Extra Credit 18

Q17.2.7

Why is a salt bridge necessary in galvanic cells like the one below?

A17.2.7

The purpose of using a salt bridge in galvanic cells is to facilitate the balance of charge throughout the cell in order to maintain its neutrality because the electrons are moving from one half of the cell to the other half. The following oxidation reaction that occurs at the anode:

\[\ce{Cu}(s)⟶\ce{Cu^{2+}}(aq)+{2e^-}\]

This reaction generates both positively-charged ions, \(\ce{Cu^{2+}}(aq)\), and electrons, the latter of which will flow through the wire to the cathodic side of the cell. Without a salt bridge, a build up of negative charge (due to electrons) will occur at the \(\ce{Ag}\) cathode, whilst a build up of positive charge (due to the positively-charged \(\ce{Cu^{2+}}(aq)\) ions) will occur at the \(\ce{Cu}\) anode, therefore bringing the electrode reaction to a halt as a result of the charge imbalance. The salt bridge, usually composed of an inert electrolyte such as sodium chloride, potassium nitrate, or sodium nitrate as in the cell shown above, allows the migration of positively and negatively charged ions to opposite sides of the cell to balance its overall charge. In the case of this specific cell, negatively charged nitrate ions migrate towards the anode, whilst positively-charged sodium ions migrate towards the cathode.

Q19.1.16

Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.)

1. \(\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\)
2. \(\ce{CoO}(s)+\ce{O2}(g)⟶\)
3. \(\ce{La}(s)+\ce{O2}(g)⟶\)
4. \(\ce{V}(s)+\ce{VCl4}(s)⟶\)
5. \(\ce{Co}(s)+xs\ce{F2}(g)⟶\)
6. \(\ce{CrO3}(s)+\ce{CsOH}(aq)⟶\)

A19.1.16

There is a myriad of reactions that can occur, which include: single replacement, double replacement, combustion, acid-base/neutralization, decomposition or synthesis. The first step to determine the products of a reaction is to identify the type of reaction. From then on, the next steps you take to predict the products will vary based on the reaction type.

1. This reaction is a double displacement reaction, in which the cations and anions of the reactants switch places to form new compounds. Writing out the equation in terms of it's aqueous ions will help you visualize what exactly is getting moved around:
   \[2H^{+}(aq)+2I^{-}(aq)\rightarrow 2H^{+}(aq)+CO_{3}^{2-}(aq)+Mn^{2+}(aq)+2I^{-}(aq)\]
   In this case, the hydrogen cations will recombine with carbonate anions whilst manganese cations will recombine with iodide anions giving us the following equation:
   \[\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{H2CO3}(aq)\]
   This is still not the final answer however, as carbonic acid is unstable and decomposes to carbon dioxide and water under standard conditions. Taking this into account, our final equation is:
   \[\ce{MnCO3}(s)+\ce{2HI}(aq)⟶\ce{MnI2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\]
2. This reaction is a synthesis reaction, in which two or more reactants combine to form a more complex compound. In this case we are also reacting a metal oxide with oxygen which would result in another metal oxide as the product. The resulting product would be the mixed valence oxide Co₃O₄ in which one cobalt atom has a +2 oxidation state whilst the other two have a +3 oxidation state. Now all is left is to balance the equation:
   \[\ce{6CoO}(s)+\ce{O2}(g)⟶\ce{2Co3O4}(s)\]
3. Like equation 2, this reaction is also a synthesis reaction involving a metal and oxygen which should result in the formation of a metal oxide. It is a matter now of balancing the oxidation states to attain a neutral compound. Oxygen will always hold a -2 oxidation state in compounds whilst lanthanum will always exhibit a +3 oxidation state. As such, a combination of 2 lanthanum atoms with a +3 oxidation state and 3 oxygen atoms with a -2 oxidation state will give us a molecule with an overall charge of 0 (3(-2)+2(+3)=0). We know our product now, La₂O₃, and now just need to balance the overall equation, giving us:
   \[\ce{4La}(s)+\ce{3O2}(g)⟶\ce{2La2O3}(s)\]
4. This reaction is slightly harder to define as it encapsulates both the properties of synthesis and decomposition reactions, wherein vanadium reacts with vanadium tetrachloride to produce vanadium trichloride. This reaction however is primarily a synthesis reaction since we are combining two reactants to produce one complex compound. With vanadium trichloride as our product, we can balance the equation:
   \[\ce{V}(s)+\ce{3VCl4}(s)⟶\ce{4VCl3}(s)\]
5. This is another synthesis reaction combining cobalt and fluorine. This equation includes the "xs" notation in front of fluorine which is short for 'excess', meaning more fluorine than actually required is present in the reactants, ensuring the reaction goes to completion. Finding the products if a simple matter of combining the cobalt and fluorine into one molecule, which already leaves us with a balanced equation:
   \[\ce{Co}(s)+xs\ce{F2}(g)⟶\ce{CoF2}(s)\]
6. It may not be obvious here, but the reaction we've been given here is actually an acid-base/neutralization reaction, with chromium trioxide acting as the acid and cesium hydroxide as the base. Chromium trioxide is referred to as an acidic oxide which means that it will react with water to form an acid. Note that this reaction can still proceed even if the reactants aren't in the same phases. The basic rule for acid-base/neutralization reactions is they form a salt (salt being the general term for any ionic compound formed from acid-base reactions) and water. Since we know water is one of our products, our other product must be a salt composed of cesium, chromium and oxygen. Thus, our other product should be cesium chromate, and you can now balance the equation accordingly:
   \[\ce{CrO3}(s)+\ce{2CsOH}(aq)⟶\ce{Cs2CrO4}(aq)+\ce{H2O}\]

Q19.3.8

For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the t_2g orbitals increases. Which complex in each of the following pairs of complexes is more stable?

1. [Fe(H₂O)₆]²⁺ or [Fe(CN)₆]⁴⁻
2. [Co(NH₃)₆]³⁺ or [CoF₆]³⁻
3. [Mn(CN)₆]⁴⁻ or [MnCl₆]⁴⁻

A19.3.8

As we know, the d-orbitals of an octahedral complex is split into two sets of orbitals with differing energy levels, t_2g and e_g, with the difference in energy being denoted by crystal field splitting energy, Δ_o. When we analyze stability, we are actually analyzing the magnitude of Δ_o. With metal ion complexes, stability (i.e. the magnitude of Δ_o) is primarily dependent on 4 factors: the identity of the metal ion, its respective oxidation state in the complex, the overall geometry of the complex, and the nature of the attached ligands. In our question, each pair of complexes both have the same metal ion, oxidation state, and overall geometry, so the primary factor we need to consider is the nature of the ligands. So how exactly does the nature of the ligand affect stability?

Ligands can be classified as either strong or weak field, which can be determined by analyzing the spectrochemical series:

\[\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}\]

Weak field ligands (on the right) induce high spin on a complex, meaning that the energy required to place electrons together in an orbital, called the spin-pairing energy (P) is greater than the magnitude of Δ_o. As a result, the lowest energy arrangement has both t_2g orbitals and the e_g orbitals filled singly with electrons before pairing. Strong field ligands (on the left) induce low spin, meaning Δ_o is greater than P, and therefore only t_2g orbitals are completely filled first before e_g orbitals. As the question states above, stability increases as the number of electron in t_2g orbitals increases, suggesting that weak field ligands that induce high spin are more likely to create a stable complex since they allow both sets of orbitals to be filled first.

1. [Fe(CN)₆]⁴⁻ - CN^- is higher on the spectrochemical series and therefore, a stronger field ligand than H₂O, meaning that this complex is more stable.
2. [Co(NH₃)₆]³⁺ - NH₃ is a stronger field ligand than F^-, meaning that this complex is more stable.
3. [Mn(CN)₆]⁴⁻ - Again, CN^- is a stronger field ligand than Cl^-, meaning that this complex is more stable.

Q12.4.8

What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is 8.0 × 10⁻⁸ L/mol/s.

A12.4.8

The half-life of a reaction, t_1/2, is the amount of time that is required for a reactant concentration to decrease by half compared to its initial concentration. When solving for the half-life of a reaction, we should first consider the order of reaction to determine it's rate law. In this case, we are told that this reaction is second-order, so we know that the integrated rate law is given as:

\[\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0­}\]

Isolating for time, we find that:

\[t_{1/2} = \dfrac{1}{k[A]_0­}\]

Now it is just a matter of substituting the information we have been given to calculate \(t_{1/2}\), where the rate constant, \({k}\), is equal to 8.0 × 10⁻⁸ L/mol/s and initial concentration, \({[A]_0}\), is equal to 0.15M:

\[t_{1/2} = \dfrac{1}{(8.0×10^{-8})(0.15)} = {8.33×10^7 seconds}\]

Q21.2.3

For the following isotopes that have missing information, fill in the missing information to complete the notation

1. \(\ce{^{34}_{14}X}\)
2. \(\ce{^{36}_{X}P}\)
3. \(\ce{^{57}_{X}Mn}\)
4. \(\ce{^{121}_{56}X}\)

A21.2.3

Standard notation for elements and and their isotopes are as follows:

\[\ce{^{A}_{Z}X}\]

Where \({Z}\) corresponds to the atomic number (number of protons), \({A}\) corresponds to the atomic mass number (total number of protons and neutrons), and \({X}\) corresponds to the letters assigned to our given element. It's important to remember that an element can have a number of isotopes with varying numbers of neutrons, however the atomic number (number of protons) is unique for each and every element and will dictate its identity.

1. \(\ce{^{34}_{14}Si}\) - The atomic (proton) number (lower number) dictates the identity of the element. In this case, an atomic number of 14 corresponds to silicon.
2. \(\ce{^{36}_{15}P}\) - Again, each element has its own unique atomic number. The atomic number for phosphorus is 15.
3. \(\ce{^{57}_{25}Mn}\) - Looking at the periodic table, the atomic number for Mn is 25.
4. \(\ce{^{121}_{56}Ba}\) - Finally, an atomic number of 56 corresponds to barium according to the periodic table.

Q21.5.6

In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary.

A21.5.6

Q20.4.4

If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?

A20.4.4

A galvanic, or voltaic, cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place in a cell and involve the movement of electrons. Since we know that galvanic cells operate based on spontaneous redox reactions, we have to consider the spontaneity of the redox reactions associated with the given elements that we are working with. To find this information, we can refer to a table of standard reduction potentials, from which we find the following information for aluminum and bromine:

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

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

With the information above, we find that the reduction potential for bromine is higher than that of aluminum, which therefore tells us that the reduction of bromine is more spontaneous. This means that in a galvanic cell with aluminum and bromine, bromine will undergo reduction whilst aluminum will undergo oxidation. We are interested in finding the direction of electron flow, so we need to consider where electrons are being produced and where they are going. Also consider that oxidation always occurs at the anode whilst reduction always occurs at the cathode. Oxidation results in the loss of electrons, so in our case, electrons are being produced at the aluminum anode, and will therefore flow through the wire towards the bromine cathode.

Q20.7.2

Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?

A20.7.2

In a lead-acid battery, two types of lead are acted upon electrochemically by an electrolytic solution. When a lead-acid battery is discharged, the electrolyte divides into H₂ and SO₄.

The mechanics of a lead-acid battery relies on the following redox reaction:

\[{Pb}(s)+{PbO_2}(s)+{2H_2SO_4}(aq)→{2PbSO_4}(s)+{2H_2O}(l)\]

Two lead plates, one coated with lead dioxide, act as the electrodes in the battery along with sulfuric acid and water. As the battery is discharged, the lead dioxide-coated cathode facilitates the reduction reaction with the ionized sulfuric acid and hydrogen ions to produce solid lead sulfate and water, whilst the other lead anode facilitates the oxidation reaction with ionized sulfuric acid to produce hydrogen ions. The density of sulfuric acid (1.84 g/cm³) is higher than that of water (1.00 g/cm³), so as the battery is discharged removing sulfuric acid and producing water, the density of the fluid in these batteries will begin to decrease.