{NH_3}\) and \( \mathrm{F^-}\) respectively. On the spectrochemical series, \( \mathrm{NH_3}\) is ranked as a stronger field ligand than \( \mathrm{F^-}\). Therefore, \( \mathrm{[Co(NH_3)_6]^{3+}}\) is more stable.

c. \( \mathrm{[Mn(CN)_6]^{4-}}\) or \( \mathrm{[Mn(Cl)_6]^{4-}}\)

As you can see, the procedure you are meant to follow is quite easy once you understand that all you have to do is look at the nature of the ligand. In this case, the two complexes have the ligands \( \mathrm{CN^-}\) and \( \mathrm{Cl^-}\) respectively. Using the spectrochemical series, you can clearly see that \( \mathrm{CN^-}\) is ranked as a stronger field ligand than \( \mathrm{Cl^-}\). Therefore, \( \mathrm{[Mn(CN)_6]^{4-}}\) is more stable.

S12.4.8

The half-life of a reaction, \( \mathrm{t_{1/2}}\), is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. It is important to note that the half-life is varied between different type of reactions. Therefore, the first step in attacking problems involving half life is identifying the order of the reaction. In this case, we are dealing with a second order reaction. The half life formula that corresponds to a second order reaction is:

\( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{k[A]_o})}\)

where k refers to the rate constant of the reaction and \({[A]_o}\) refers to the initial concentration of the reactant in question.

From here, we take the information we are given in the problem and break it down into more straightforward terms:

• rate constant (k) = 8.0 × 10⁻⁸ L/mol/s
• [NOCl]= 0.15 M. Since we are dealing with the decomposition of NOCl, this corresponds to our "\({[A]_o}\)," as this is how much reactant is initially present.

Finally, we simply substitute the provided information into the formula:

\( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{k[A]_o})}\)

\( \mathrm{t_{1/2}} = \mathrm{(\frac{1}{(8.0*10^{-8})(.15)})}\)

\( \mathrm{t_{1/2}} = \mathrm{8.3*10^{7}}\) s

S21.2.3

The protons and neutrons that make up the nucleus of an atom are called nucleons, and an atom with a particular number of protons and neutrons is called a nuclide. Each element can be represented by nuclide notation:

\(\ce{^{A}_{Z}X}\)

where A, the mass number, is the sum of the number of protons and the number of neutrons, Z, the atomic number, is the number of protons, and X is the corresponding element.

In the problems given, either X (elemental symbol) or Z (atomic number) is the missing aspect of the notation. You should be able to recognize that these are directly and obviously related. Therefore, in the questions where the identity of the element is missing, the simplest way to go about figuring out which element it is, is to first look at Z, the atomic number, and recognize that this refers to the number of protons, which corresponds to the number that element is on the periodic table. In the questions where the atomic number is left missing, looking at the given element on the periodic table will give you the atomic number, Z.

a. \(\ce{^{34}_{14}Si}\)

The atomic number reveals the identity of the element. In this example, the atomic number is 14 and by looking at the periodic table, the element with the atomic number, 14, is silicon, Si.

b. \(\ce{^{36}_{15}P}\)

As previously stated, each element has its own specific atomic number. The atomic number for phosphorus is 15.

c. \(\ce{^{57}_{25}Mn}\)

The atomic number for manganese is 25.

d. \(\ce{^{121}_{56}Ba}\)

The element with an atomic number of 56 is barium.

S21.5.6

Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. Nuclear chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor. Any nuclear reactor that produces energy via the fission of uranium or plutonium by bombardment with neutrons has a variety of different, very necessary components, two of them being a nuclear moderator and control rods, which we will discuss below in greater detail.

Moderator:

Control Rods:

Nuclear reactors use control rods to control the fission rate of the nuclear fuel. They do so by adjusting the number of slow neutrons present in order to keep the rate of the chain reaction at a safe level. Control rods are made of elements that are known to absorb neutrons, such as boron, cadmium, or hafnium. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:

\( \mathrm{\ce{^{10}_{5}B}}+\mathrm{\ce{^{1}_{0}n}}\rightarrow\mathrm{\ce{^{7}_{3}Li}} +\mathrm{\ce{^{4}_{2}He}}\)

When control rods are inserted into the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, you can fully insert all of the control rods into the nuclear core between the fuel rods in order to completely shut down the chain reaction.

S20.4.4

A redox reaction is a type of chemical reaction that involves the transfer of electrons between two species, which results in a change in the oxidation numbers of each species. In order to harness the energy that is released and actually use it to perform useful work instead of losing it as heat, the reaction must be split into two separate half reactions: the oxidation and reduction reactions. The reactions are put into two different containers and a wire connecting the two compartments is used to drive the electrons from one side to the other. This creates the galvanic cell. In essence, galvanic cells, also known as voltaic cells, are electrochemical cells that are driven by spontaneous chemical reactions (i.e. redox reactions that take place within the cell) that produce an electric current through an outside circuit. To reiterate, in order for galvanic cells to actually work and release energy, the redox reaction taking place must be spontaneous. For the reaction to be spontaneous, electron flow must be from the element that is willing to give up electrons to the element that is willing to gain electrons. In technical terms, electrons must flow from the anode (where oxidation occurs) to the cathode (where reduction occurs).

In order to figure out which element is being oxidized and which element is being reduced, you need to refer to a table of standard reduction potentials (SRPs). This table is a list of, obviously, standard reduction potentials, which is the tendency for a chemical species to be reduced and is measured in volts at standard conditions. The more positive the potential is for an element, the more likely it will be reduced.

Applying this logic to the question at hand, you must look up the standard reduction potentials for both aluminum and bromine in the table and compare them to one another:

\( \mathrm{Al^{3+}}(aq)+\mathrm{3e^-}\rightarrow\mathrm{Al}(s);\textrm{-1.676 V}\)

\( \mathrm{Br_2}(l)+\mathrm{2e^-}\rightarrow\mathrm{2Br^{-}}(aq);\textrm{+1.065 V}\)

Bromine clearly has a much higher positive SRP, which tells us that it is more likely to be reduced (i.e. more willing to gain electrons --> stronger oxidizing agent). Therefore, electrons will flow from aluminum to bromine.

S20.7.2

The lead–acid battery is an example of a rechargeable battery and is used to provide the starting power in virtually every automobile and marine engine on the market. The overall redox reaction that occurs in a lead-acid battery is as follows:

\(\mathrm{Pb}(s)+\mathrm{PbO_2}(s)+\mathrm{2HSO^-_4}(aq)+\mathrm{2H^+}\rightarrow\mathrm{2PbSO_4}(s)+\mathrm{2H_2O}(l)\), with an \(\mathrm{E_{cell}}=2.041v\)

The anode of each cell in a lead storage battery is a plate of lead metal, and the cathode is a similar grid containing powdered lead dioxide. They are both immersed in a pool of electrolyte, which is usually a solution of sulfuric acid in water. As the cell is discharged, a powder of \(\mathrm{PbSO_4}\) forms on the electrodes. More importantly, sulfuric acid is consumed and water is produced (therefore, there is more water present in comparison to sulfuric acid as the reaction proceeds), which directly decreases the density of the electrolyte as the density of sulfuric acid is 1.28 g/ml and the density of water is 1 g/ml.